chapter 9 infinite series - meyers'...
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C H A P T E R 9Infinite Series
Section 9.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . 233
Section 9.2 Series and Convergence . . . . . . . . . . . . . . . . . . . 247
Section 9.3 The Integral Test and p-Series . . . . . . . . . . . . . . . . 260
Section 9.4 Comparisons of Series . . . . . . . . . . . . . . . . . . . . 270
Section 9.5 Alternating Series . . . . . . . . . . . . . . . . . . . . . . 277
Section 9.6 The Ratio and Root Tests . . . . . . . . . . . . . . . . . . 286
Section 9.7 Taylor Polynomials and Approximations . . . . . . . . . . 298
Section 9.8 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . 308
Section 9.9 Representation of Functions by Power Series . . . . . . . 321
Section 9.10 Taylor and Maclaurin Series . . . . . . . . . . . . . . . . 329
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354
C H A P T E R 9Infinite Series
Section 9.1 Sequences
233
1.
a5 � 25 � 32
a4 � 24 � 16
a3 � 23 � 8
a2 � 22 � 4
a1 � 21 � 2
an � 2n 2.
a5 �35
5!�
243120
�8140
a4 �34
4!�
8124
�278
a3 �33
3!�
276
�92
a2 �32
2!�
92
a1 �31!
� 3
an �3n
n!3.
a5 � ��12�
5
� �132
a4 � ��12�
4
�1
16
a3 � ��12�
3
� �18
a2 � ��12�
2
�14
a1 � ��12�
1
� �12
an � ��12�
n
4.
a5 � �32243
a4 �1681
a3 � �827
a2 �49
a1 � �23
an � ��23�
n
5.
a5 � sin 5�
2� 1
a4 � sin 2� � 0
a3 � sin 3�
2� �1
a2 � sin � � 0
a1 � sin �
2� 1
an � sin n�
26.
a5 �108
�54
a4 �87
a3 �66
� 1
a2 �45
a1 �24
�12
an �2n
n � 3
7.
a5 ���1�15
52 � �125
a4 ���1�10
42 �1
16
a3 ���1�6
32 �19
a2 ���1�3
22 � �14
a1 ���1�1
12 � �1
an ���1�n�n�1��2
n2 8.
a5 �25
a4 � �24
� �12
a3 �23
a2 � �22
� �1
a1 �21
� 2
an � ��1�n�1�2n� 9.
a5 � 5 �15
�1
25�
12125
a4 � 5 �14
�1
16�
7716
a3 � 5 �13
�19
�439
a2 � 5 �12
�14
�194
a1 � 5 � 1 � 1 � 5
an � 5 �1n
�1n2
234 Chapter 9 Infinite Series
10.
a5 � 10 �25
�6
25�
26625
a4 � 10 �12
�38
�878
a3 � 10 �23
�23
�343
a2 � 10 � 1 �32
�252
a1 � 10 � 2 � 6 � 18
an � 10 �2n
�6n2 11.
� 2�10 � 1� � 18
a5 � 2�a4 � 1�
� 2�6 � 1� � 10
a4 � 2�a3 � 1�
� 2�4 � 1� � 6
a3 � 2�a2 � 1�
� 2�3 � 1� � 4
a2 � 2�a1 � 1�
a1 � 3, ak�1 � 2�ak � 1� 12.
a5 � �4 � 12 �a4 � 30
a4 � �3 � 12 �a3 � 12
a3 � �2 � 12 �a2 � 6
a2 � �1 � 12 �a1 � 4
a1 � 4, ak�1 � �k � 12 �ak
13.
a5 �12
a4 �12
�4� � 2
a4 �12
a3 �12
�8� � 4
a3 �12
a2 �12
�16� � 8
a2 �12
a1 �12
�32� � 16
a1 � 32, ak�1 �12
ak 14.
a5 �13
a42 �
13
�768�2 � 196,608
a4 �13
a32 �
13
�482� � 768
a3 �13
a22 �
13
�122� � 48
a2 �13
a12 �
13
�62� � 12
a1 � 6, ak�1 �13
ak2 15.
decreases to 0; matches (f).
an �8
n � 1, a1 � 4, a2 �
83
,
16.
increases towards 8; matches (a).
an �8n
n � 1, a1 � 4, a2 �
163
, 17.decreases to 0; matches (e).an � 4�0.5�n�1, a1 � 4, a2 � 2, 18.
eventually approaches 0; matches (b).
an �4n
n!, a1 � 4, a2 � 8,
19. etc.; matches (d).a3 � �1,an��1�n, a1 � �1, a2 � 1, 20. etc.; matches (c).a3 � �13
,an ���1�n
n, a1 � �1, a2 �
12
,
21.
an �2
3n, n � 1, . . . , 10
−1
−1
12
8 22.
n � 1, . . . , 10an � 2 �4n
,
−1 12
−3
4 23.
an � 16��0.5�n�1, n � 1, . . . , 10
12−1
−10
18
24.
an �2n
n � 1, n � 1, 2, . . . , 10
12−1
−1
3 25.
Add 3 to preceding term.
a6 � 3�6� � 1 � 17
a5 � 3�5� � 1 � 14
an � 3n � 1 26.
a6 �6 � 6
2� 6
a5 �5 � 6
2�
112
an �n � 6
2
Section 9.1 Sequences 235
27.
a6 � 2�80� � 160
a5 � 2�40� � 80
a1 � 5 an�1 � 2an, 28.
a5 �1
16, a6 � �
132
an � �12
an�1, a1 � 1 29.
Multiply the preceding term by �12.
a6 �3
��2�5 � �332
a5 �3
��2�4 �3
16
an �3
��2�n�1
30.
a5 �8116
, a6 � �24332
an � �32
an�1, a1 � 1 31.
� �9��10� � 90
10!8!
�8!�9��10�
8!32.
� �24��25� � 600
25!23!
�23!�24��25�
23!
33. �n � 1�!
n!�
n!�n � 1�n!
� n � 1 34.
� �n � 1��n � 2�
�n � 2�!
n!�
n!�n � 1��n � 2�n!
35.
�1
2n�2n � 1�
�2n � 1�!�2n � 1�! �
�2n � 1�!�2n � 1�!�2n��2n � 1�
36.
� �2n � 1��2n � 2�
�2n � 2�!
�2n�! ��2n�!�2n � 1��2n � 2�
�2n�!37. lim
n→�
5n2
n2 � 2� 5 38. lim
n→� �5 �
1n2� � 5 � 0 � 5
39. limn→�
2n
�n2 � 1� lim
n→�
2�1 � �1�n2�
�21
� 2 40. limn→�
5n
�n2 � 4� lim
n→�
5�1 � �4�n2�
�51
� 5
41. limn→�
sin�1n� � 0 42. lim
n→� cos�2
n� � 1
43.
The graph seems to indicate that the sequence convergesto 1. Analytically,
limn→�
an � limn→�
n � 1
n� lim
x→� x � 1
x� lim
x→� 1 � 1.
−1
−1
12
3 44.
The graph seems to indicate that the sequence convergesto 0. Analytically,
limn→�
an � limn→�
1
n3�2 � limx→�
1
x3�2 � 0.
−1 12
−1
2
45.
The graph seems to indicate that the sequence diverges.Analytically, the sequence is
Hence, does not exist.limn→�
an
�an� � �0, �1, 0, 1, 0, �1, . . .�.
12−1
−2
2 46.
The graph seems to indicate that the sequence convergesto 3. Analytically,
limn→�
an � limn→�
�3 �12n� � 3 � 0 � 3.
−1 12
−1
4
236 Chapter 9 Infinite Series
47.
does not exist (oscillates betweenand 1), diverges.�1
limn→�
��1�n� nn � 1� 48.
does not exist, (alternates between0 and 2), diverges.
limn→�
1 � ��1�n 49. convergeslimn→�
3n2 � n � 4
2n2 � 1�
32
,
50. convergeslimn→�
3�n
3�n � 1� 1, 51.
Thus, converges.limn→�
an � 0,
�1
2n�
32n
�52n
. . . 2n � 1
2n<
12n
an �1 � 3 � 5 . . . �2n � 1�
�2n�n
52. The sequence diverges. To prove this analytically, we use mathematical induction to show that Clearly, Assume that Then,
Since
we have which shows that for all n.an ≥ �32�n�1
ak�1 ≥ �32�k
,
2k � 1k � 1
≥32
,
� ak
2k � 1k � 1
≥ �32�
k�1�2k � 1k � 1 �. ak�1 �
1 � 3 � 5 . . . �2k � 1��2k � 1��k � 1�!
ak ≥ �32�k�1
.a1 � 1 ≥ �32�0
� 1.an ≥ �32�n�1
.
53. convergeslimn→�
1 � ��1�n
n� 0, 54. convergeslim
n→� 1 � ��1�n
n2 � 0,
55.
converges
(L’Hôpital’s Rule)
� limn→�
32 �
1n� � 0,
limn→�
ln�n3�
2n� lim
n→� 32
ln�n�
n56.
converges
(L’Hôpital’s Rule)
� limn→�
1
2n� 0,
limn→�
ln�n
n� lim
n→� 1�2 ln n
n
57. convergeslimn→�
�34�
n
� 0, 58. convergeslimn→�
�0.5�n � 0, 59.
� �, diverges
limn→�
�n � 1�!
n!� lim
n→� �n � 1�
60. convergeslimn→�
�n � 2�!
n!� lim
n→�
1n�n � 1� � 0, 61.
converges � limn→�
1 � 2nn2 � n
� 0,
limn→�
�n � 1n
�n
n � 1� � limn→�
�n � 1�2 � n2
n�n � 1�
62.
converges
limn→�
� n2
2n � 1�
n2
2n � 1� � limn→�
�2n2
4n2 � 1� �
12
, 63. converges
�p > 0, n ≥ 2�
limn→�
n p
en � 0,
Section 9.1 Sequences 237
64.
Let
(L’Hôpital’s Rule)
or,
Therefore limn→�
n sin 1n
� 1.limx→�
sin�1�x�
1�x� lim
y→0� sin� y�
y� 1.
limx→�
x sin 1x
� limx→�
sin�1�x�
1�x� lim
x→� ��1�x2� cos�1�x�
�1�x2 � limx→�
cos 1x
� cos 0 � 1
f �x� � x sin 1x.
an � n sin 1n
65.
where convergesu � k�n,
limn→�
�1 �kn�
n
� limu→0
�1 � u�1�uk � ek
an � �1 �kn�
n
66. convergeslimn→�
21�n � 20 � 1,
67.
converges (because isbounded)
�sin n�
limn→�
sin n
n� lim
n→� �sin n�1
n� 0, 68. convergeslim
n→� cos �n
n2 � 0, 69. an � 3n � 2
70. an � 4n � 1 71. an � n2 � 2 72. an ���1�n�1
n2 73. an �n � 1n � 2
74. an ���1�n�1
2n�275. an � 1 �
1n
�n � 1
n76.
�2n�1 � 1
2n
an � 1 �2n � 1
2n 77. an �n
�n � 1��n � 2�
81. an � �2n�!, n � 1, 2, 3, . . . 82. an � �2n � 1�!, n � 1, 2, 3, . . . 83.
monotonic; bounded�an� < 4,
an � 4 �1n
< 4 �1
n � 1� an�1,
84. Let Then
Thus, f is increasing which implies is increasing.
bounded�an� < 3,
�an�
f � �x� �6
�x � 2�2.f �x� �3x
x � 2.
78. an �1n! 79.
���1�n�1 2nn!
�2n�!
an ���1�n�1
1 � 3 � 5 . . . �2n � 1�80. an �
xn�1
�n � 1�!
238 Chapter 9 Infinite Series
85.
Hence,
True; monotonic; bounded�an� ≤ 18,
an ≥ an�1.
n
2n�2 ≥ n � 1
2�n�1��2
2n�3n ≥ 2n�2�n � 1�
2n ≥ n � 1
n ≥ 1
n ≥ 1
2n ≥?
n � 1
2n�3n ≥?
2n�2�n � 1�
n
2n�2 ≥? n � 1
2�n�1��2 86.
Not monotonic; bounded�an� ≤ 0.7358,
a3 � 0.6694
a2 � 0.7358
a1 � 0.6065
an � ne�n�2
87.
Not monotonic; bounded�an� ≤ 1,
a3 � �13
a2 �12
a1 � �1
an � ��1�n �1n� 88.
Not monotonic; bounded�an� ≤ 23,
a3 � �827
a2 �49
a1 � �23
an � ��23�
n
89.
Monotonic; bounded�an� ≤ 23,
an � �23�
n
> �23�
n�1
� an�1
90.
Monotonic; notbounded
limn→�
an � �,
an � �32�
n
< �32�
n�1
� an�1 91.
Not monotonic; bounded�an� ≤ 1,
a4 � 0.8660
a3 � 1.000
a2 � 0.8660
a1 � 0.500
an � sin�n�
6 � 92.
Not monotonic; bounded�an� ≤ 1,
a3 � cos�3�
2 � � 0
a2 � cos � � �1
a1 � cos �
2� 0
an � cos�n�
2 �
93.
Not monotonic; bounded�an� ≤ 1,
a4 � �0.1634
a3 � �0.3230
a2 � �0.2081
a1 � 0.5403
an �cos n
n94.
Not monotonic, bounded�an� ≤ 1,
a6 �sin�6�
6� �0.0466a3 �
sin�3�3
� 0.0470
a5 �sin�5�
5� �0.1918a2 �
sin�2�2
� 0.4546
a4 �sin�4�
4� �0.1892a1 �
sin�1�1
� 0.8415
an �sin�n
n
Section 9.1 Sequences 239
95. (a)
Therefore, converges.
(b)
limn→�
�5 �1n� � 5
−1
−1
12
7
�an�
� an�1 ⇒ �an�, monotonic
an � 5 �1n
> 5 �1
n � 1
�5 �1n� ≤ 6 ⇒ �an�, bounded
an � 5 �1n
96. (a)
Therefore, converges.
(b)
limn→�
�4 �3n� � 4
−10
12
8
�an�
an � 4 �3n
< 3 �4
n � 1� an�1 ⇒ monotonic
�4 �3n� < 4 ⇒ bounded
an � 4 �3n
97. (a)
Therefore, converges.
(b)
limn→�
13�1 �
13n�� �
13
−1
−0.1
12
0.4
�an�
� an�1 ⇒ �an�, monotonic
an �13�1 �
13n� <
13�1 �
13n�1�
�13�1 �13n�� <
13
⇒ �an�,bounded
an �13�1 �
13n� 98. (a)
Therefore, converges.
(b)
limn→�
�4 �12n� � 4
−1 12
−1
6
�an�
� an�1 ⇒ �an�, monotonic
an � 4 �12n > 4 �
12n�1
�4 �12n� ≤ 4.5 ⇒ �an�, bounded
an � 4 �12n
99. has a limit because it is a bounded, monotonicsequence. The limit is less than or equal to 4, and greaterthan or equal to 2.
2 ≤ limn→�
an ≤ 4
�an� 100. The sequence could converge or diverge. If is increasing, then it converges to a limit less than orequal to 1. If is decreasing, then it could converge(example: ) or diverge (example: ).an � �nan � 1�n
�an�
�an��an�
101.
(a) No, the sequence diverges,
The amount will grow arbitrarily large over time.
limn→�
An � �.�An�
An � P�1 �r
12�n
(b)
A10 � 9421.11A5 � 9208.15
A9 � 9378.13A4 � 9166.14
A8 � 9335.34A3 � 9124.32
A7 � 9292.75A2 � 9082.69
A6 � 9250.35A1 � 9041.25
P � 9000, r � 0.055, An � 9000�1 �0.055
12 �n
240 Chapter 9 Infinite Series
102. (a)
(b) (c) A240 � 32,912.28A60 � 6480.83
A6 � 605.27
A5 � 503.76
A4 � 402.51
A3 � 301.50
A2 � 200.75
A1 � 100.25
A0 � 0
An � 100�401��1.0025n � 1� 103. (a) A sequence is a function whose domain is the set ofpositive integers.
(b) A sequence converges if it has a limit. See thedefinition.
(c) A sequence is monotonic if its terms are nondecreas-ing, or nonincreasing.
(d) A sequence is bounded if it is bounded belowfor some and bounded above
for some M�.�an ≤ MN��an ≥ N
104. The first sequence because everyother point is below the x-axis.
105. an � 10 �1n
106. Impossible. The sequenceconverges by Theorem 9.5.
107. an �3n
4n � 1108. Impossible. An unbounded sequence diverges.
109. (a)
(b)
(c) limn→�
�0.8�n�2.5� � 0
A4 � $1.024 billion
A3 � $1.28 billion
A2 � $1.6 billion
A1 � $2 billion
An � �0.8�n �2.5� billion 110.
P5 � $19,938.91
P4 � $19,080.30
P3 � $18,258.66
P2 � $17,472.40
P1 � $16,720.00
Pn � 16,000�1.045�n
111. (a)
(b) In 2008, and endangered species.a18 � 979n � 18
1500
13500
5
an � �6.60n2 � 151.7n � 387 112. (a)
(b) For 2008, and million.a18 � 7646n � 18
7000
130
2
an � 244.2n � 3250
113.
(a)
(b) Decreasing
(c) Factorials increase more rapidly than exponentials.
�1,562,500
567
�1,000,000,000
362,880
a9 � a10 �109
9!
an �10n
n!114.
limn→�
�1 �1n�
n
� e
a6 � 2.5216
a5 � 2.4883
a4 � 2.4414
a3 � 2.3704
a2 � 2.2500
a1 � 2.0000
an � �1 �1n�
n
Section 9.1 Sequences 241
115.
Let
Since we have Therefore,lim
n→� n�n � 1.
y � e0 � 1.ln y � 0,
ln y � limn→�
�1n
ln n� � limn→�
ln n
n� lim
n→� 1�nn
� 0
y � limn→�
n1�n.
a6 � 6�6 � 1.3480
a5 � 5�5 � 1.3797
a4 � 4�4 � 1.4142
a3 � 3�3 � 1.4422
a2 � �2 � 1.4142
a1 � 11�1 � 1
�an� � � n�n � � �n1�n� 116. Since
there exists for each
an integer N such that for every
Let and we have,
or
for each n > N.
0 < sn < 2L�L < sn � L < L,�sn � L� < L,
� � L > 0
n > N.�sn � L� < �
� > 0,
limn→�
sn � L > 0,
117. True 118. True 119. True 120. True
121.
(a)
(b)
b10 �8955
b5 �85
b9 �5534
b4 �53
b8 �3421
b3 �32
b7 �2113
b2 �21
� 2
b6 �138
b1 �11
� 1
bn �an�1
an
, n ≥ 1
a12 � 89 � 55 � 144 a6 � 5 � 3 � 8
a11 � 55 � 34 � 89 a5 � 3 � 2 � 5
a10 � 34 � 21 � 55 a4 � 2 � 1 � 3
a9 � 21 � 13 � 34 a3 � 1 � 1 � 2
a8 � 13 � 8 � 21 a2 � 1
a7 � 8 � 5 � 13 a1 � 1
an�2 � an � an�1
(c)
(d) If then
Since we have
Since and thus is positive, �1 ��5
2� 1.6180.bn,an,
�1 ± �1 � 4
2�
1 ± �52
0 � 2 � � 1
� 1 � 2
1 � �1�� � .
limn→�
bn � limn→�
bn�1,
limn→�
�1 �1
bn�1� � .lim
n→� bn � ,
�an � an�1
an
�an�1
an
� bn
� 1 �an�1
an
1 �1
bn�1� 1 �
1an�an�1
122.
The limit of the sequence appears to be In fact, this sequence is Newton’s Method applied to f �x� � x2 � 2.�2.
x10 � 1.414214 x5 � 1.414214
x9 � 1.414214 x4 � 1.414214
x8 � 1.414114 x3 � 1.414216
x7 � 1.414214 x2 � 1.41667
x6 � 1.414214 x1 � 1.5
x0 � 1, xn �12
xn�1 �1
xn�1, n � 1, 2, . . .
242 Chapter 9 Infinite Series
123. (a) (b)
(c) We first use mathematical induction to show that clearly So assume Then
Now we show that is an increasing sequence. Since and
Since is a bounding increasing sequence, it converges to some number by Theorem 9.5.
⇒ �L � 2��L � 1� � 0 ⇒ L � 2 �L �1�
limn→�
an � L ⇒ �2 � L � L ⇒ 2 � L � L2 ⇒ L2 � L � 2 � 0
L,�an�
an ≤ an�1.
an ≤ �an � 2
an2 ≤ an � 2
an2 � an � 2 ≤ 0
�an � 2��an � 1� ≤ 0
an ≤ 2,an ≥ 0�an�
ak�1 ≤ 2.
�ak � 2 ≤ 2
ak � 2 ≤ 4
ak ≤ 2.a1 ≤ 2.an ≤ 2;
a5 � �2 � �2 � �2 � �2 � �2 � 1.9976
a4 � �2 � �2 � �2 � �2 � 1.9904
a3 � �2 � �2 � �2 � 1.9616
a2 � �2 � �2 � 1.8478
an � �2 � an�1, n ≥ 2, a1 � �2a1 � �2 � 1.4142
124. (a) (b)
(c) We first use mathematical induction to show that clearly So assume Then
Now we show that is an increasing sequence. Since and
Since is a bounded increasing sequence, it converges to some number Thus,
⇒ �L � 3��L � 2� � 0 ⇒ L � 3 �L �2�
�6 � L � L ⇒ 6 � L � L2 ⇒ L2 � L � 6 � 0
L: limn→�
an � L.�an�
an ≤ an�1.
an ≤ �an � 6
an2 ≤ an � 6
an2 � an � 6 ≤ 0
�an � 3��an � 2� ≤ 0
an ≤ 3,an ≥ 0�an�
ak�1 ≤ 3.
�6 � ak ≤ 3
6 � ak ≤ 9
ak ≤ 3.a1 ≤ 3.an ≤ 3;
a5 � �6 � �6 � �6 � �6 � �6 � 2.9996
a4 � �6 � �6 � �6 � �6 � 2.9974
a3 � �6 � �6 � �6 � 2.9844
a2 � �6 � �6 � 2.9068
an � �6 � an�1, n ≥ 2, a1 � �6a1 � �6 � 2.4495
Section 9.1 Sequences 243
125. (a) We use mathematical induction to show that
[Note that if then and if then ] Clearly,
Before proceeding to the induction step, note that
So assume Then
is increasing because
an ≤ an�1.
an ≤ �an � k
an2 ≤ an � k
an2 � an � k ≤ 0
�an �1 � �1 � 4k
2 ��an �1 � �1 � 4k
2 � ≤ 0
�an�
an�1 ≤1 � �1 � 4k
2.
�an � k ≤ �1 � �1 � 4k2
� k
an � k ≤1 � �1 � 4k
2� k
an ≤1 � �1 � 4k
2.
�1 � �1 � 4k2
� k �1 � �1 � 4k
2.
1 � �1 � 4k
2� k � 1 � �1 � 4k
2 �2
1 � �1 � 4k
2� k �
1 � 2�1 � 4k � 1 � 4k4
2 � 2�1 � 4k � 4k � 2 � 2�1 � 4k � 4k
a1 � �k ≤�1 � 4k
2≤
1 � �1 � 4k2
.
an ≤ 3.k � 6,an ≤ 3,k � 2,
an ≤1 � �1 � 4k
2.
(b) Since is bounded and increasing, it has a limit
(c) implies that
Since L > 0, L �1 � �1 � 4k
2.
⇒ L �1 ± �1 � 4k
2.
⇒ L2 � L � k � 0
L � �k � L ⇒ L2 � k � L
limn→�
an � L
L.�an�
126. (a)
The terms of are decreasing, and those of are increasing. They both seem to approach the same limit.
—CONTINUED—
�bn��an�
b5 � �a4b4 � 5.9777a5 �a4 � b4
2� 5.9777
b4 � �a3b3 � 5.9777a4 �a3 � b3
2� 5.9777
b3 � �a2b2 � 5.9777a3 �a2 � b2
2� 5.9777
b2 � �a1b1 � 5.9667a2 �a1 � b1
2� 5.9886
b1 � �a0b0 � �10�3� � 5.4772a1 �a0 � b0
2�
10 � 32
� 6.5
a0 � 10, b0 � 3
244 Chapter 9 Infinite Series
126. —CONTINUED—
(b) For we need to show Since
Since
Since Thus, we have shown that
Now assume Since
Since
Since
Thus, we have shown that
(c) converges because it is decreasing and bounded below by 0. converges because it is increasing and bounded above by
(d) Let and Then A �A � B
2 ⇒ A � B.lim
n→� bn � B.lim
n→� an � A
a0.�bn��an�
ak�1 > ak�2 > bk�2 > bk�1.
ak�1bk�1 > bk�12 ⇒ �ak�1bk�1 > bk�1 ⇒ bk�2 > bk�1.ak�1 > bk�1,
⇒ ak�2 > bk�2.
⇒ ak�1 � bk�1 > 2�ak�1bk�1
⇒ �ak�1 � bk�1�2 > 4ak�1bk�1
ak�12 � 2ak�1bk�1 � bk�1
2 > 0 ⇒ ak�12 � 2ak�1bk�1 � bk�1
2 > 4ak�1bk�1
�ak�1 � bk�1�2 > 0,
2ak�1 > ak�1 � bk�1 ⇒ ak�1 >ak�1 � bk�1
2� ak�2.
ak�1 > bk�1,ak > ak�1 > bk�1 > bk.
a0 > a1 > b1 > b0.a0 > b0, a0b0 > b02 ⇒ �a0b0 > b0 ⇒ b1 > b0.
⇒ �a0 � b0�2 > 4a0b0 ⇒ a0 � b0 > 2�a0b0 ⇒ a1 > b1.
a02 � 2a0b0 � b0
2 > 0 ⇒ a02 � 2a0b0 � b0
2 > 4a0b0
�a0 � b0�2 > 0,
2a0 > a0 � b0 ⇒ a0 >a0 � b0
2� a1.a0 > b0,a0 > a1 > b1 > b0.n � 0,
127. (a)
(b)
� limn→�
an
� limn→�
nf �1n�
� limn→�
f �1�n��1�n�
� limh→0�
f �h�h
f ��0� � limh→0�
f �0 � h� � f �0�
h
limn→�
an � limn→�
n sin 1n
� limn→�
sin�1�n�
�1�n� � 1 � f ��0�
f ��x� � cos x, f ��0� � 1
f �x� � sin x, an � n sin 1n
128.
(a)
(b) diverges
(c) diverges
(d) If diverges. If
The sequence converges for �r� < 1.
→ 1
�r�n ln�r� ��rn
ln�r� → 0.
nrn �n
r�n “��
”
�r� < 1,�r� ≥ 1,
an � n�32�
2
,r �32
:
an � n,r � 1:
an � n�12�
n
�n2n → 0r �
12
:
an � nrn
Section 9.1 Sequences 245
129. (a)
� ln1 � 2 � 3 . . . n � ln�n!�
�n
1ln x dx < ln 2 � ln 3 � . . . � ln n
x
y
2 43
0.5
1.0
1.5
2.0 y = lnx
x
y
y = lnx
... n
(b)
� ln�n!�
�n�1
1ln x dx > ln 2 � ln 3 � . . . � ln n
x
y
2 3 4
0.5
1.0
1.5
2.0
n + 1...
y = lnx
n + 1...x
y
n + 1...
y = lnx
n + 1...
(c)
From part (a):
From part (b):
�n � 1�n�1
en > n!
eln�n�1�n�1�n > n!
ln�n � 1�n�1 � n > ln�n!�
�n�1
1 ln x dx � �n � 1� ln�n � 1� � �n � 1� � 1 � ln�n � 1�n�1 � n
nn
en�1 < n!
eln nn�n�1 < n!
ln nn � n � 1 < ln�n!�
�n
1ln x dx � n ln n � n � 1 � ln nn � n � 1
�ln x dx � x ln x � x � C
(d)
By the Squeeze Theorem, limn→�
n�n!n
�1e.
� �1�1e
�1e
limn→�
�n � 1�1��1�n�
ne� lim
n→� �n � 1�
n �n � 1�1�n
e
limn→�
1
e1��1�n� �1e
1
e1��1�n� <n�n!n
<�n � 1�1��1�n�
ne
n
e1��1�n� < n�n! <�n � 1��n�1��n
e
nn
en�1 < n! <�n � 1�n�1
en (e)
1e
� 0.3679
100�100!100
� 0.3799n � 100:
50�50!50
� 0.3897n � 50:
20�20!20
� 0.4152n � 20:
246 Chapter 9 Infinite Series
130.
(a)
(b) limn→�
an � limn→�
1n
�n
k�1
11 � �k�n� � �1
0
11 � x
dx � ln�1 � x��1
0� ln 2
a5 �15
11 � �1�5� �
11 � �2�5� �
11 � �3�5� �
11 � �4�5� �
11 � 1� �
16272520
� 0.6456
a4 �14
11 � �1�4� �
11 � �2�4� �
11 � �3�4� �
11 � 1� �
533840
� 0.6345
a3 �13
11 � �1�3� �
11 � �2�3� �
11 � 1� �
3760
� 0.6167
a2 �12
11 � �1�2� �
11 � 1� �
12
23
�12� �
712
� 0.5833
a1 �1
1 � 1�
12
� 0.5
an �1n
�n
k�1
11 � �k�n�
131. For a given we must find such that
whenever That is,
or
So, let be given. Let be an integer satisfyingFor we have
Thus, limn→�
1n3 � 0.
� >1n3 ⇒ � 1
n3 � 0� < �.
n3 >1�
n > �1��
1�3
n > M,M > �1���1�3.M� > 0
n > �1��
1�3.n3 >
1�
n > M.
�an � L� � � 1n3� < �
M > 0� > 0, 132. For a given we must find such thatwhenever That is,
or
(since for ).
So, let be given. Let be an integer satisfying
For we have
Thus, for �1 < r < 1.limn→�
rn � 0
�rn � 0� < �.
�r�n < �
ln�r�n < ln���
n ln�r� < ln���
n >ln���ln�r�
n > M,
M >ln���ln�r� .
M� > 0
�r� < 1ln�r� < 0n >ln���ln�r�
n ln�r� < ln���n > M.�an � L� � �r n��
M > 0� > 0,
133. If is bounded, monotonic and nonincreasing,then Then
is a bounded, monotonic, nondecreasing sequence whichconverges by the first half of the theorem. Since converges, then so does �an�.
��an�
�a1 ≤ �a2 ≤ �a3 ≤ . . . ≤ �an ≤ . . .a1 ≥ a2 ≥ a3 ≥ . . . ≥ an ≥ . . . .
�an� 134. Define
Hence, Thus,
xn�1 � an xn � xn�1 � axn � xn�1.
a1 � a2 � . . . � a.
an � an�1
xn�1 � xn�1
xn
�xn�2 � xn
xn�1
xn�1�xn�1 � xn�1� � xn�xn � xn�2�
xn�12 � xn xn�2 � 1 � xn
2 � xn�1xn�1 ⇒
an �xn�1 � xn�1
xn
, n ≥ 1.
Section 9.2 Series and Convergence 247
135.
We use mathematical induction to verify the formula.
Assume Then
By mathematical induction, the formula is valid for all n.
� �k � 1�! � 2k�1.
� �k2 � 5k � 4k � 4 � 4��k � 1�! � 8 � 2k�2
� ��k � 5��k��k � 1� � 4�k � 1��k � 1� � 4�k � 1���k � 2�! � ��k � 5�4 � 8�k � 1� � 4�k � 1��2k�2
� �k � 5��k! � 2k� � 4�k � 1���k � 1�! � 2k�1� � �4k � 4���k � 2�! � 2k�2�
Tk�1 � �k � 1 � 4�Tk � 4�k � 1�Tk�1 � �4�k � 1� � 8�Tk�2
Tk � k! � 2k.
T2 � 2 � 4 � 6
T1 � 1 � 2 � 3
T0 � 1 � 1 � 2
Tn � n! � 2n
Section 9.2 Series and Convergence
1.
S5 � 1 �14 �
19 �
116 �
125 � 1.4636
S4 � 1 �14 �
19 �
116 � 1.4236
S3 � 1 �14 �
19 � 1.3611
S2 � 1 �14 � 1.2500
S1 � 1 2.
S5 �16 �
16 �
320 �
215 �
542 � 0.7357
S4 �16 �
16 �
320 �
215 � 0.6167
S3 �16 �
16 �
320 � 0.4833
S2 �16 �
16 � 0.3333
S1 �16 � 0.1667
3.
S5 � 3 �92 �
274 �
818 �
24316 � 10.3125
S4 � 3 �92 �
274 �
818 � �4.875
S3 � 3 �92 �
274 � 5.25
S2 � 3 �92 � �1.5
S1 � 3 4.
S5 � 1 �13 �
15 �
17 �
19 � 1.7873
S4 � 1 �13 �
15 �
17 � 1.6762
S3 � 1 �13 �
15 � 1.5333
S2 � 1 �13 � 1.3333
S1 � 1
5.
S5 � 3 �32 �
34 �
38 �
316 � 5.8125
S4 � 3 �32 �
34 �
38 � 5.625
S3 � 3 �32 �
34 � 5.250
S2 � 3 �32 � 4.5
S1 � 3 6.
S5 � 1 �12 �
16 �
124 �
1120 � 0.6333
S4 � 1 �12 �
16 �
124 � 0.6250
S3 � 1 �12 �
16 � 0.6667
S2 � 1 �12 � 0.5
S1 � 1
7.
Geometric series
Diverges by Theorem 9.6
r �32
> 1
��
n�0 3�3
2�n
8.
Geometric series
Diverges by Theorem 9.6
r �43
> 1
��
n�0�4
3�n
9.
Geometric series
Diverges by Theorem 9.6
r � 1.055 > 1
��
n�01000�1.055�n
248 Chapter 9 Infinite Series
10.
Geometric series
Diverges by Theorem 9.6
r � 1.03 > 1
��
n�02��1.03�n 11.
Diverges by Theorem 9.9
limn→�
n
n � 1� 1 � 0
��
n�1
nn � 1
12.
Diverges by Theorem 9.9
limn→�
n
2n � 3�
12
� 0
��
n�1
n2n � 3
13.
Diverges by Theorem 9.9
limn→�
n2
n2 � 1� 1 � 0
��
n�1
n2
n2 � 114.
Diverges by Theorem 9.9
limn→�
n
n2 � 1� lim
n→�
11 � �1�n2�
� 1 � 0
��
n�1
nn2 � 1
15.
Diverges by Theorem 9.9
limn→�
2n � 12n�1 � lim
n→� 1 � 2�n
2�
12
� 0
��
n�1 2n � 12n�1
16.
Diverges by Theorem 9.9
limn→�
n!2n � �
��
n�1 n!2n
17.
Matches graph (c). Analytically, the series is geometric:
��
n�0 �9
4��14�
n
�9�4
1 � 1�4�
9�43�4
� 3
S0 �94
, S1 �94
�54
�4516
, S2 �94
�2116
� 2.95, . . .
��
n�0 94�
14�
n
�94�1 �
14
�1
16� . . . 18.
Matches graph (b). Analytically, the series is geometric:
��
n�0 �2
3�n
�1
1 � 2�3�
11�3
� 3
S0 � 1, S1 �53
, S2 � 2.11, . . . .
��
n�0�2
3�n
� 1 �23
�49
� . . .
19.
Matches graph (a). Analytically, the series is geometric:
��
n�0 154 ��
14�
n
�15�4
1 � ��1�4� �15�45�4
� 3
S0 �154
, S1 �4516
, S2 � 3.05, . . .
��
n�0 154 ��
14�
n
�154 �1 �
14
�1
16� . . . 20.
Matches (d). Analytically, the series is geometric:
��
n�0 173
��89�
n
�17�3
1 � ��8�9� �17�317�9
� 3
S0 �173
, S1 � 0.63, S3 � 5.1, . . .
��
n�0 173 ��
89�
n
�173
�1 �89
�6481
� . . .
21.
Matches (f). The series is geometric:
��
n�0 173 ��
12�
n�
173
1
1 � ��1�2� �173
23
�349
S0 �173
, S1 �176
, . . .
��
n�0 173 ��
12�
n�
173 �1 �
12
�14
� . . . 22.
Matches (e). The series is geometric:
��
n�0 �2
5�n
�1
1 � �2�5� �53
S0 � 1, S1 �75
, . . .
��
n�0 �2
5�n
� 1 �25
�4
25� . . .
23.
��
n�1
1n�n � 1� � lim
n→� Sn � lim
n→� �1 �
1n � 1� � 1
��
n�1
1n�n � 1� � �
�
n�1�1
n�
1n � 1� � �1 �
12� � �1
2�
13� � �1
3�
14� � �1
4�
15� � . . ., Sn � 1 �
1n � 1
Section 9.2 Series and Convergence 249
24.
��
n�1
1n�n � 2� � lim
n→� Sn � lim
n→� �1
2�
14
�1
2�n � 1� �1
2�n � 2� �12
�14
�34
� �12
�16� � �1
4�
18� � �1
6�
110� � �1
8�
112� � � 1
10�
114� � . . . �
�
n�1
1n�n � 2� � �
�
n�1� 1
2n�
12�n � 2��
25.
Geometric series with
Converges by Theorem 9.6
r �34 < 1
��
n�0 2�3
4�n
26.
Geometric series with
Converges by Theorem 9.6
r � �12 < 1
��
n�0 2��
12�
n
27.
Geometric series with
Converges by Theorem 9.6
r � 0.9 < 1
��
n�0�0.9�n 28.
Geometric series with
Converges by Theorem 9.6
r � �0.6 < 1
��
n�0��0.6�n
29. (a)
(b) (c)
(d) The terms of the series decrease in magnitude slowly. Thus,the sequence of partial sums approaches the sum slowly.
0 110
5
� 2�1 �12
�13 �
113
� 3.667
�Sn � 2�1 �12
�13
� � 1n � 1
�1
n � 2�
1n � 3� �
� 2��1 �14� � �1
2�
15� � �1
3�
16� � �1
4�
17� � . . .
��
n�1
6
n�n � 3�� 2 �
�
n�1�1
n�
1
n � 3�
n 5 10 20 50 100
2.7976 3.1643 3.3936 3.5513 3.6078Sn
30. (a)
(b) (c)
(d) The terms of the series decrease in magnitude slowly. Thus,the sequence of partial sums approaches the sum slowly. 0
0 11
4
� 1 �12
�13
�14
�2512
� 2.0833
� �1 �15� � �1
2�
16� � �1
3�
17� � �1
4�
18� � �1
5�
19� � �1
6�
110� � . . .
��
n�1
4n�n � 4� � �
�
n�1�1
n�
1n � 4�
n 5 10 20 50 100
1.5377 1.7607 1.9051 2.0071 2.0443Sn
250 Chapter 9 Infinite Series
31. (a) (c)
(b)
(d) The terms of the series decrease in magnitude slowly. Thus,the sequence of partial sums approaches the sum slowly.
00
11
22��
n�1 2�0.9�n�1 � �
�
n�0 2�0.9�n �
21 � 0.9
� 20
n 5 10 20 50 100
8.1902 13.0264 17.5685 19.8969 19.9995Sn
32. (a) (Geometric series) (c)
(b)
(d) The terms of the series decrease in magnitude slowly. Thus,the sequence of partial sums approaches the sum slowly.
00
11
24��
n�13�0.85�n�1 �
31 � 0.85
� 20
n 5 10 20 50 100
11.1259 16.0625 19.2248 19.9941 19.999998Sn
33. (a) (c)
(b)
(d) The terms of the series decrease in magnitude rapidly. Thus,the sequence of partial sums approaches the sum rapidly.
1100
15��
n�110�0.25�n�1 �
101 � 0.25
�403
� 13.3333
n 5 10 20 50 100
13.3203 13.3333 13.3333 13.3333 13.3333Sn
34. (a) (c)
(b)
(d) The terms of the series decrease in magnitude rapidly. Thus,the sequence of partial sums approaches the sum rapidly.
00 11
7��
n�1 5��
13�
n�1
�5
1 � ��1�3� �154
� 3.75
n 5 10 20 50 100
3.7654 3.7499 3.7500 3.7500 3.7500Sn
35.
�12�1 �
12� �
34
�12��1 �
13� � �1
2�
14� � �1
3�
15� � �1
4�
16� � . . .
��
n�2
1n2 � 1
� ��
n�2� 1�2
n � 1�
1�2n � 1� �
12
��
n�2� 1
n � 1�
1n � 1�
36. ��
n�1
4n�n � 2� � 2 �
�
n�1�1
n�
1n � 2� � 2��1 �
13� � �1
2�
14� � �1
3�
15� � . . . � 2�1 �
12� � 3
37. ��
n�1
8�n � 1��n � 2� � 8 �
�
n�1� 1
n � 1�
1n � 2� � 8��1
2�
13� � �1
3�
14� � �1
4�
15� � . . . � 8�1
2� � 4
38. ��
n�1
1�2n � 1��2n � 3� �
12
��
n�1� 1
2n � 1�
12n � 3� �
12��
13
�15� � �1
5�
17� � �1
7�
19� � . . . �
12�
13� �
16
Section 9.2 Series and Convergence 251
39. ��
n�0�1
2�n
�1
1 � �1�2� � 2 40.
(Geometric)
��
n�06�4
5�n
�6
1 � �4�5� � 30 41. ��
n�0��
12�
n
�1
1 � ��1�2� �23
42. ��
n�0 2��
23�
n
�2
1 � ��2�3� �65
43. ��
n�0� 1
10�n
�1
1 � �1�10� �109
44. ��
n�0 8�3
4�n
�8
1 � �3�4� � 32
45. ��
n�0 3��
13�
n
�3
1 � ��1�3� �94
46. ��
n�0 4��
12�
n
�4
1 � ��1�2� �83
47.
� 2 �32
�12
�1
1 � �1�2� �1
1 � �1�3�
��
n�0� 1
2n �13n� � �
�
n�0�1
2�n
� ��
n�0�1
3�n
48.
�103
� 10 � 2 �343
�1
1 � �7�10� �1
1 � �9�10� � 2
��
n�1 ��0.7�n � �0.9�n� � �
�
n�0� 7
10�n
� ��
n�0� 9
10�n
� 2
49. Note that The series is geometric with Thus,
��
n�1�sin�1��n � sin�1� �
�
n�0�sin�1��n �
sin�1�1 � sin�1� � 5.3080.
r � sin�1� < 1.��
n�1�sin�1��nsin�1� � 0.8415 < 1.
50.
limn→�
Sn � limn→�
13�
12
�1
3n � 2 �16
�13�
12
�1
3n � 2
�13��
12
�15� � �1
5�
18� � �1
8�
111� � . . . � � 1
3n � 1�
13n � 2�
� �n
k�1� 1
9k � 3�
19k � 6 �
13
�n
k�1� 1
3k � 1�
13k � 2
Sn � �n
k�1
19k2 � 3k � 2
� �n
k�1
1�3k � 1��3k � 2�
51. (a)
(b) Geometric series with and
S �a
1 � r�
4�101 � �1�10� �
49
r �110a �
410
0.4 � ��
n�0
410�
110�
n
52. (a)
(b) 0.9 �9
10
11 � �1�10� � 1
� ��
n�1 9� 1
10�n
�9
10 ��
n�0� 1
10�n
0.9 �9
10�
9100
�9
1000� . . .
53. (a)
(b) Geometric series with and
S �a
1 � r�
81�1001 � �1�100� �
8199
�9
11
r �1
100a �81100
0.81 � ��
n�0
81100
� 1100�
n
54. (a)
(b) 0.01 �1
100�
11 � �1�100� �
1100
�10099
�1
99
0.01 � ��
n�1� 1
100�n
�1
100 ��
n�0� 1
100�n
252 Chapter 9 Infinite Series
55. (a)
(b) Geometric series with and
S �a
1 � r�
3�4099�100
�5
66
r �1
100a �340
0.075 � ��
n�0
340�
1100�
n
56. (a)
(b) Geometric series with and
S �15
�a
1 � r�
15
�3�200
99�100�
71330
r �1
100a �3
200
0.215 �15
� ��
n�0
3200�
1100�
n
57.
Diverges by Theorem 9.9
limn→�
n � 1010n � 1
�1
10� 0
��
n�1
n � 1010n � 1
58.
Diverges by Theorem 9.9
limn→�
n � 12n � 1
�12
� 0
��
n�1
n � 12n � 1
59. converges��
n�1�1
n�
1n � 2� � �1 �
13� � �1
2�
14� � �1
3�
15� � �1
4�
16� � . . . � 1 �
12
�32
,
60.
�13�1 �
12
�13� �
1118
, converges
�13��1 �
14� � �1
2�
15� � �1
3�
16� � �1
4�
17� � �1
5�
18� � �1
6�
19� � . . .
��
n�1
1n�n � 3� �
13
��
n�1�1
n�
1n � 3�
61.
Diverges by Theorem 9.9
limn→�
3n � 12n � 1
�32
� 0
��
n�1 3n � 12n � 1
62.
(by L’Hôpital’s Rule); diverges by Theorem 9.9
� limn→�
�ln 2�23n
6n� lim
n→�
�ln n�33n
6� �
limn→�
3n
n3 � limn→�
�ln 2�3n
3n2
��
n�1 3n
n3
63.
Geometric series with
Converges by Theorem 9.6
r �12
� �
n�0 42n � 4 �
�
n�0�1
2�n
64.
Geometric series with
Converges by Theorem 9.6
r �14
��
n�0 14n
65.
Geometric series with
Diverges by Theorem 9.6
r � 1.075
��
n�0�1.075�n 66.
Geometric series with
Diverges by Theorem 9.6
r � 2
��
n�1
2n
100
67. Since the terms
do not approach 0 as Hence, the series
diverges.��
n�2
nln�n�
n →�.
an �n
ln�n�n > ln�n�, 68.
Since diverges, diverges.��
n�1ln�1
n�limn→�
Sn
� 0 � ln 2 � ln 3 � . . . �ln�n�
Sn � �n
k�1ln�1
k� � �n
k�1�ln�k�
Section 9.2 Series and Convergence 253
72.
Diverges
� ln�n � 1� � ln�1� � ln�n � 1�
� �ln 2 � ln 1� � �ln 3 � ln 2� � . . . � �ln�n � 1� � ln n�
� ln�21� � ln�3
2� � . . . � ln�n � 1n �
Sn � �n
k�1ln�k � 1
k �
69. For
For
Hence, diverges.��
n�1�1 �
kn
n
limn→�
�1 � 0�n � 1 � 0.k � 0,
� ek � 0.
limn→�
�1 �kn�
n� lim
n→���1 �
kn�
n�k k
k � 0, 70. converges since
it is geometric with
r �1e
< 1.
��
n�1e�n � �
�
n�1�1
e�n
71.
Hence, diverges.��
n�1arctan n
limn→�
arctan n ��
2� 0
73. See definition on page 606.
74. means that the limit of the sequence is 5.
means that the limit of the
partial sums is 5.
��
n�1an � a1 � a2 � . . . � 5
�an�limn→�
an � 5 75. The series given by
is a geometric series with ratio r. When the series converges to The series diverges if
r ≥ 1.a��1 � r�.
0 < r < 1,
��
n�0arn � a � ar � ar2 � . . . � arn � . . . , a � 0
76. If then diverges.��
n�1anlim
n→� an � 0,
77.
diverges because limn→�
an � 0.��
n�1an
limn→�
an � limn→�
n � 1
n� 1 ��an� converges�
an �n � 1
n78. (a)
(b)
These are the same. The third series is different,unless is constant.
(c) ��
n�1ak � ak � ak � . . .
a1 � a2 � . . . � a
��
k�1ak � a1 � a2 � a3 � . . .
��
n�1an � a1 � a2 � a3 � . . .
79.
Geometric series: converges for or
�x2
1
1 � �x�2� �x2
2
2 � x�
x2 � x
, x < 2
f �x� �x2
��
n�0�x
2�n
x < 2x2 < 1
��
n�1 xn
2n � ��
n�1�x
2�n
�x2
��
n�0�x
2�n
80.
Geometric series: converges for
f �x� � �3x� ��
n�0�3x�n � �3x� 1
1 � 3x�
3x1 � 3x
, x <13
3x < 1 ⇒ x <13
��
n�1�3x�n � �3x� �
�
n�0�3x�n
254 Chapter 9 Infinite Series
81.
Geometric series: converges for
� �x � 1� 11 � �x � 1� �
x � 12 � x
, 0 < x < 2
f �x� � �x � 1� ��
n�0�x � 1�n
x � 1 < 1 ⇒ 0 < x < 2
��
n�1�x � 1�n � �x � 1� �
�
n�0�x � 1�n 82.
Geometric series: converges for
�4
�4 � x � 3��4�
167 � x
, �1 < x < 7
f �x� � ��
n�04�x � 3
4 �n�
41 � ��x � 3��4�
x � 34 < 1 ⇒ x � 3 < 4 ⇒ �1 < x < 7
��
n�04�x � 3
4 �n
83.
Geometric series: converges for
f �x� � ��
n�0��x�n �
11 � x
, �1 < x < 1
�x < 1 ⇒ x < 1 ⇒ �1 < x < 1
��
n�0��1�nxn � �
�
n�0��x�n 84.
Geometric series: converges for
f �x� � ��
n�0��x2�n �
11 � ��x2� �
11 � x2, �1 < x < 1
�x2 < 1 ⇒ �1 < x < 1
��
n�0��1�nx2n � �
�
n�0��x2�n
85.
Geometric series: converges if
or
x > 1 or x < �1f �x� � ��
n�0�1
x�n
�1
1 � �1�x� �x
x � 1,
x > 1⇒ x > 1 ⇒ x < �1
1x < 1
��
n�0�1
x�n
86.
Geometric series: converges for
f �x� �x2
x2 � 4�
1
1 �x2
x2 � 4
�x2
x2 � 4 x2 � 4
4�
x2
4
⇒ x2 < x2 � 4 ⇒ converges for all x
x2
x2 � 4 < 1
��
n�1� x2
x2 � 4�n
�x2
x2 � 4 �
�
n�0� x2
x2 � 4�n
87. (a) Yes, the new series will still diverge.
(b) Yes, the new series will converge.
88. Neither statement is true. The formula
holds for �1 < x < 1.
1x � 1
� 1 � x � x2 � . . .
89. (a) is the common ratio.
(b) 1 � x � x2 � . . . � ��
n�0xn �
11 � x
, x < 1
x (c)
y3 � S5 � 1 � x � x2 � x3 � x 4
y2 � S3 � 1 � x � x2
3
1.5−1.50
f
S3
S5
y1 �1
1 � x
90. (a) is the common ratio.
(b)
�2
2 � x, x < 2
�1
1 � ��x�2�
1 �x2
�x2
4�
x3
8� . . . � �
�
n�0��
x2�
n
��x2� (c)
y3 � S5 � 1 �x2
�x2
4�
x3
8�
x 4
16
y2 � S3 � 1 �x2
�x2
4
f
S3
S5
5
5
−5
−5
y1 �2
2 � x
Section 9.2 Series and Convergence 255
91.
Horizontal asymptote:
The horizontal asymptote is the sum of the series. is the partial sum.nthf �n�
S �3
1 � �1�2� � 6
��
n�0 3�1
2�n
y � 6
−2 10
−1
7y = 6f �x� � 3�1 � 0.5x
1 � 0.5 92.
Horizontal asymptote:
The horizontal asymptote is the sum of the series. is the partial sum.nthf �n�
S �2
1 � �4�5� � 10
��
n�0 2�4
5�n
y � 10
−2 16
−2
y = 1012
f �x� � 2�1 � 0.8x
1 � 0.8
93.
Choosing the positive value for n we have The first term that is less than 0.0001 is
This inequality is true when This series convergesat a faster rate.
n � 5.
10,000 < 8n
�18�
n
< 0.0001
n � 100.n � 99.5012.
n ��1 ± 12 � 4�1���10,000�
2
0 < n2 � n � 10,000
10,000 < n2 � n
1
n�n � 1� < 0.0001 94.
This inequality is true when
This inequality is true when This series convergesat a faster rate.
n � 5.
10,000 < 10n
�0.01�n < 0.0001
n � 14.
10,000 < 2n
12n < 0.0001
95.
� 80,000�1 � 0.9n�, n > 0
�n�1
i�0 8000�0.9�i �
8000�1 � �0.9��n�1��1�1 � 0.9
96.
V�5� � �0.7�5�225,000� � $37,815.75
V�t� � 225,000�1 � 0.3�n � �0.7�n�225,000�
97.
Sum � 400 million dollars
� 400�1 � 0.75n� million dollars
�n�1
i�0 100�0.75�i �
100�1 � 0.75�n�1��1�1 � 0.75
98.
million dollars
Sum � 250 million dollars
� 250�1 � 0.6n�
�n�1
i�0100�0.60�i �
100�1 � 0.6n�1 � 0.6
99.
� 152.42 feet
� �16 � ��
n�0 32�0.81�n � �16 �
321 � 0.81
D � 16 � 32�0.81� � 32�0.81�2 � . . .
�
D3 � 16�0.81�2 � 16�0.81�2 � 32�0.81�2
D2 � 0.81�16� � 0.81�16� � 32�0.81�
D1 � 16
up down
� �
100. The ball in Exercise 99 takes the following times for each fall.
Beginning with the ball takes the same amount of time to bounce up as it takes to fall. The total elapsed time before the ball comes to rest is
� �1 �2
1 � 0.9� 19 seconds.
t � 1 � 2 ��
n�1�0.9�n � �1 � 2 �
�
n�0�0.9�n
s2,
sn � 0 if t � �0.9�n�1 sn � �16t 2 � 16�0.81�n�1
��
s3 � 0 if t � �0.9�2s3 � �16t 2 � 16�0.81�2
s2 � 0 if t � 0.9 s2 � �16t 2 � 16�0.81�
s1 � 0 if t � 1 s1 � �16t 2 � 16
256 Chapter 9 Infinite Series
101.
��
n�0 12�
12�
n
�1�2
1 � �1�2� � 1
P�2� �12�
12�
2
�18
P�n� �12�
12�
n
102.
��
n�0 13�
23�
n
�1�3
1 � �2�3� � 1
P�2� �13�
23�
2
�4
27
P�n� �13�
23�
n
103. (a)
(b) No, the series is not geometric.
(c) ��
n�1n�1
2�n
� 2
�12
1
�1 � �1�2�� � 1
��
n�1�1
2�n
� ��
n�0
12
�12�
n
104. Person 1:
Person 2:
Person 3:
Sum:47
�27
�17
� 1
123 �
126 �
129 � . . . �
18
��
n�0�1
8�n
�18
1
1 � �1�8� �17
122 �
125 �
128 � . . . �
14
��
n�0�1
8�n
�14
1
1 � �1�8� �27
12
�124 �
127 � . . . �
12
��
n�0�1
8�n
�12
1
1 � �1�8� �47
105. (a)
(b)
Note: This is one-half of the area of the original square!
��
n�0 64�1
2�n
�64
1 � �1�2� � 128 in.216 in.
16 in.
64 � 32 � 16 � 8 � 4 � 2 � 126 in.2
106. (a) (b) If and then
Total: z sin � � z sin2 � � z sin3 � � . . . � zsin �
1 � sin �
sin � �x1y2x1y1 ⇒ x1y2 � x1y1 sin � � z sin3 �
sin � �x1y1Yy1 ⇒ x1y1 � Yy1 sin � � z sin2 �
total �1�2
1 � �1�2� � 1.� ��
6,z � 1sin � �
Yy1z
⇒ Yy1 � z sin �
107.
� 573,496.06
�50,0001.06 �1 � 1.06�20
1 � 1.06�1 , �n � 20, r � 1.06�1�
�20
n�150,000� 1
1.06�n
�50,0001.06
�19
i�0� 1
1.06�i
108. Surface area � 4� �1�2 � 9�4� �13�2� � 92 � 4� �1
9�2� . . . � 4�� � � � . . .� � �
109.
(a) When
(b) When
(c) When n � 31: w � $21,474,836.47
n � 30: w � $10,737,418.23
n � 29: w � $5,368,709.11
w � �n�1
i�0 0.01�2�i �
0.01�1 � 2n�1 � 2
� 0.01�2n � 1�
Section 9.2 Series and Convergence 257
110.
�12t�1
n�0 P�er�12�n �
P�1 � �er�12�12t�1 � er�12 �
P�ert � 1�er�12 � 1
� P�12r ���1 �
r12�
12t
� 1 � P��12r ���1 � �1 �
r12�
12t
�12t�1
n�0 P�1 �
r12�
n
�
P�1 � �1 �r
12�12t
1 � �1 �
r12�
113.
(a)
(b) A �100�e0.04�40� � 1�
e0.04�12 � 1� $118,393.43
A � 100� 120.04���1 �
0.0412 �
12�40�� 1 � $118,196.13
P � 100, r � 0.04, t � 40 114.
(a)
(b) A �20�e0.06�50� � 1�
e0.06�12 � 1� $76,151.45
A � 20� 120.06���1 �
0.0612 �
12�50�� 1 � $75,743.82
P � 20, r � 0.06, t � 50
115. (a)
6500
3500132
an � 3484.1363�1.0502�n � 3484.1363e0.04897n (b) Adding the ten values you obtain
or $50,809,000,000.
(c) or $50,803,000,000
(Answers will vary.)
�12
n�33484.1363e0.04897n � 50,803,
�12
n�3an � 50,809,
a3, a4, . . . , a12
116.
� 40,000�1 � 1.0440
1 � 1.04 � � $3,801,020 � �39
n�040,000�1.04�n
T � 40,000 � 40,000�1.04� � . . . � 40,000�1.04�39 117. False. but diverges.��
n�1 1n
limn→�
1n
� 0,
111.
(a)
(b) A �50�e0.03�20� � 1�
e0.03�12 � 1� $16,421.83
A � 50� 120.03���1 �
0.0312 �
12�20�� 1 � $16,415.10
P � 50, r � 0.03, t � 20 112.
(a)
(b) A �75�e0.05�25� � 1�
e0.05�12 � 1� $44,732.85
A � 75 � 120.05���1 �
0.0512 �
12�25�� 1 � $44,663.23
P � 75, r � 0.05, t � 25
118. True 119. False;
The formula requires that the geo-metric series begins with n � 0.
��
n�1 arn � � a
1 � r� � a 120. True
limn→�
n
1000�n � 1� �1
1000� 0
121. True
� 0.74 �1
100� 0.75
� 0.74 �9
103 �109
� 0.74 �9
103 �1
1 � �1�10�
� 0.74 �9
103 ��
n�0� 1
10�n
0.74999 . . . � 0.74 �9
103 �9
104 � . . .
122. True
258 Chapter 9 Infinite Series
123. By letting we have
Thus,
� ��
n�1��c � Sn�1� � �c � Sn��.
� ��
n�1�Sn � Sn�1�c � c�
��
n�1 an � �
�
n�1�Sn � Sn�1�
an � �n
k�1 ak � �
n�1
k�1 ak � Sn � Sn�1.
S0 � 0, 124. Let be the sequence of partial sums for theconvergent series
Then and since
we have
� limn→�
L � limn→�
Sn � L � L � 0.
limn→�
Rn � limn→�
�L � Sn�
Rn � ��
k�n�1ak � L � Sn,
limn→�
Sn � L��
n�1 an � L.
�Sn�
125. Let and
Both are divergent series.
��an � bn� � ��
n�0�1 � ��1�� � �
�
n�0�1 � 1� � 0
� bn � ��
n�0 ��1�.� an � �
�
n�0 1 126. If converged, then
would converge, which is a contradiction. Thus,diverges.� �an � bn�
� bn
� �an � bn� � � an �� �an � bn�
127. Suppose, on the contrary, that converges. Because
converges. This is a contradiction since diverged.Hence, diverges.� can
� can
��1c�can � � an
c � 0,� can 128. If converges, then Hence,
which implies that diverges.� 1�an
limn→�
1an
� 0,
limn→�
an � 0.� an
129. (a)
(b)
��
n�0
1an�1an�3
� limn→�
Sn � limn→�
�1 �1
an�2an�3 � 1
� 1 �1
an�2an�3
�1
a1a2�
1an�2an�3
� � 1a1a2
�1
a2a3 � � 1
a2a3�
1a3a4
� . . . � � 1an�1an�2
�1
an�2an�3
� �n
k�0� 1
ak�1ak�2�
1ak�2ak�3
Sn � �n
k�0
1ak�1ak�3
�1
an�1an�3
�an�2
an�1an�2an�3
1
an�1an�2�
1an�2an�3
�an�3 � an�1
an�1an�2an�3
Section 9.2 Series and Convergence 259
130.
As the two geometric series converge for
limn→�
S2n �1
1 � x2 � 2x� 11 � x2� �
1 � 2x1 � x2 , x < 1
x < 1:n →�,
� �n
k�0�x2�k � 2x �
n
k�0�x2�k
� �1 � x2 � x 4 � . . . � x2n� � �2x � 2x3 � . . . � 2x2n�1�
S2n � 1 � 2x � x2 � 2x3 � . . . � x2n � 2x2n�1
131.
This is a geometric series which converges if
1r < 1 ⇔ r > 1.
�since 1r < 1� �1
r � 1
�1�r
1 � �1�r�
1r
�1r 2 �
1r 3 � . . . � �
�
n�0 1r�
1r�
n
132. The entire rectangle has area 2 because the height is 1and the base is The squares alllie inside the rectangle, and the sum of their areas is
Thus, ��
n�1 1n2 < 2.
1 �122 �
132 �
142 � . . . .
1 �12 �
14 � . . . � 2.
133. Let H represent the half-life of the drug. If a patient receives n equal doses of P units each of this drug,administered at equal time interval of length t, the total amount of the drug in the patient’s system at the time the last dose is administered is given by where One time interval after the last dose is administered is given by Two time intervals after the last dose is administered is given by and so on. Since as where s is an integer.s →�,k < 0, Tn�s→ 0
Tn�2 � Pe2kt � Pe3kt � Pe4kt � . . . � Pe�n�1�kt
Tn�1 � Pekt � Pe2kt � Pe3kt � . . . � Penkt.k � ��ln 2��H.Tn � P � Pekt � Pe2kt � . . . � Pe�n�1�kt
134. The series is telescoping:
limn→�
Sn � 3 � 1 � 2
� 3 �3n�1
3n�1 � 2n�1
� �n
k�1� 3k
3k � 2k �3k�1
3k�1 � 2k�1
Sn � �n
k�1
6k
�3k�1 � 2k�1��3k � 2k�
135.
In general:
(See below for a proof of this.)
and are either both odd or both even. If both even, then
If both odd,
Proof by induction that the formula for is correct. It is true for Assume that the formula is valid for
If is even, then and
The argument is similar if is odd.k
�k2 � 2k
4�
�k � 1�2 � 14
.
f �k � 1� � f �k� �k2
�k2
4�
k2
f �k� � k2�4kk.n � 1.
f �n�
� xy.
f �x � y� � f �x � y� ��x � y�2 � 1
4�
�x � y�2 � 14
f �x � y� � f �x � y� ��x � y�2
4�
�x � y�2
4� xy.
x � yx � y
f �n� � �n2�4,
�n2 � 1��4, n even
n odd.
f �1� � 0, f �2� � 1, f �3� � 2, f �4� � 4, . . .
Section 9.3 The Integral Test and p-Series
260 Chapter 9 Infinite Series
1.
Let
f is positive, continuous, and decreasing for
Diverges by Theorem 9.10
��
1
1x � 1
dx � �ln�x � 1��1
�� �
x ≥ 1.
f �x� �1
x � 1.
��
n�1
1n � 1
2.
Let
f is positive, continuous, and decreasing for
Diverges by Theorem 9.10
��
1
23x � 5
dx � �23
ln�3x � 5���
1� �
x ≥ 1.
f �x� �2
3x � 5.
��
n�1
2
3n � 5
3.
Let
f is positive, continuous, and decreasing for
Converges by Theorem 9.10
��
1 e�x dx � ��e�x�
�
1�
1e
x ≥ 1.
f �x� � e�x.
��
n�1e�n 4.
Let
f is positive, continuous, and decreasing for since
for
Converges by Theorem 9.10
��
3xe�x�2 dx � ��2�x � 2�e�x�2�
�
3� 10e�3�2
x ≥ 3.f � �x� �2 � x2e x�2 < 0
x ≥ 3
f �x� � xe�x�2.
��
n�1ne�n�2
5.
Let
is positive, continuous, and decreasing for
Converges by Theorem 9.10
��
1
1x2 � 1
dx � �arctan x��
1�
�
4
x ≥ 1.f
f �x� �1
x2 � 1.
��
n�1
1n2 � 1
6.
Let
f is positive, continuous, and decreasing for
Diverges by Theorem 9.10
��
1
12x � 1
dx � �ln �2x � 1��
1� �
x ≥ 1.
f �x� �1
2x � 1.
��
n�1
12n � 1
7.
Let
f is positive, continuous, and decreasing for since
Diverges by Theorem 9.10
��
1 ln�x � 1�
x � 1 dx � �ln�x � 1� 2
2 ��
1� �
f��x� �1 � ln�x � 1�
�x � 1�2 < 0 for x ≥ 2.
x ≥ 2
f �x� �ln�x � 1�
x � 1.
��
n�1 ln�n � 1�
n � 18.
Let
is positive, continuous, and decreasing for
diverges
Hence, the series diverges by Theorem 9.10.
��
2 ln x�x
dx � �2�x�ln x � 2���
2� �,
x > e2 � 7.4.f
f �x� �ln x�x
, f ��x� �2 � ln x
2x3�2 .
��
n�2 ln n�n
Section 9.3 The Integral Test and p-Series 261
9.
Let
is positive, continuous, and decreasing for
diverges
Hence, the series diverges by Theorem 9.10.
��
1
1�x��x � 1� dx � �2 ln��x � 1��
�
1� �,
x ≥ 1.f
f �x� �1
�x��x � 1�, f ��x� � �1 � 2�x
2x3�2��x � 1�2 < 0.
��
n�1
1�n��n � 1�
10.
Let
is positive, continuous, and decreasing for since
Diverges by Theorem 9.10
��
1
xx2 � 3
dx � �ln�x2 � 3��
1� �
f��x� �3 � x2
�x2 � 3� < 0 for x ≥ 2.
x ≥ 2f �x�
f �x� �x
x2 � 3.
��
n�1
nn2 � 3
11.
Let
is positive, continuous, and decreasing for
diverges
Hence, the series diverges by Theorem 9.10.
��
1
1�x � 1
dx � �2�x � 1��
1� �,
x ≥ 1.f
f �x� �1
�x � 1, f ��x� �
�12�x � 1�3�2 < 0.
��
n�1
1�n � 1
12.
Let
is positive, continuous, and decreasing for
Hence, the series converges by Theorem 9.10.
�2 ln 2 � 1
16, converges
��
2 ln xx2 dx � ��
�2 ln x � 1�4x 4 �
�
2
x > 2.f
f �x� �ln xx3 , f ��x� �
1 � 3 ln xx4 .
��
n�2
ln nn3
13.
Let
is positive, continuous, and decreasing for
converges
Hence, the series converges by Theorem 9.10.
��
1 ln xx2 dx � ���ln x � 1�
x ��
1� 1,
x > e1�2 � 1.6.f
f �x� �ln xx2 , f ��x� �
1 � 2 ln xx3 .
��
n�1 ln nn2 14.
Let
is positive, continuous, and decreasing for
diverges
Hence, the series diverges by Theorem 9.10.
��
2
1
x�ln x dx � �2�ln x�
�
2� �,
x ≥ 2.f
f �x� �1
x�ln x, f ��x� � �
2 ln x � 12x2�ln x�3�2.
��
n�2
1
n�ln n
15.
Let for
is positive, continuous, and decreasing for
converges
Hence, the series converges by Theorem 9.10.
��
1 arctan xx2 � 1
dx � ��arctan x�2
2 ��
1�
3� 2
32,
x ≥ 1.f
x ≥ 1.f �x� �arctan xx2 � 1
, f ��x� �1 � 2x arctan x
�x2 � 1�2 < 0
��
n�1 arctan nn2 � 1
16.
Let
is positive, continuous, and decreasing for
diverges
Hence, the series diverges by Theorem 9.10.
��
3
1x ln x ln�ln x� dx � �ln�ln�ln x���
�
3� �,
x ≥ 3.f
f ��x� ���ln x � 1� ln�ln x� � 1
x2�ln x�2�ln�ln x��2 .
f �x� �1
x ln x ln�ln x�,
��
n�3
1n ln n ln�ln n�
262 Chapter 9 Infinite Series
17.
Let for
is positive, continuous, and decreasing for
diverges
Hence, the series diverges by Theorem 9.10.
��
1
2xx2 � 1
dx � �ln�x2 � 1���
1� �,
x > 1.f
x > 1.f �x� �2x
x2 � 1, f ��x� � 2
1 � x2
�x2 � 1�2 < 0
��
n�1
2nn2 � 1
18.
Let for
is positive, continuous, and decreasing for
converges
Hence, the series converges by Theorem 9.10.
��
1
xx4 � 1
dx � �12
arctan�x2���
1�
�
8,
x > 1.f
x > 1.f �x� �x
x 4 � 1, f ��x� �
1 � 3x 4
�x 4 � 1�2 < 0
��
n�1
nn4 � 1
19.
Let
is positive, continuous, and decreasing forsince
Diverges by Theorem 9.10
��
1
xk�1
xk � c dx � �1
k ln�xk � c��
�
1� �
for x > k�c�k � 1�.
f � �x� �xk�2c�k � 1� � xk
�xk � c�2 < 0
x > k�c�k � 1�f
f �x� �xk�1
xk � c.
��
n�1
nk�1
nk � c20.
Let
f is positive, continuous, and decreasing for since
for
We use integration by parts.
Converges by Theorem 9.10
�1e
�ke
�k�k � 1�
e� . . . �
k!e
��
1 x ke�x dx � ��x ke�x�
�
1� k ��
1 x k�1e�x dx
x > k.f � �x� �x k�1�k � x�
ex < 0
x > k
f �x� �xk
ex .
��
n�1 n ke�n
21. Let
The function is not positive for x ≥ 1.f
f �x� ���1�x
x, f �n� � an.
22. Let
The function is not positive for x ≥ 1.f
f �x� � e�x cos x, f �n� � an.
23. Let
The function is not decreasing for x ≥ 1.f
f �x� �2 � sin x
x, f �n� � an. 24. Let
The function is not decreasing for x ≥ 1.f
f �x� � �sin xx 2
, f �n� � an.
25.
Let
f is positive, continuous, and decreasing for
Converges by Theorem 9.10
��
1
1x3 dx � ��
12x2�
�
1�
12
x ≥ 1.
f �x� �1x3.
��
n�1 1n3 26.
Let
is positive, continuous, and decreasing for
Diverges by Theorem 9.10
��
1
1x1�3 dx � �3
2 x2�3�
�
1� �
x ≥ 1.f
f �x� �1
x1�3.
��
n�1
1n1�3
Section 9.3 The Integral Test and p-Series 263
27.
Let for
is positive, continuous, and decreasing for
diverges
Hence, the series diverges by Theorem 9.10.
��
1
1�x
dx � �2�x��
1� �,
x ≥ 1.f
x ≥ 1.f �x� �1�x
, f ��x� ��1
2x3�2 < 0
��
n�1
1�n
28.
Let for
is positive, continuous, and decreasing for
converges
Hence, the series converges by Theorem 9.10.
��
1
1x2 dx � ��
1x�
�
1� 1,
x ≥ 1.f
x ≥ 1.f �x� �1x2, f ��x� �
�2x3 < 0
��
n�1 1n2
29.
Divergent p-series with p �15 < 1
��
n�1
15�n
� ��
n�1
1n1�5 30.
Convergent p-series with p �53 > 1
��
n�1
3n5�3 31.
Divergent p-series with p �12 < 1
��
n�1
1n1�2
32.
Convergent p-series with p � 2 > 1
��
n�1 1n2 33.
Convergent p-series with p �32 > 1
��
n�1
1n3�2 34.
Divergent p-series with p �23 < 1
��
n�1
1n2�3
35.
Convergent p-series withp � 1.04 > 1
��
n�1
1n1.04 36.
Convergent p-series withp � � > 1
��
n�1
1n� 37.
Matches (c), diverges
S4 � 4.7740
S3 � 4.0666
S2 � 3.1892
S1 � 2
��
n�1
2n3�4
38.
Matches (f), diverges
S4 � 4.1667
S3 � 3.6667
S2 � 3
S1 � 2
��
n�1 2n
39.
Matches (b), converges
Note: The partial sums for 39 and41 are very similar because��2 � 3�2.
S4 � 3.2560
S3 � 3.0293
S2 � 2.6732
S1 � 2
��
n�1
2�n�
� ��
n�1
2n��2 40.
Matches (a), diverges
S3 � 4.8045
S2 � 3.5157
S1 � 2
��
n�1
2n2�5
41.
Matches (d), converges
Note: The partial sums for 39 and 41 are very similarbecause ��2 � 3�2.
S4 � 3.3420
S3 � 3.0920
S2 � 2.7071
S1 � 2
��
n�1
2
n�n� �
�
n�1
2n3�2 42.
Matches (e), converges
S3 � 2.7222
S2 � 2.5
S1 � 2
��
n�1 2n2
49. The area under the rectangles is greater than the area under the graph of
Since diverges, the series diverges.��
n�1
1�n
��
1
1�x
dx � �2�x��
1� �
��
n�1
1�n
> ��
1
1�x
dx
y � 1��x, x ≥ 1.
x
y
1 2 3 4 5 n
0.5
1.0
264 Chapter 9 Infinite Series
43. (a)
The partial sums approach the sum 3.75 very rapidly.
(b)
The partial sums approach the sum slower than the series in part (a).
� 2�6 � 1.6449 00
11
5
00
11
8
n 5 10 20 50 100
3.7488 3.75 3.75 3.75 3.75Sn
n 5 10 20 50 100
1.4636 1.5498 1.5962 1.6251 1.635Sn
44.
(a)
�N
n�1 1n
� 1 �12
�13
�14
� . . . �1N
> M
(b) No. Since the terms are decreasing (approaching zero),more and more terms are required to increase the partialsum by 2.
M 2 4 6 8
N 4 31 227 1674
45. Let f be positive, continuous, and decreasing for and Then,
and
either both converge or both diverge (Theorem 9.10). See Example 1, page 618.
��
1f �x� dx�
�
n�1an
an � f �n�.x ≥ 1 46. A series of the form is a p-series,
The p-series converges if and diverges if0 < p ≤ 1.
p > 1
p > 0.��
n�1
1np
47. Your friend is not correct. The series
is the harmonic series, starting with the term,and hence diverges.
10,000th
��
n�10,000 1n
�1
10,000�
110,001
� . . .
48. No. Theorem 9.9 says that if the series converges, then the terms tend to zero. Some of the series in Exercises37–42 converge because the terms tend to 0 very rapidly.
an
51.
If then the series diverges by the Integral Test. If
Converges for �p � 1 < 0 or p > 1
��
2
1x�ln x� p dx � ��
2�ln x��p
1x dx � ��ln x��p�1
�p � 1 ��
2.
p � 1,p � 1,
��
n�2
1n�ln n� p50.
1 2 3 4 5 6 7
1
x
y
�6
n�1an ≥ �7
1f �x� dx ≥ �
7
n�2an
Section 9.3 The Integral Test and p-Series 265
52.
If then the series diverges by the Integral Test. If
(Use integration by parts.)
Converges for �p � 1 < 0 or p > 1
��
2 ln xx p dx � ��
2x�p ln x dx � � x�p�1
��p � 1�2 �1 � ��p � 1� ln x��
2.
p � 1,p � 1,
��
n�2 ln nn p
53.
If diverges (see Example 1). Let
For a fixed is eventually negative. is positive, continuous, and eventually decreasing.
For this integral converges. For it diverges.0 < p < 1,p > 1,
��
1
x�1 � x2� p dx � � 1
�x2 � 1� p�1�2 � 2p���
1
fp > 0, p � 1, f ��x�
f ��x� �1 � �2p � 1�x2
�1 � x2� p�1 .
f �x� �x
�1 � x2� p , p � 1
��
n�1
n1 � n2p � 1,
��
n�1
n�1 � n2� p
54.
Since the series diverges for all values of p.p > 0,
��
n�1n�1 � n2� p 55. converges for
Hence, diverges.��
n�2
1n ln n
,
p > 1.��
n�2
1n�ln n� p
56. converges for
Hence, diverges.��
n�2
1
n 3��ln n�2� �
�
n�2
1n�ln n�2�3,
p > 1.��
n�2
1n�ln n� p 57. converges for
Hence, converges.��
n�2
1n�ln n�2,
p > 1.��
n�2
1n�ln n� p
58. converges for
�12
��
n�2
1n ln�n�, diverges
Hence, ��
n�2
1n ln�n2� � �
�
n�2
12n ln n
p > 1.��
n�2
1n�ln n� p 59. Since f is positive, continuous, and decreasing for
and we have,
Also,
Thus, 0 ≤ RN ≤ ��
N
f �x� dx.
≤ ��
N
f �x� dx.
RN � S � SN � ��
n�N�1
an ≤ aN�1 � ��
N�1 f �x� dx
RN � S � SN � ��
n�1 an � �
N
n�1 an � �
�
n�N�1 an > 0.
an � f �n�,x ≥ 1
266 Chapter 9 Infinite Series
60. From Exercise 59, we have:
�N
n�1 an ≤ S ≤ �
N
n�1 an � ��
N
f �x� dx
SN ≤ S ≤ SN � ��
N
f �x� dx
0 ≤ S � SN ≤ ��
N
f �x� dx
61.
1.0811 ≤ ��
n�1 1n4 ≤ 1.0811 � 0.0015 � 1.0826
R6 ≤ ��
6 1x4 dx � ��
13x3�
�
6� 0.0015
S6 � 1 �124 �
134 �
144 �
154 �
164 � 1.0811
62.
1.0363 ≤ ��
n�1 1n5 ≤ 1.0363 � 0.0010 � 1.0373
R4 ≤ ��
4 1x5 dx � ��
14x 4�
�
4� 0.0010
S4 � 1 �125 �
135 �
145 � 1.0363
63.
0.9818 ≤ ��
n�1
1n2 � 1
≤ 0.9818 � 0.0997 � 1.0815
R10 ≤ ��
10
1x2 � 1
dx � �arctan x��
10�
�
2� arctan 10 � 0.0997
� 1
50�
165
�1
82�
1101
� 0.9818 S10 �12
�15
�1
10�
117
�1
26�
137
64.
1.9821 ≤ ��
n�1
1�n � 1�ln�n � 1�3 ≤ 1.9821 � 0.0870 � 2.0691
R10 ≤ ��
10
1�x � 1�ln�x � 1�3 dx � ��
12ln�x � 1�2�
�
10�
12�ln 11�3 � 0.0870
S10 �1
2�ln 2�3 �1
3�ln 3�3 �1
4�ln 4�3 � . . . �1
11�ln 11�3 � 1.9821
65.
0.4049 ≤ ��
n�1 ne�n2
≤ 0.4049 � 5.6 � 10�8
R4 ≤ ��
4 xe�x2
dx � ��12
e�x2��
4�
e�16
2� 5.6 � 10�8
S4 �1e
�2e4 �
3e9 �
4e16 � 0.4049 66.
0.5713 ≤ ��
n�0 e�n ≤ 0.5713 � 0.0183 � 0.5896
R4 ≤ ��
4 e�x dx � ��e�x�
�
4� 0.0183
S4 �1e
�1e2 �
1e3 �
1e4 � 0.5713
67.
N ≥ 7
N > 6.93
N 3 > 333.33
1
N 3 < 0.003
0 ≤ RN ≤ ��
N
1x4 dx � ��
13x3�
�
N�
13N 3 < 0.001 68.
N ≥ 4,000,000
�N > 2000
N�1�2 < 0.0005
0 ≤ RN ≤ ��
N
1
x3�2 dx � ��2
x1�2��
N�
2�N
< 0.001
Section 9.3 The Integral Test and p-Series 267
69.
N ≥ 2
N > 1.0597
N > ln 200
5
5N > ln 200
e5N > 200
1
e5N < 0.005
RN ≤ ��
N
e�5x dx � ��15
e�5x��
N �
e�5N
5 < 0.001 70.
N ≥ 16
N > 2 ln 2000 � 15.2
N2
> ln 2000
eN�2 > 2000
2
eN�2 < 0.001
RN ≤ ��
N
e�x�2 dx � ��2e�x�2��
N�
2eN�2 < 0.001
71.
N ≥ 1004
N > tan 1.5698
arctan N > 1.5698
�arctan N < �1.5698
��
2� arctan N < 0.001
RN ≤ ��
N
1
x2 � 1 dx � �arctan x�
�
N72.
N ≥ 2004
N�5
> tan 1.56968
1.56968 < arctan� N�5
�
2� arctan� N
�5 < 0.001118
�2�5��
2� arctan� N
�5 < 0.001
Rn ≤ ��
N
2x2 � 5
dx � 2� 1�5
arctan� x�5 �
�
N
73. (a) This is a convergent p-series with a divergent series. Use the Integral Test.
is positive, continuous, and decreasing for
(b)
For the terms of the convergent series seem to be larger than those of the divergent series!
(c)
This inequality holds when Or, Then ln e40 � 40 < �e40�0.1 � e4 � 55.n > e40. n ≥ 3.5 � 1015.
ln n < n0.1
n ln n < n1.1
1
n1.1 < 1
n ln n
n ≥ 4,
�6
n�2
1n ln n
�1
2 ln 2�
13 ln 3
�1
4 ln 4�
15 ln 5
�1
6 ln 6� 0.7213 � 0.3034 � 0.1803 � 0.1243 � 0.0930
�6
n�2
1n1.1 �
121.1 �
131.1 �
141.1 �
151.1 �
161.1 � 0.4665 � 0.2987 � 0.2176 � 0.1703 � 0.1393
��
2
1x ln x
dx � �ln�ln x���
2� �
x ≥ 2.f �x� �1
x ln x
��
n�2
1n ln n
isp � 1.1 > 1.��
n�2
1n1.1 .
74. (a)
(c) The horizontal asymptote is As n increases, theerror decreases.
y � 0.
��
10 1x p dx � � x�p�1
�p � 1��
10�
1�p � 1�10 p�1, p > 1 (b)
≤ Area under the graph of f over the interval 10, ��
R10� p� � ��
n�11
1n p
f �x� �1x p
268 Chapter 9 Infinite Series
75. (a) Let f is positive, continuous, and decreasing on
Hence, Similarly,
Thus,
(b) Since we have Also, since is an increasing function,for Thus, and the sequence is bounded.
(c)
Thus, and the sequence is decreasing.
(d) Since the sequence is bounded and monotonic, it converges to a limit,
(e) (Actually ) � 0.577216.a100 � S100 � ln 100 � 0.5822
.
an ≥ an�1
an � an�1 � Sn � ln n � Sn�1 � ln�n � 1� � �n�1
n
1x dx �
1n � 1
≥ 0
�an�0 ≤ Sn � ln n ≤ 1n ≥ 1.ln�n � 1� � ln n > 0ln xln�n � 1� � ln n ≤ Sn � ln n ≤ 1.ln�n � 1� ≤ Sn ≤ 1 � ln n,
ln�n � 1� ≤ Sn ≤ 1 � ln n.
Sn ≥ �n�1
1 1x dx � ln�n � 1�.
Sn ≤ 1 � ln n.
Sn � 1 ≤ ln n
Sn � 1 ≤ �n
1 1x dx
1 2 3 ...n−1
n n+1x
1
12
y1, ��.f �x� � 1�x.
76.
� �ln 7 � ln 5 � 2 ln 6� � �ln 8 � ln 6 � 2 ln 7� � �ln 9 � ln 7 � 2 ln 8� � . . . � �ln 2
� ln 3 � ln 1 � 2 ln 2� � �ln 4 � ln 2 � 2 ln 3� � �ln 5 � ln 3 � 2 ln 4� � �ln 6 � ln 4 � 2 ln 5�
��
n�2 ln�1 �
1n2 � �
�
n�2 ln�n2 � 1
n2 � ��
n�2 ln
�n � 1�(n � 1�n2 � �
�
n�2 ln�n � 1� � ln�n � 1� � 2 ln n
77.
(a) diverges
(b) diverges��
n�2�1
e ln n
� ��
n�2e�ln n � �
�
n�2 1n
,x �1e
:
��
n�21ln n � �
�
n�21,x � 1:
��
n�2xln n
(c) Let be given, Put
This series converges for p > 1 ⇒ x <1e.
��
n�2xln n � �
�
n�2e�p ln n � �
�
n�2n�p � �
�
n�2
1n p
x � e�p ⇔ ln x � �p.x > 0.x
78.
Converges for by Theorem 9.11x > 1
�x� � ��
n�1 n�x � �
�
n�1 1nx
79.
Let
f is positive, continuous, and decreasing for
Diverges by Theorem 9.10
��
1
12x � 1
dx � �ln �2x � 1��
1� �
x ≥ 1.
f �x� �1
2x � 1.
��
n�1
12n � 1
80.
Let
f is positive, continuous, and decreasing for
Converges by Theorem 9.10
��
2
1
x�x2 � 1 dx � �arcsec x�
�
2�
�
2�
�
3
x ≥ 2.
f �x� �1
x�x2 � 1.
��
n�2
1
n�n2 � 1
Section 9.3 The Integral Test and p-Series 269
83.
Geometric series with
Converges by Theorem 9.6
r �23
��
n�0 �2
3 n
84.
Geometric series with
Diverges by Theorem 9.6
r � 1.075
��
n�0 �1.075�n
89.
Let
f is positive, continuous and decreasing for
Converges by Theorem 9.10. See Exercise 51.
��
2
1x�ln x�3 dx � ��
2 �ln x��3
1x dx � ��ln x��2
�2 ��
2� ��
12�ln x�2��
2�
12�ln 2�2
x ≥ 2.
f �x� �1
x�ln x�3.
��
n�2
1n�ln n�3
90.
Let
f is positive, continuous, and decreasing for since for
(Use integration by parts.)
Converges by Theorem 9.10. See Exercise 30.
��
2 ln xx3 dx � ��
ln x2x2��
2�
12
��
2 1x3 dx �
ln 28
� ��1
4x2��
2�
ln 28
�1
16
x ≥ 2.f��x� �1 � 3 ln x
x4 < 0x ≥ 2
f �x� �ln xx3 .
��
n�2 ln nn3
81.
p-series with
Converges by Theorem 9.11
p �54
��
n�1
1
n 4�n� �
�
n�1
1n5�4 82.
p-series with
Diverges by Theorem 9.11
p � 0.95
3 ��
n�1
1n0.95
85.
Diverges by Theorem 9.9
limn→�
n
�n2 � 1� lim
n→�
1�1 � �1�n2�
� 1 � 0
��
n�1
n�n2 � 1
86.
Since these are both convergent p-series, the difference is convergent.
��
n�1 � 1
n2 �1n3 � �
�
n�1 1n2 � �
�
n�1 1n3
87.
Fails nth-Term Test
Diverges by Theorem 9.9
limn→�
�1 �1n
n
� e � 0
��
n�1 �1 �
1n
n
88.
Diverges by Theorem 9.9
limn→�
ln�n� � �
��
n�2 ln�n�
Section 9.4 Comparisons of Series
270 Chapter 9 Infinite Series
1. (a)
(b) The first series is a p-series. It converges
(c) The magnitude of the terms of the other two series are less than the corresponding terms at the convergent p-series. Hence, the other two series converge.
(d) The smaller the magnitude of the terms, the smaller the magnitude of the termsof the sequence of partial sums.
n
2
2
4
4
6
6 8
8
10
10
12
6k3/2Σ
k = 1
n
Σ 6k3/2 + 3k = 1
n
Σ 6
k2+ 0.5kk = 1
n
Sn
� p �32 > 1�.
��
n�1
6
n�n2 � 0.5�
6
1�1.5�
62�4.5
� . . . ; S1 �6
�1.5� 4.9
��
n�1
6n3�2 � 3
�64
�6
23�2 � 3� . . . ; S1 �
32
n
2
1
2
4
3
4
6
5
6 8 10
6n3/2
an =
6n3/2 + 3
an =
6
n2+ 0.5an =
n
an��
n�1
6n3�2 �
61
�6
23�2 � . . . ; S1 � 6
2. (a)
(b) The first series is a p-series. It diverges
(c) The magnitude of the terms of the other two series are greater than the corresponding terms of the divergent p-series. Hence, the other two series diverge.
(d) The larger the magnitude of the terms, the larger the magnitude of the terms of thesequence of partial sums.
n4
4
8
8
10
12
16
20
62
n + 0.54
2n
n − 0.52Σ
Σ
Σ
Sn
� p �12 < 1�.
��
n�1
4�n � 0.5
�4
�1.5�
4�2.5
� . . . S1 � 3.3
��
n�1
2�n � 0.5
�2
0.5�
2�2 � 0.5
� . . . S1 � 4
n
2
2
4
4 6 8 10
1
3an =
n + 0.54
an =n − 0.5
2
an = 2n
an��
n�1
2�n
� 2 �2�2
� . . . S1 � 2
3.
Therefore,
converges by comparison with theconvergent p-series
��
n�1 1n2.
��
n�1
1n2 � 1
0 <1
n2 � 1 <
1n2 4.
Therefore,
converges by comparison with theconvergent p-series
13
��
n�1 1n2.
��
n�1
13n2 � 2
13n2 � 2
< 1
3n2 5. for
Therefore,
diverges by comparison with thedivergent p-series
��
n�2 1n
.
��
n�2
1n � 1
n ≥ 21
n � 1 >
1n
> 0
Section 9.4 Comparisons of Series 271
6. for
Therefore,
diverges by comparison with thedivergent p-series
��
n�2
1�n
.
��
n�2
1�n � 1
n ≥ 21
�n � 1 >
1�n
7.
Therefore,
converges by comparison with theconvergent geometric series
��
n�0 �1
3�n
.
��
n�0
13n � 1
0 <1
3n � 1 <
13n 8.
Therefore,
converges by comparison with theconvergent geometric series
��
n�0�3
4�n
.
��
n�0
3n
4n � 5
3n
4n � 5< �3
4�n
9. For
Therefore,
diverges by comparison with thedivergent series
Note: diverges by the
Integral Test.
��
n�1
1n � 1
��
n�1
1n � 1
.
��
n�1
ln nn � 1
ln nn � 1
> 1
n � 1> 0.n ≥ 3, 10.
Therefore,
converges by comparison with theconvergent p-series
��
n�1
1n3�2.
��
n�1
1�n3 � 1
1�n3 � 1
< 1
n3�2 11. For
Therefore,
converges by comparison with theconvergent p-series
��
n�1 1n2.
��
n�0 1n!
1n2 >
1n!
> 0.n > 3,
12.
Therefore,
diverges by comparison with thedivergent p-series
14
��
n�1
14�n
.
��
n�1
1
4 3�n � 1
1
4 3�n � 1>
1
4 4�n13.
Therefore,
converges by comparison with theconvergent geometric series
��
n�0 �1
e�n
.
��
n�0
1en2
0 <1
en2 ≤ 1en 14.
Therefore,
diverges by comparison with thedivergent geometric series
��
n�1 �4
3�n
.
��
n�1
4n
3n � 1
4n
3n � 1 >
4n
3n
15.
Therefore,
diverges by a limit comparison with the divergent p-series
��
n�1 1n
.
��
n�1
nn2 � 1
limn→�
n��n2 � 1�
1�n� lim
n→�
n2
n2 � 1� 1 16.
Therefore,
converges by a limit comparison with the convergentgeometric series
��
n�1�1
3�n
.
��
n�1
23n � 5
limn→�
2��3n � 5�
1�3n � limn→�
2 � 3n
3n � 5� 2
272 Chapter 9 Infinite Series
17.
Therefore,
diverges by a limit comparison with the divergent p-series
��
n�1 1n
.
��
n�0
1�n2 � 1
limn→�
1��n2 � 1
1�n� lim
n→�
n�n2 � 1
� 1 18.
Therefore,
diverges by a limit comparison with the divergentharmonic series
��
n�3 1n
.
��
n�3
3�n2 � 4
limn→�
3��n2 � 4
1�n� lim
n→�
3n�n2 � 4
� 3
19.
Therefore,
converges by a limit comparison with the convergent p-series
��
n�1 1n3.
��
n�1
2n2 � 13n5 � 2n � 1
limn→�
2n2 � 13n5 � 2n � 1
1�n3 � limn→�
2n5 � n3
3n5 � 2n � 1�
23
20.
Therefore,
diverges by a limit comparison with the divergent p-series
��
n�1 1n
.
��
n�1
5n � 3n2 � 2n � 5
limn→�
5n � 3n2 � 2n � 5
1�n� lim
n→�
5n2 � 3nn2 � 2n � 5
� 5
21.
Therefore,
diverges by a limit comparison with the divergent p-series
��
n�1 1n
.
��
n�1
n � 3n�n � 2�
limn→�
n � 3n�n � 2�
1�n� lim
n→� n2 � 3nn2 � 2n
� 1 22.
Therefore,
converges by a limit comparison with the convergent p-series
��
n�1 1n3.
��
n�1
1n�n2 � 1�
limn→�
1n�n2 � 1�
1�n3 � limn→�
n3
n3 � n� 1
23.
Therefore,
converges by a limit comparison with the convergent p-series
��
n�1 1n2.
��
n�1
1
n�n2 � 1
limn→�
1��n�n2 � 1�
1�n2 � limn→�
n2
n�n2 � 1� 1 24.
Therefore,
converges by a limit comparison with the convergentgeometric series
��
n�1�1
2�n�1
.
��
n�1
n�n � 1�2n�1
limn→�
n��n � 1�2n�1
1��2n�1� � limn→�
n
n � 1� 1
Section 9.4 Comparisons of Series 273
25.
Therefore,
diverges by a limit comparison with the divergent p-series
��
n�1 1n
.
��
n�1
nk�1
nk � 1
limn→�
�nk�1���nk � 1�
1�n� lim
n→�
nk
nk � 1� 1 26.
Therefore,
diverges by a limit comparison with the divergentharmonic series
��
n�1 1n
.
��
n�1
5
n � �n2 � 4
limn→�
5��n � �n2 � 4�
1�n� lim
n→�
5n
n � �n2 � 4�
52
27.
Therefore,
diverges by a limit comparison with the divergent p-series
��
n�1 1n
.
��
n�1 sin�1
n�
� limn→�
cos�1n� � 1
limn→�
sin�1�n�
1�n� lim
n→� ��1�n2� cos�1�n�
�1�n2 28.
Therefore,
diverges by a limit comparison with the divergent p-series
��
n�1 1n
.
��
n�1 tan�1
n�
� limn→�
sec2�1n� � 1
limn→�
tan�1�n�
1�n� lim
n→� ��1�n2� sec2�1�n�
�1�n2
29.
Diverges
p-series with p �12
��
n�1 �nn
� ��
n�1
1�n
30.
Converges
Geometric series with r � �15
��
n�0 5��
15�
n
31.
Converges
Direct comparison with ��
n�1�1
3�n
��
n�1
13n � 2
32.
Converges
Limit comparison with ��
n�1 1n2
��
n�4
13n2 � 2n � 15
33.
Diverges; nth-Term Test
limn→�
n
2n � 3�
12
� 0
��
n�1
n2n � 3
34.
Converges; telescoping series
��
n�1� 1
n � 1�
1n � 2� � �1
2�
13� � �1
3�
14� � �1
4�
15� � . . . �
12
35.
Converges; Integral Test
��
n�1
n�n2 � 1�2 36.
Converges; telescoping series
��
n�1 �1
n�
1n � 3�
��
n�1
3n�n � 3�
274 Chapter 9 Infinite Series
37. By given conditions
is finite and nonzero. Therefore,
diverges by a limit comparison with the p-series
��
n�1 1n
.
��
n�1an
limn→�
nanlimn→�
an
1�n� lim
n→� nan. 38. If then The p-series with
converges and since
the series
converges by the Limit Comparison Test. Similarly, ifthen which implies that
diverges by the Limit Comparison Test.
��
n�1 P�n�Q�n�
k � j ≤ 1j ≥ k � 1,
��
n�1 P�n�Q�n�lim
n→� P�n��Q�n�
1�nk� j � L > 0,
p � k � jk � j > 1.j < k � 1,
39.
which diverges since the degree of the numerator is onlyone less than the degree of the denominator.
12
�25
�3
10�
417
�5
26� . . . � �
�
n�1
nn2 � 1
, 40.
which converges since the degree of the numerator is twoless than the degree of the denominator.
13
�18
�1
15�
124
�1
35� . . . � �
�
n�2
1n2 � 1
,
41.
converges since the degree of the numerator is three lessthan the degree of the denominator.
��
n�1
1n3 � 1
42.
diverges since the degree of the numerator is only one lessthan the degree of the denominator.
��
n�1
n2
n3 � 1
43.
Therefore, diverges.��
n�1
n3
5n4 � 3
limn→�
n� n3
5n4 � 3� � limn→�
n4
5n4 � 3�
15
� 0 44.
Therefore, diverges.��
n�2
1ln n
limn→�
n
ln n� lim
n→�
11�n
� limn→�
n � � � 0
45.
diverges, (harmonic)
1200
�1
400�
1600
� . . . � ��
n�1
1200n
46.
diverges
1200
�1
210�
1220
� . . . � ��
n�0
1200 � 10n
47.
converges
1201
�1
204�
1209
�1
216� �
�
n�1
1200 � n2
48.
converges
1201
�1
208�
1227
�1
264� . . . � �
�
n�1
1200 � n3
49. Some series diverge or converge very slowly. You cannot decide convergence or divergence of a series by comparing the first few terms.
50. See Theorem 9.12, page 624. One example is
converges because and
converges ( p-series).��
n�1 1n2
1n2 � 1
<1n2�
�
n�1
1n2 � 1
51. See Theorem 9.13, page 626. One example is
diverges because and
diverges ( p-series).��
n�2
1�n
limn→�
1��n � 1
1��n� 1�
�
n�2
1�n � 1
52. This is not correct. The beginning terms do not affect theconvergence or divergence of a series. In fact,
diverges (harmonic)
and converges ( p-series).1 �14
�19
� . . . � ��
n�1 1n2
11000
�1
1001� . . . � �
�
n�1000 1n
Section 9.4 Comparisons of Series 275
53.
For Hence, the lower terms are those of � an2.0 < an < 1, 0 < an
2 < an < 1.
0.2
0.4
0.6
0.8
1.0
n4 8 12 16 20
Σan∞
n = 1
Σan∞
n = 1
2
Terms of
Terms of
54. (a)
converges since the degree of the numerator is two less than the degree of the denominator. (See Exercise 38.)
(c) ��
n�3
1�2n � 1�2 �
� 2
8� S2 � 0.1226
��
n�1
1�2n � 1�2 � �
�
n�1
14n2 � 4n � 1
(b)
(d) ��
n�10
1�2n � 1�2 �
� 2
8� S9 � 0.0277
n 5 10 20 50 100
1.1839 1.02087 1.2212 1.2287 1.2312Sn
55. False. Let and and both and converge.��
n�1 1n2�
�
n�1 1n30 < an ≤ bnbn �
1n2.an �
1n3
58. False. Let Then,
but converges.��
n�1cnan ≤ bn � cn,
an � 1�n, bn � 1�n, cn � 1�n2. 59. True
56. True 57. True
60. False. could converge or diverge.
For example, let which diverges.
and diverges, but
and converges.��
n�1 1n20 <
1n2 <
1�n
��
n�1 1n
0 <1n
<1�n
��
n�1bn � �
�
n�1
1�n
,
��
n�1an 61. Since converges, There exists N such
that for Thus, for and
converges by comparison to the convergent
series ��
i�1 an .
��
n�1 an bn
n > Nanbn < ann > N.bn < 1
limn→�
bn � 0.��
n�1 bn
62. Since converges, then
converges by Exercise 61.
��
n�1an an � �
�
n�1 an
2��
n�1 an 63. and both converge, and hence so does
�� 1n2�� 1
n3� � � 1n5.
� 1n3�
1n2
64. converge, and hence so does �� 1n2�2
� � 1n4.�
1n2
65. Suppose and converges.
From the definition of limit of a sequence, there existssuch that
whenever Hence, for From theComparison Test, converges.� an
n > M.an < bnn > M.
�an
bn
� 0� < 1
M > 0
� bnlimn→�
an
bn
� 0
276 Chapter 9 Infinite Series
66. Suppose and diverges. From the definition of limit of a sequence, there exists such that
for Thus, for By the Comparison Test, diverges.� ann > M.an > bnn > M.
an
bn
> 1
M > 0� bnlimn→�
an
bn
� �
67. (a) Let and converges.
By Exercise 65, converges.��
n�1
1�n � 1�3
limn→�
an
bn
� limn→�
1���n � 1�3�
1��n2� � limn→�
n2
�n � 1�3 � 0
� bn � � 1n2,� an � �
1�n � 1�3, (b) Let and converges.
By Exercise 65, converges.��
n�1
1�n� n
limn→�
an
bn
� limn→�
1���n� n�
1��� n� � limn→�
1�n
� 0
� bn � � 1
� n,� an � � 1
�n� n,
68. (a) Let and diverges.
By Exercise 66, diverges.��
n�1 ln n
n
limn→�
an
bn
� limn→�
�ln n��n
1�n� lim
n→� ln n � �
� bn � � 1n
,� an � � ln n
n, (b) Let and diverges.
By Exercise 66, diverges.� 1
ln n
limn→�
an
bn
� limn→�
n
ln n� �
� bn � � 1n
,� an � � 1
ln n,
69. Since the terms of are positive for
sufficiently large Since
and
converges, so does � sin�an�.
� anlimn→�
sin�an�
an
� 1
n.
� sin�an�limn→�
an � 0, 70.
Since converges, and
converges. � 1
1 � 2 � . . . � n
limn→�
2��n�n � 1��
1��n2� � limn→�
2n2
n�n � 1� � 2,
� 1�n2
� ��
n�1
2n�n � 1�
��
n�1
11 � 2 � . . . � n
� ��
n�1
1�n�n � 1���2
71. The series diverges. For
Since diverges, so does � 1
n�n�1��n.� 12n
1
n�n�1��n >12n
1
n1�n >12
n1�n < 2
n < 2n
n > 1, 72. Consider two cases:
If then and
If then and
combining,
Since converges, so does by
the Comparison Test.
��
n�1an
n��n�1���
n�1�2an �
12n
ann��n�1� ≤ 2an �
12n.
ann��n�1� ≤ � 1
2n�1n��n�1��
12n,an ≤
12n�1,
ann��n�1� �
an
an1��n�1� ≤ 2an.
an1��n�1� ≥ � 1
2n�11��n�1�
�12
,an ≥1
2n�1,
Section 9.5 Alternating Series
Section 9.5 Alternating Series 277
1.
Matches (d).
S3 � 8.1667
S2 � 7.5
S1 � 6
��
n�1 6n2 2.
Matches (f ).
S3 � 5.1667
S2 � 4.4
S1 � 6
��
n�1
��1�n�16n2 3.
Matches (a).
S3 � 5.0
S2 � 4.5
S1 � 3
��
n�1 3n!
4.
Matches (b).
S3 � 2.0
S2 � 1.5
S1 � 3
��
n�1 ��1�n�13
n!5.
Matches (e).
S2 � 6.25
S1 � 5
��
n�1 10n2n 6.
Matches (c).
S2 � 3.75
S1 � 5
��
n�1 ��1�n10
n2n
7.
(a)
(b)
(c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases.
(d) The distance in part (c) is always less than the magnitude of the next term of the series.
0.60 11
1.1
��
n�1 ��1�n�1
2n � 1�
�
4� 0.7854
n 1 2 3 4 5 6 7 8 9 10
1 0.6667 0.8667 0.7238 0.8349 0.7440 0.8209 0.7543 0.8131 0.7605Sn
8.
(a)
(b)
(c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases.
(d) The distance in part (c) is always less than the magnitude of the next series.
0 110
2
��
n�1 ��1�n�1
�n � 1�! �1e
� 0.3679
n 1 2 3 4 5 6 7 8 9 10
1 0 0.5 0.3333 0.375 0.3667 0.3681 0.3679 0.3679 0.3679Sn
n 1 2 3 4 5 6 7 8 9 10
1 0.75 0.8611 0.7986 0.8386 0.8108 0.8312 0.8156 0.8280 0.8180Sn
278 Chapter 9 Infinite Series
9.
(a)
(b)
(c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases.
(d) The distance in part (c) is always less than the magnitude of the next term in the series.
0 110.6
1.1
��
n�1 ��1�n�1
n2 ��2
12� 0.8225
10.
(a)
(b)
(c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases.
(d) The distance in part (c) is always less than the magnitude of the next series.
0 110
2
��
n�1
��1�n�1
�2n � 1�! � sin�1� � 0.8415
n 1 2 3 4 5 6 7 8 9 10
1 0.8333 0.8417 0.8415 0.8415 0.8415 0.8415 0.8415 0.8415 0.8415Sn
11.
Converges by Theorem 9.14
limn→�
1n
� 0
an�1 �1
n � 1 <
1n
� an
��
n�1 ��1�n�1
n12.
Diverges by the nth-Term Test
limn→�
n
2n � 1�
12
��
n�1 ��1�n�1 n
2n � 1
13.
Converges by Theorem 9.14
limn→�
1
2n � 1� 0
an�1 �1
2�n � 1� � 1 <
12n � 1
� an
��
n�1 ��1�n�1
2n � 114.
Converges by Theorem 9.14
limn→�
1
ln�n � 1� � 0
an�1 �1
ln�n � 2� <1
ln�n � 1� � an
��
n�1
��1�n
ln�n � 1�
Section 9.5 Alternating Series 279
15.
Thus,
Diverges by the nth-Term Test
limn→�
an � 0.limn→�
n2
n2 � 1� 1.
��
n�1 ��1�n n2
n2 � 116.
Converges by Theorem 9.14
limn→�
n
n2 � 1� 0
an�1 �n � 1
�n � 1�2 � 1 <
nn2 � 1
� an
��
n�1 ��1�n�1n
n2 � 1
17.
Converges by Theorem 9.14
limn→�
1�n
� 0
an�1 �1
�n � 1 <
1�n
� an
��
n�1 ��1�n
�n18.
Diverges by nth-Term Test
limn→�
n2
n2 � 5� 1
��
n�1
��1�n�1n2
n2 � 5
19.
Diverges by the nth-Term Test
� limn→�
�n � 1� � �limn→�
n � 1
ln�n � 1� � limn→�
1
1��n � 1�
��
n�1 ��1�n�1�n � 1�
ln�n � 1� 20.
Converges by Theorem 9.14
limn→�
ln�n � 1�
n � 1� lim
n→� 1��n � 1�
1� 0
an�1 �ln��n � 1� � 1�
�n � 1� � 1 <
ln�n � 1�n � 1
for n ≥ 2
��
n�1 ��1�n�1 ln�n � 1�
n � 1
21.
Diverges by the nth-Term Test
��
n�1 sin�2n � 1��
2 � ��
n�1 ��1�n�1 22.
Converges; (See Exercise 11.)
��
n�1 1n
sin�2n � 1��2 � �
�
n�1 ��1�n�1
n
23.
Diverges by the nth-Term Test
��
n�1 cos n� � �
�
n�1 ��1�n 24.
Converges; (See Exercise 11.)
��
n�1 1n
cos n� � ��
n�1 ��1�n
n25.
Converges by Theorem 9.14
limn→�
1n!
� 0
an�1 �1
�n � 1�! < 1n!
� an
��
n�0 ��1�n
n!
26.
Converges by Theorem 9.14
limn→�
1
�2n � 1�! � 0
an�1 �1
�2n � 3�! <1
�2n � 1�! � an
��
n�0
��1�n
�2n � 1�! 27.
Converges by Theorem 9.14
limn→�
�n
n � 2� 0
<�n
n � 2 for n ≥ 2
an�1 ��n � 1
�n � 1� � 2
��
n�1 ��1�n�1�n
n � 228.
Diverges by the nth-Term Test
limn→�
n1�2
n1�3 � limn→�
n1�6 � �
��
n�1 ��1�n�1�n
3�n
280 Chapter 9 Infinite Series
29.
Converges by Theorem 9.14
� limn→�
233
�45
�57
. . . n
2n � 3 �1
2n � 1� 0
� limn→�
1 � 2 � 3 . . . n
1 � 3 � 5 . . . �2n � 1�
limn→�
an � limn→�
n!
1 � 3 � 5 . . . �2n � 1�
an�1 ��n � 1�!
1 � 3 � 5 . . . �2n � 1��2n � 1� �n!
1 � 3 � 5 . . . �2n � 1� �n � 12n � 1
� an� n � 12n � 1� < an
��
n�1
��1�n�1n!1 � 3 � 5 . . . �2n � 1�
30.
Converges by Theorem 9.14
limn→�
an � limn→�
354
�77
�910
. . . 2n � 13n � 5
13n � 2
� 0
an�1 �1 � 3 � 5 . . . �2n � 1��2n � 1�1 � 4 � 7 . . . �3n � 2��3n � 1� � an�2n � 1
3n � 1� < an
��
n�1��1�n�1
1 � 3 � 5 . . . �2n � 1�1 � 4 � 7 . . . �3n � 2�
31.
Let Then
Thus, is decreasing. Therefore, and
The series converges by Theorem 9.14.
limn→�
2en
e2n � 1� lim
n→� 2en
2e2n � limn→�
1en � 0.
an�1 < an ,f �x�
f��x� ��2ex �e2x � 1�
�e2x � 1�2 < 0.
f �x� �2ex
e2x � 1.
��
n�1 ��1�n�1�2�
en � e�n � ��
n�1 ��1�n�1�2en�
e2n � 132.
Let Then
for
Thus, is decreasing for which implies
The series converges by Theorem 9.14.
limn→�
2en
e2n � 1� lim
n→� 2en
2e2n � limn→�
1en � 0
an�1 < an.x > 0f �x�
x > 0.f��x� �2e2x �1 � e2x�
�e2x � 1�2 < 0
f �x� �2ex
e2x � 1.
��
n�1 2��1�n�1
en � e�n � ��
n�1 ��1�n�1�2en�
e2n � 1
33.
2.3713 ≤ S ≤ 2.4937
R6 � S � S6 ≤ a7 �3
49� 0.0612
2.4325 � 0.0612 ≤ S ≤ 2.4325 � 0.0612
S6 � �6
n�1
3��1�n�1
n2� 2.4325 34.
0.7831 ≤ S ≤ 4.6303
R6 � S � S6 ≤ a7 �4
ln 8� 1.9236
S6 � �6
n�1
4��1�n�1
ln�n � 1� � 2.7067
35.
0.7305 ≤ S ≤ 0.7361
R6 � S � S6 ≤ a7 �26!
� 0.002778
0.7333 � 0.002778 ≤ S ≤ 0.7333 � 0.002778
S6 � �5
n�0
2��1�n
n!� 0.7333 36.
0.1328 ≤ S ≤ 0.2422
R6 � S � S6 ≤ a7 �727 � 0.05469
S6 � �6
n�1
��1�n�1n2n � 0.1875
Section 9.5 Alternating Series 281
37.
(a) By Theorem 9.15,
This inequality is valid when
(b) We may approximate the series by
(7 terms. Note that the sum begins with )n � 0.
� 0.368.
�6
n�0 ��1�n
n!� 1 � 1 �
12
�16
�1
24�
1120
�1
720
N � 6.
RN ≤ aN�1 �1
�N � 1�! < 0.001.
��
n�0 ��1�n
n!38.
(a) By Theorem 9.15,
This inequality is valid when
(b) We may approximate the series by
(5 terms. Note that the sum begins with )n � 0.
�4
n�0 ��1�n
2n n!� 1 �
12
�18
�1
48�
1348
� 0.607.
N � 4.
Rn ≤ aN�1 �1
2N�1�N � 1�! < 0.001.
��
n�0 ��1�n
2n n!
39.
(a) By Theorem 9.15,
This inequality is valid when
(b) We may approximate the series by
(3 terms. Note that the sum begins with )n � 0.
�2
n�0
��1�n
�2n � 1�! � 1 �16
�1
120� 0.842.
N � 2.
RN ≤ aN�1 �1
�2�N � 1� � 1�! < 0.001.
��
n�0
��1�n
�2n � 1�! 40.
(a) By Theorem 9.15,
This inequality is valid when
(b) We may approximate the series by
(4 terms. Note that the sum begins with )n � 0.
�3
n�0 ��1�n
�2n�! � 1 �12
�1
24�
1720
� 0.540.
N � 3.
RN ≤ aN�1 �1
�2N � 2�! < 0.001.
��
n�0 ��1�n
�2n�!
41.
(a) By Theorem 9.15,
This inequality is valid when
(b) We may approximate the series by
(1000 terms)
� 0.693.
�1000
n�1 ��1�n�1
n� 1 �
12
�13
�14
� . . . �1
1000
N � 1000.
RN ≤ aN�1 �1
N � 1 < 0.001.
��
n�1 ��1�n�1
n42.
(a) By Theorem 9.15,
This inequality is valid when
(b) We may approximate the series by
(3 terms)
�3
n�1 ��1�n�1
4n n�
14
�1
32�
1192
� 0.224.
N � 3.
RN ≤ aN�1 �1
4N�1�N � 1� < 0.001.
��
n�1 ��1�n�1
4n n
43.
By Theorem 9.15,
Use 10 terms.
⇒ �N � 1�3 > 1000 ⇒ N � 1 > 10.
RN ≤ aN�1 �1
�N � 1�3 < 0.001
��
n�1 ��1�n�1
n3 44.
By Theorem 9.15,
Use 31 terms.
⇒ �N � 1�2 > 1000 ⇒ N � 31.
RN ≤ aN�1 �1
�N � 1�2 < 0.001
��
n�1 ��1�n�1
n2
282 Chapter 9 Infinite Series
45.
By Theorem 9.15,
This inequality is valid when Use 7 terms.N � 7.
RN ≤ aN�1 �1
2�N � 1�3 � 1 < 0.001.
��
n�1 ��1�n�1
2n3 � 146.
By Theorem 9.15,
This inequality is valid when N � 5.
RN ≤ aN�1 �1
�N � 1�4 < 0.001.
��
n�1 ��1�n�1
n4
47.
converges by comparison to the p-series
Therefore, the given series converges absolutely.
��
n�1 1n2.
��
n�1
1�n � 1�2
��
n�1 ��1�n�1
�n � 1�2 48.
The given series converges by the Alternating Series Test,but does not converge absolutely since the series
diverges by the Integral Test. Therefore, the seriesconverges conditionally.
��
n�1
1n � 1
��
n�1 ��1�n�1
n � 1
49.
The given series converges by the Alternating Series Test,but does not converge absolutely since
is a divergent p-series. Therefore, the series convergesconditionally.
��
n�1
1�n
��
n�1 ��1�n�1
�n50.
which is a convergent p-series.
Therefore, the given series converges absolutely.
��
n�1
1
n�n� �
�
n�1
1n3�2
��
n�1 ��1�n�1
n�n
51.
Therefore, the series diverges by the nth-Term Test.
limn→�
n2
�n � 1�2 � 1
��
n�1 ��1�n�1 n2
�n � 1�2 52.
Therefore, the series diverges by the nth-Term Test.
limn→�
2n � 3n � 10
� 2
��
n�1 ��1�n�1�2n � 3�
n � 10
53.
The given series converges by the Alternating Series Testbut does not converge absolutely since the series
diverges by comparison to the harmonic series
Therefore, the series converges conditionally.��
n�1 1n
.
��
n �2
1ln n
��
n�2 ��1�n
ln n54.
converges by a comparison to
the convergent geometric series Therefore, the
given series converges absolutely.
��
n�0 �1
e�n
.
��
n�0
1e n2
��
n�0 ��1�n
en2
55.
converges by a limit comparison to
the convergent p-series Therefore, the given series
converges absolutely.
��
n�2 1n2.
��
n�2
nn3 � 1
��
n�2 ��1�n nn3 � 1
56.
is a convergent p-series.
Therefore, the given series converges absolutely.
��
n�1
1n1.5
��
n�1 ��1�n�1
n1.5
Section 9.5 Alternating Series 283
57.
is convergent by comparison to the convergent geometricseries
since
Therefore, the given series converges absolutely.
1�2n � 1�! <
12n for n > 0.
��
n�0 �1
2�n
��
n�0
1�2n � 1�!
��
n�0
��1�n
�2n � 1�! 58.
The given series converges by the Alternating Series Test,but
diverges by a limit comparison to the divergent p-series
Therefore, the given series converges conditionally.
��
n�1
1�n
.
��
n�0
1�n � 4
��
n�0
��1�n
�n � 4
59.
The given series converges by the Alternating Series Test,but
diverges by a limit comparison to the divergent harmonicseries,
therefore the series
converges conditionally.
limn→�
cos n� ��n � 1�1�n
� 1,
��
n�1 1n
.
��
n�0 cos n�
n � 1� �
�
n�0
1n � 1
��
n�0 cos n�
n � 1� �
�
n�0 ��1�n
n � 160.
Therefore, the series diverges by the nth-Term Test.
limn→�
arctan n ��
2� 0
��
n�1 ��1�n�1 arctan n
61.
is a convergent p-series.
Therefore, the given series converges absolutely.
��
n�1 1n2
��
n�1 cos n�
n2 � ��
n�1 ��1�n
n2 62.
The given series converges by the Alternating Series Test,but
is a divergent p-series. Therefore, the series convergesconditionally.
��
n�1 sin��2n � 1���2�
n � ��
n�1 1n
��
n�1 sin��2n � 1���2�
n� �
�
n�1 ��1�n�1
n
63. An alternating series is a series whose terms alternate insign. See Theorem 9.14.
64. (Theorem 9.15) S � Sn � Rn ≤ an�1
65. is absolutely convergent if converges. is conditionally convergent if diverges, but converges.
� an� an � an� an � an 66. (b). The partial sums alternate above and below the
horizontal line representing the sum.
67. False. Let
Then converges and converges.
But, diverges.� an � � 1n
� ��an�� an
an ���1�n
n. 68. True. If converged, then so would by
Theorem 9.16.� an� an
284 Chapter 9 Infinite Series
69. True.
Since the next term is negative, is anoverestimate of the sum.
S100�1
101
S100 � �1 �12
�13
� . . . �1
10070. False. Let
Then both converge by the Alternating Series Test. But,
which diverges.�anbn � � 1n
,
�an � �bn � � ��1�n
�n.
71.
If then diverges.
If then diverges.
If then and
Therefore, the series converges for p > 0.
an�1 �1
�n � 1� p <1n p � an.
limn→�
1n p � 0p > 0,
��
n�1��1�nn�pp < 0,
��
n�1��1�np � 0,
��
n�1��1�n
1n p 72.
Assume that so that aredefined for all For all
Therefore, the series converges for all p.
an�1 �1
n � 1 � p<
1n � p
< an.
limn→�
an � limn→�
1
n � p� 0
p,n.an � 1��n � p�n � p � 0
��
n�1��1�n 1
n � p
73. Since
converges we have Thus, there must exist an such that for all
and it follows that for all Hence, by the Comparison Test,
converges. Let to see that the converse is false.an � 1�n
��
n�1 an
2
n > N.an2 ≤ �an�
n > N�aN� < 1N > 0limn→�
�an� � 0.
��
n�1�an�
74. converges, but diverges.��
n�1 1n�
�
n�1 ��1�n�1
n75. converges, hence so does �
�
n�1 1n4.�
�
n�1 1n2
76. (a)
converges absolutely (by comparison) for since
and
is a convergent geometric series for
(b) When we have the convergent alternating series
When we have the divergent harmonic series Therefore,
converges conditionally for x � �1.��
n�1 xn
n
1�n.x � 1,
��
n�1 ��1�n
n.
x � �1,
�1 < x < 1.
� xn�xn
n � < �xn�
�1 < x < 1,
��
n�1 xn
n77. (a) No, the series does not satisfy for all For
example,
(b) Yes, the series converges.
As
S2n → 11 � �1�2� �
11 � �1�3� � 2 �
32
�12
.
n →�,
� �1 �12
� . . . �12n� � �1 �
13
� . . . �13n�
� �12
� . . . �12n� � �1
3� . . . �
13n�
S2n �12
�13
� . . . �12n �
13n
19 < 1
8.n.an�1 ≤ an
Section 9.5 Alternating Series 285
78. (a) No, the series does not satisfy
and
18
<1�3
.
��
n�1��1�n�1an � 1 �
18
�1�3
�1
64� . . .
an�1 ≤ an: (b) No, the series diverges because diverges.� 1�n
88. Diverges by comparison toDivergent Harmonic Series:
for n ≥ 3ln n
n>
1n
89. The first term of the series is zero,not one. You cannot regroup seriesterms arbitrarily.
90. Rearranging the terms of an alternating series can change the sum.
79.
convergent p-series
��
n�1
10n3�2 � 10 �
�
n�1
1n3�2, 80.
converges by limit comparison toconvergent p-series
� 1n2.
��
n�1
3n2 � 5
81. Diverges by nth-Term Test
limn→�
an � �
82. Converges by limit comparison to
convergent geometric series � 12n.
83. Convergent geometric series
�r �78
< 1�84. Diverges by nth-Term Test
limn→�
an �32
85. Convergent geometric seriesor Integral Test�r � 1��e �
86. Converges (conditionally) byAlternating Series Test
87. Converges (absolutely) byAlternating Series Test
91.
(i)
Adding:
(ii) (In fact, )
since
Thus, S � s.
S � limn→�
S3n � s4n �12
s2n � s �12
s �32
s
s >12
.s � 0
s � ln 2.limn→�
sn � s
s4n �12
s2n � 1 �13
�12
�15
�17
�14
� . . . �1
4n � 3�
14n � 1
�1
2n� S3n
12
s2n �12
�14
�16
�18
�1
10� . . . �
14n � 2
�1
4n
s4n � 1 �12
�13
�14
�15
�16
� . . . �1
4n � 1�
14n
S � 1 �13
�12
�15
�17
�14
�19
�1
11�
16
� . . .
s � 1 �12
�13
�14
� . . .
Section 9.6 The Ratio and Root Tests
286 Chapter 9 Infinite Series
1.
� �n � 1��n��n � 1�
�n � 1�!�n � 2�! �
�n � 1��n��n � 1��n � 2�!�n � 2�!
2.
�1
�2k��2k � 1�
�2k � 2�!
�2k�! ��2k � 2�!
�2k��2k � 1��2k � 2�!
3. Use the Principle of Mathematical Induction. When the formula is valid since Assume that
and show that
To do this, note that:
The formula is valid for all n ≥ 1.
��2n � 2�!
2n�1�n � 1�!
��2n�!�2n � 1��2n � 2�
2n�1n!�n � 1�
��2n�!�2n � 1�
2n n!�
�2n � 2�2�n � 1�
��2n�!2n n!
� �2n � 1� �Induction hypothesis�
1 � 3 � 5 . . . �2n � 1��2n � 1� � �1 � 3 � 5 . . . �2n � 1���2n � 1�
1 � 3 � 5 . . . �2n � 1��2n � 1� ��2n � 2�!
2n�1�n � 1�!.
1 � 3 � 5 . . . �2n � 1� ��2n�!2n n!
1 ��2�1��!21 � 1!
.k � 1,
4. Use the Principle of Mathematical Induction. When the formula is valid since Assume that
and show that
To do this, note that:
The formula is valid for all n ≥ 3.
�2n�1�n � 1�!�2n � 1��2n � 1�
�2n � 2�!
�2n �2��n � 1�n!�2n � 1��2n � 1�
�2n�!�2n � 1��2n � 2�
�2n n!�2n � 1�
�2n�! ��2n � 1��2n � 2��2n � 1��2n � 2�
�2n n!�2n � 3��2n � 1�
�2n�! �1
�2n � 3�
1
1 � 3 � 5 . . . �2n � 5��2n � 3� �1
1 � 3 � 5 . . . �2n � 5� �1
�2n � 3�
11 � 3 � 5 . . . �2n � 5��2n � 3� �
2n�1�n � 1�!�2n � 1��2n � 1��2n � 2�! .
11 � 3 � 5 . . . �2n � 5� �
2n n!�2n � 3��2n � 1��2n�!
11
�233!�3��5�
6!� 1.k � 3,
Section 9.6 The Ratio and Root Tests 287
5.
Matches (d).
S1 �34
, S2 � 1.875
��
n�1 n�3
4�n
� 1�34� � 2� 9
16� � . . . 6.
Matches (c).
S1 �34
, S2 � 1.03
��
n�1 �3
4�n
� 1n!� �
34
�9
16 �1
2� � . . . 7.
Matches (f).
S1 � 9
��
n�1 ��3�n�1
n!� 9 �
33
2� . . .
8.
Matches (b).
S1 � 2
��
n�1 ��1�n�1 4
(2n�! �42
�4
24� . . . 9.
Matches (a).
S1 � 2, S2 � 3.31
��
n�1� 4n
5n � 3�n
�4
2� �8
7�2
� . . . 10.
Matches (e).
S1 � 4
��
n�0
4e�n � 4 �4
e� . . .
11. (a) Ratio Test: Converges
(b)
(c)
(d) The sum is approximately 19.26.
(e) The more rapidly the terms of the series approach 0, the more rapidly the sequence of the partial sums approaches the sumof the series.
00
12
20
� limn→�
�n � 1n �
2
58
�58
< 1.limn→�
an�1
an � limn→�
�n � 1�2�58�n�1
n2�58�n
n 5 10 15 20 25
9.2104 16.7598 18.8016 19.1878 19.2491Sn
12. (a) Ratio Test: Converges
(b)
(c)
0 110
10
� limn→�
�n2 � 2n � 2n2 � 1 �� 1
n � 1� � 0 < 1.limn→�
an�1
an � limn→�
�n � 1�2 � 1�n � 1�!n2 � 1
n!
n 5 10 15 20 25
7.0917 7.1548 7.1548 7.1548 7.1548Sn
(d) The sum is approximately 7.15485
(e) The more rapidly the terms of the series approach 0, the more rapidly thesequence of the partial sums approaches the sum of the series.
13.
Therefore, by the Ratio Test, the series diverges.
� limn→�
n � 1
3� �
limn→�
an�1
an � limn→�
�n � 1�!3n�1 �
3n
n!��
n�0 n!3n 14.
Therefore, by the Ratio Test, the series converges.
� limn→�
3
n � 1� 0
limn→�
an�1
an � limn→�
3n�1
�n � 1�! �n!3n
��
n�0 3n
n!
288 Chapter 9 Infinite Series
15.
Therefore, by the Ratio Test, the series converges.
� limn→�
3�n � 1�4n �
34
limn→�
an�1
an � limn→�
�n � 1��34�n�1
n�34�n ��
n�1n�3
4�n
16.
Therefore, by the Ratio Test, the series diverges.
� limn→�
3�n � 1�
2n�
32
limn→�
an�1
an � limn→�
�n � 1�3n�1
2n�1 �2n
n3n��
n�1 n�3
2�n
17.
Therefore, by the Ratio Test, the series converges.
� limn→�
n � 1
2n�
12
limn→�
an�1
an � limn→�
n � 12n�1 �
2n
n ��
n�1 n2n
19.
Therefore, by the Ratio Test, the series diverges.
� limn→�
2n2
�n � 1�2 � 2
limn→�
an�1
an � limn→�
2n�1
�n � 1�2 �n2
2n��
n�1 2n
n2
18.
Therefore, by the Ratio Test, the series converges.
� limn→�
�n � 1�3
2n3 �12
limn→�
an � 1
an � limn→�
�n � 1�32n�1
n32n ��
n�1
n3
2n
20.
Therefore, by Theorem 9.14, the series converges.
Note: The Ratio Test is inconclusive since
The series converges conditionally.
limn→� an�1
an � 1.
limn→�
n � 2
n�n � 1� � 0
an�1 �n � 3
�n � 1��n � 2� ≤ n � 2
n�n � 1� � an
��
n�1 ��1�n�1�n � 2�
n�n � 1�
21.
Therefore, by the Ratio Test, the series converges.
� limn→�
2
n � 1� 0
limn→�
an�1
an � limn→�
2n�1
�n � 1�! �n!2n
��
n�0 ��1�n 2n
n!
23.
Therefore, by the Ratio Test, the series diverges.
� limn→�
n3
� �
limn→�
an�1
an � limn→�
�n � 1�!�n � 1�3n�1 �
n3n
n! ��
n�1
n!n3n
22.
Therefore, by the Ratio Test, the series diverges.
� limn→�
3n2
2�n2 � 2n � 1� �32
> 1
limn→�
an�1
an � limn→�
�32�n�1
n2 � 2n � 1�
n2
�32�n��
n�1 ��1�n�1�32�n
n2
24.
Therefore, by the Ratio Test, the series diverges.
� limn→�
�2n � 2��2n � 1�n5
�n � 1�5 � �
limn→�
an�1
an � limn→�
�2n � 2�!�n � 1�5 �
n5
�2n�!��
n�1 �2n�!
n5
Section 9.6 The Ratio and Root Tests 289
25.
Therefore, by the Ratio Test, the series converges.
� limn→�
4
n � 1� 0
limn→�
an�1
an � limn→�
4n�1
�n � 1�! �n!4n
��
n�0 4n
n!26.
Therefore, by the Ratio Test, the series diverges.
� limn→�
�n � 1n �
n
� e > 1
� limn→�
�n � 1��n � 1�n n!
�n � 1�n!nn
limn→�
an�1
an � limn→�
�n � 1�n�1
�n � 1�! �n!nn
��
n�1 nn
n!
27.
To find let Then,
by L’Hôpital’s Rule
Therefore, by the Ratio Test, the series converges.
y � e�1 �1e.
ln y � limn→�
��1��n � 1�� � ��1��n � 2��
��1n2� � �1
ln y � limn→�
n ln�n � 1n � 2� � lim
n→� ln��n � 1��n � 2��
1n�
00
y � limn→�
�n � 1n � 2�
n
.limn→�
�n � 1n � 2�
n
,
limn→�
an�1
an � limn→�
3n�1
�n � 2�n�1 ��n � 1�n
3n � limn→�
3�n � 1�n
�n � 2�n�1 � limn→�
3
n � 2 �n � 1
n � 2�n
� �0��1e� � 0
��
n�0
3n
�n � 1�n
28.
Therefore, by the Ratio Test, the series converges.
� limn→�
�n � 1�2
�3n � 3��3n � 2��3n � 1� � 0
limn→�
an�1
an � limn→�
��n � 1�!�2
�3n � 3�! ��3n�!�n!�2
��
n�0 �n!�2
�3n�! 29.
Therefore, by the Ratio Test, the series diverges.
� limn→�
4�1 � 13n�3 � 13n �
43
� limn→�
4�3n � 1�3n�1 � 1
limn→�an�1
an � limn→� 4n�1
3n�1 � 1�
3n � 14n
��
n�0
4n
3n � 1
30.
Therefore, by the Ratio Test, the series converges.
limn→�
an�1
an � limn→�
24n�4
�2n � 3�! ��2n � 1�!
24n � limn→�
24
�2n � 3��2n � 2� � 0
��
n�0 ��1�n 24n
�2n � 1�!
31.
Therefore, by the Ratio Test, the series converges.
Note: The first few terms of this series are �1 �1
1 � 3�
2!1 � 3 � 5
�3!
1 � 3 � 5 � 7� . . ..
limn→�
an�1
an � limn→�
�n � 1�!1 � 3 � 5 . . . �2n � 1��2n � 3� �
1 � 3 � 5 . . . �2n � 1�n! � lim
n→�
n � 12n � 3
�12
��
n�0
��1�n�1n!1 � 3 � 5 . . . �2n � 1�
290 Chapter 9 Infinite Series
32.
Therefore, by the Ratio Test, the series converges.
Note: The first few terms of this series are �22
�2 � 42 � 5
�2 � 4 � 62 � 5 � 8
� . . ..
limn→�
an�1
an � limn→�
2 � 4 . . . 2n�2n � 2�2 � 5 . . . �3n � 1��3n � 2� �
2 � 5 . . . �3n � 1�2 � 4 . . . 2n � lim
n→� 2n � 23n � 2
�23
��
n�1 ��1�n 2 � 4 � 6 . . . 2n2 � 5 � 8 . . . �3n � 1�
33.
Ratio Test is inconclusive.
� limn→� �
nn � 1�
32
� 1
limn→�
an�1
an � limn→� 1
�n � 1�32 �n32
1 ��
n�1
1n32 34.
Ratio Test is inconclusive.
� limn→�
� nn � 1�
12
� 1
limn→�
an�1
an � limn→�
1�n � 1�12 �
n12
1 ��
n�1
1n12
35.
limn→�
an�1
an � limn→�
1�n � 1�4 �
n4
1 � limn→�
� nn � 1�
4
� 1
��
n�1 1n4 36.
limn→�
an�1
an � limn→�
1�n � 1� p �
n p
1 � limn→�
� nn � 1�
p
� 1
��
n�1 1np
37.
Therefore, by the Root Test, the series converges.
� limn→�
n
2n � 1�
12
limn→�
n�an � limn→�
n�� n2n � 1�
n
��
n�1 � n
2n � 1�n
38.
Therefore, by the Root Test, the series diverges.
� limn→�
2n
n � 1� 2
limn→�
n�an � limn→�
n�� 2nn � 1�
n
��
n�1 � 2n
n � 1�n
39.
Since the series diverges.2 > 1,
� limn→�
�2n � 1n � 1 � � 2
limn→�
n�an � limn→�
n��2n � 1n � 1 �
n
��
n�2 �2n � 1
n � 1 �n
40.
Since the series diverges.2 > 1,
� limn→�
4n � 32n � 1
� 2
limn→�
n�an � limn→�
n��4n � 32n � 1�
n
��
n�1 �4n � 3
2n � 1�n
41.
Therefore, by the Root Test, the series converges.
� limn→�
1
ln n � 0
limn→�
n�an � limn→�
n���1�n
�ln n�n��
n�2 ��1�n
�ln n�n 42.
Therefore, by the Root Test, the series diverges.
� limn→�
� 3n2n � 1�
3
� �32�
3
�278
limn→�
n�an � limn→�
n�� �3n2n � 1�3n
��
n�1� �3n
2n � 1�3n
Section 9.6 The Ratio and Root Tests 291
43.
To find let Then
Thus, so and Therefore, by the Root Test, the series diverges.limn→�
�2 n�n � 1� � 2�1� � 1 � 3.y � e0 � 1ln y � 0,
ln y � limn→�
�ln n�n � � limn→�
1n
ln n � limn→�
ln n
n� lim
n→� 1n1
� 0.
y � limn→�
n�n .limn→�
n�n,
limn→�
n�an � limn→�
n��2 n�n � 1�n � limn→�
�2 n�n � 1�
��
n�1 �2 n�n � 1�n
44.
Therefore, by the Root Test, the series converges.
limn→�
n�an � limn→�
n� 1en �
1e
��
n�0 e�n 45.
Since the series converges.
Note: You can use L’Hôpital’s Rule to show
Let
Hence, ln y → 0 ⇒ y � n1n → 1.
limn→�
ln nn
� limn→�
1n1
� 0.
y � n1n ⇒ ln y �ln n
n
limn→�
n1n � 1.
14 < 1,
� limn→�
n1n
4�
14
limn→�
n�an � limn→�
n� n4n
��
n�1 n4n
46.
The series diverges.
� limn→�
� n500� � �
limn→�
n�an � lim
n→� n�� n
500�n
��
n�1 � n
500�n
47.
Hence, the series converges.
� limn→�
�1n
�1n2� � 0 � 0 � 0 < 1
limn→�
n�an � limn→�
n��1n
�1n2�
n
��
n�1 �1
n�
1n2�
n
48.
By the Root Test, the series converges.
� limn→�
ln n
n� 0 < 1
limn→�
n�an � limn→�
n��ln nn �
n
��
n�1 �ln n
n �n
49.
Since the series converges by the Root Test.0 < 1,
� limn→�
n1n
ln n� 0
limn→�
n�an � limn→�
n� n�ln n�n
��
n�2
n�ln n�n
50.
The series diverges.
� limn→�
n!n2 � �
limn→�
n�an � limn→�
n��n!��n2�n
��
n�1 �n!�n
�nn�2 � ��
n�1 �n!�n
�n2�n 51.
Therefore, by the Alternating Series Test, the seriesconverges (conditional convergence).
limn→�
5n
� 0
an�1 �5
n � 1 <
5n
� an
��
n�1 ��1�n�1 5
n
292 Chapter 9 Infinite Series
52.
This is the divergent harmonic series.
��
n�1 5n
� 5 ��
n�1 1n
53.
This is convergent p-series.
��
n�1
3
n�n� 3 �
�
n�1
1n32
55.
This diverges by the nth-Term Test for Divergence.
limn→�
2n
n � 1� 2 � 0
��
n�1
2nn � 1
59.
Therefore, the series converges by a Limit ComparisonTest with the geometric series
��
n�0 �1
2�n
.
limn→�
�10n � 3�n2n
12n � limn→�
10n � 3
n� 10
��
n�1 10n � 3
n2n
57.
Since this is a divergent geometric series.r �32 > 1,
� ��
n�1 19��
32�
n
��
n�1 ��1�n 3n�2
2n � ��
n�1 ��1�n 3n 3�2
2n56.
This series diverges by limit comparison to the divergentharmonic series
��
n�1 1n
.
limn→�
n�2n2 � 1�
1n� lim
n→�
n2
2n2 � 1�
12
> 0
��
n�1
n2n2 � 1
58.
Therefore, the series converges by a Limit ComparisonTest with the p-series
��
n�1
1n32.
limn→�
103n32
1n32 �103
��
n�1
10
3�n3
60.
Therefore, the series diverges by the nth-Term Test forDivergence.
� limn→�
�ln 2�22n
8� � lim
n→�
2n
4n2 � 1� lim
n→� �ln 2�2n
8n
��
n�1
2n
4n2 � 1
62.
Therefore, by the Alternating Series Test, the seriesconverges.
limn→�
1
n ln�n� � 0
an�1 �1
�n � 1� ln�n � 1� ≤ 1
n ln�n� � an
��
n�2 ��1�n
n ln n
54.
Since this is convergent geometric series.�4 < 1,
��
n�1��
4�n
63.
Therefore, by the Ratio Test, the series converges.
� limn→�
7n
� 0limn→�
an�1
an � limn→�
�n � 1�7n�1
�n � 1�! �n!
n7n��
n�1 n7n
n!
61.
Therefore, the series
converges by comparison with the geometric series
��
n�0 �1
2�n
.
��
n�1 cos n
2n cos n
2n ≤ 12n
��
n�1 cos n
2n
Section 9.6 The Ratio and Root Tests 293
64.
Therefore, the series converges by comparison with the p-series
��
n�1
1n32.
ln�n�n2 ≤
1n32
��
n�1 ln�n�
n265.
Therefore, by the Ratio Test, the series converges.
(Absolutely)
limn→�
an�1
an � limn→�
3n
�n � 1�! �n!
3n�1 � limn→�
3
n � 1� 0
��
n�1 ��1�n 3n�1
n!
66.
Therefore, by the Ratio Test, the series diverges.
limn→�
an�1
an � limn→�
3n�1
�n � 1�2n�1 �n2n
3n � limn→�
3n
2�n � 1� �32
��
n�1 ��1�n 3n
n2n
68.
Therefore, by the Ratio Test, the series converge.
limn→�
an�1
an � limn→�
3 � 5 � 7 . . . �2n � 1��2n � 3�18n�1�2n � 1��2n � 1�n!
�18n �2n � 1�n!
3 � 5 � 7 . . . �2n � 1� � limn→�
�2n � 3��2n � 1�
18�2n � 1��2n � 1� �1
18
��
n�1 3 � 5 � 7 . . . �2n � 1�
18n �2n � 1�n!
67.
Therefore, by the Ratio Test, the series converges.
limn→�
an�1
an � limn→�
��3�n�1
3 � 5 � 7 . . . �2n � 1��2n � 3� �3 � 5 � 7 . . . �2n � 1�
��3�n � limn→�
3
2n � 3� 0
��
n�1
��3�n
3 � 5 � 7 . . . �2n � 1�
69. (a) and (c)
� 5 ��2��5�2
2!�
�3��5�3
3!�
�4��5�4
4!� . . .
��
n�1 n5n
n!� �
�
n�0 �n � 1�5n�1
�n � 1�!
71. (a) and (b) are the same.
70. (b) and (c)
� 1 � 2�34� � 3�3
4�2
� 4�34�
3
� . . .
��
n�0 �n � 1��3
4�n
� ��
n�1 n �3
4�n�1
72. (a) and (b) are the same.
��
n�1 ��1�n�1
n2n �12
�1
2 � 22 �1
3 � 23 � . . .
��
n�2
��1�n
�n � 1�2n�1 �12
�1
2 � 22 �1
3 � 23 � . . .
73. Replace n with
��
n�1 n4n � �
�
n�0 n � 14n�1
n � 1. 75. Since
use 9 terms.
�9
k�1 ��3�k
2k k!� �0.7769
310
210 10!� 1.59 � 10�5,
74. Replace n with
��
n�2
2n
�n � 2�! � ��
n�0 2n�2
n!
n � 2.
294 Chapter 9 Infinite Series
76.
(See Exercise 3 and use 10 terms, .)k � 9
� 0.40967
� ��
k�0
��6�k k!�2k � 1�!
��
k�0
��3�k
1 � 3 � 5 . . . �2k � 1� � ��
k�0
��3�k 2k k!�2k�!�2k � 1�
77.
The series diverges by the Ratio Test.
� limn→�
4n � 13n � 2
�43
> 1
limn→�an�1
an � limn→��4n � 1��3n � 2�an
an
78.
The series converges by the Ratio Test.
� limn→�
2n � 15n � 4
�25
< 1
limn→�an�1
an � limn→��2n � 1��5n � 4�an
an 79.
The series converges by the Ratio Test.
� limn→�
sin n � 1�n
� 0 < 1
limn→�an�1
an � limn→��sin n � 1���n�an
an
80.
The series converges by the Ratio Test.
� limn→�
cos n � 1n
� 0 < 1
limn→�an�1
an � limn→��cos n � 1��n�an
an 81.
The Ratio Test is inconclusive.
But, so the series diverges.limn→�
an � 0,
� limn→�
�1 �1n� � 1
limn→�an�1
an � limn→��1 � �1��n��an
an
82. The series diverges because
In general, an�1 > an > 0.
a3 � �12�
13
� 0.7937
a2 � �14�
12
�12
a1 �14
limn→�
an � 0. 83.
The series converges by the Ratio Test.
� limn→�
n � 12n � 1
�12
< 1
limn→�an�1
an � limn→� 1 � 2 . . . n�n � 1�
1 � 3 . . . �2n � 1��2n � 1�
1 � 2 . . . n1 � 3 . . . �2n � 1�
84.
Let
Since so
Therefore, by the Root Test, the series converges.
limn→�
�n � 1
3�
13
.
ln y � 0, y � e0 � 1,
� limn→�
ln�n � 1�
n�
1n � 1
� 0.
� limn→�
1n
ln�n � 1�
ln y � limn→�
�ln n�n � 1 � y � lim
n→� n�n � 1
limn→�
n�an � limn→�
n�n � 13n � lim
n→�
n�n � 13
��
n�0 n � 1
3n 85.
Therefore, by the Root Test, the series converges.
limn→�
n�an � limn→�
n� 1�ln n�n � lim
n→�
1ln n
� 0
��
n�3
1�ln n�n
Section 9.6 The Ratio and Root Tests 295
93. See Theorem 9.17, page 597.
86.
The series converges by the Ratio Test.
� limn→�
2n � 1
�2n��2n � 1� � 0 < 1
limn→�an�1
an � limn→� 1 � 3 � 5 . . . �2n � 1��2n � 1�
1 � 2 � 3 . . . �2n � 1��2n��2n � 1�
1 � 3 � 5 . . . �2n � 1�1 � 2 � 3 . . . �2n � 1� 87.
For the series to converge:
For the series diverges.
For the series diverges.
Answer: �3 < x < 3
x � �3,
x � 3,
x3 < 1 ⇒ �3 < x < 3.
� limn→�
x3 � x3 limn→�an�1
an � limn→�2�x3�n�1
2�x3�n
88.
For the series to converge,
For diverges.
For diverges.
Answer: �5 < x < 3
x � �5, ��
n�0��1�n,
x � 3, ��
n�0�1�n,
⇒ �5 < x < 3.
x � 14 < 1 ⇒ �4 < x � 1 < 4
� limn→�
x � 14 � x � 1
4 limn→�
n�an � limn→�
n�x � 14 n
89.
For the series to converge,
For converges.
For diverges.
Answer: �2 < x ≤ 0
x � �2, ��
n�1
��1�n��1�n
n� �
�
n�1 1n
,
x � 0, ��
n�1
��1�n
n,
⇒ �2 < x < 0.
x � 1 < 1 ⇒ �1 < x � 1 < 1
� limn→�
nn � 1
�x � 1� � x � 1
limn→�
an�1
an � limn→��x � 1�n�1�n � 1�
xnn
90.
For the series to converge,
For diverges.
For diverges.
Answer: 0 < x < 2
x � 0, ��
n�02��1�n,
x � 2, ��
n�02�1�n,
⇒ 0 < x < 2.
x � 1 < 1 ⇒ �1 < x � 1 < 1
� limn→�
x � 1 � x � 1
limn→�
an�1
an � limn→�2x � 1n�1
2x � 1n 91.
The series converges only at x � 0.
� limn→�
�n � 1�x2 � �
limn→�
an�1
an � limn→�
�n � 1�!x2n�1
n!x2n
92.
The series converges for all x.
� limn→�
x � 1n � 1
� 0 limn→�
an�1
an � limn→�
x � 1n�1
�n � 1�! x � 1n
n!
94. See Theorem 9.18.
296 Chapter 9 Infinite Series
95. No. Let
The series diverges.��
n�1
1n � 10,000
an �1
n � 10,000. 96. One example is �
�
n�1��100 �
1n�.
97. The series converges absolutely. See Theorem 9.17.
98. Assume that
or that
Then there exists such that for all Therefore,
⇒ limn→�
an � 0 ⇒ � an diverges. an�1 > an, n > N
n > N.an�1an > 1N > 0
limn→�
an�1an � �.limn→�
an�1an � L > 1
99. First, let
and choose R such that There must
exist some such that for all
Thus, for and since the geometric
series
converges, we can apply the Comparison Test to conclude that
converges which in turn implies that converges.��
n�1 an
��
n�1 an
��
n�0 Rn
an < Rnn > N
n > N.n�an < RN > 0
0 ≤ r < R < 1.
limn→�
n�an � r < 1
Second, let
Then there must exist some such that for infinitely many Thus, for infinitely many we have which implies that which in turn implies that
diverges.��
n�1 an
limn→�
an � 0an > Rn > 1n > M,n > M.
n�an > RM > 0
limn→�
n�an � r > R > 1.
100. , p-series
Hence, the Root Test is inconclusive.
Note: because if then
and
Hence y → 1 as n →�.
pn
ln n → 0 as n →�.ln y �pn
ln n
y � npn,limn→�
npn � 1
limn→�
n�an � limn→�
n� 1np � lim
n→�
1npn � 1
��
n�1
1 np 101. Ratio Test:
inconclusive.
Root Test:
Furthermore, let
Thus, inconclusive.limn→�
1
n1n�ln n�pn � 1,
� limn→�
p
ln�n��1n� � 0 ⇒ limn→�
�ln n�pn � 1.
limn→�
ln y � limn→�
p ln �ln n�n
ln y �pn
ln �ln n�.
y � �ln n�pn ⇒ limn→�
n1n � 1.
limn→�
n�an � limn→�
n� 1n�ln n�p � lim
n→�
1n1n�ln n�pn
limn→�an�1
an � limn→�
n�ln n�p
�n � 1��ln�n � 1��p � 1,
Section 9.6 The Ratio and Root Tests 297
102. positive integer
(a) diverges
(b) converges by the Ratio Test:
(c) converges by the Ratio Test:
(d) Use the Ratio Test:
The cases were solved above. For the limit is 0. Hence, the series converges for all integers x ≥ 2.x > 3,x � 1, 2, 3
limn→�
��n � 1�!�2
�x�n � 1��! �n!�2
�xn�! � limn→�
�n � 1�2 �xn�!
�xn � x�!
limn→�
��n � 1�!�2
�3n � 3�! �n!�2
�3n�! � limn→�
�n � 1�2
�3n � 3��3n � 2��3n � 1� � 0 < 1
x � 3: � �n!�2
�3n�!
limn→�
��n � 1�!�2
�2n � 2�! �n!�2
�2n�! � limn→�
�n � 1�2
�2n � 2��2n � 1� �14
< 1
x � 2: � �n!�2
�2n�!
x � 1: � �n!�2
n!� � n!,
x��
n�1 �n!�2
�xn�!,
103. For
Taking limits as � ��
n�1an ≤ �
�
n�1an ≤ �
�
n�1an ⇒ ��n�1
an ≤ ��
n�1an.k →�,
n � 1, 2, 3, . . . ,�an ≤ an ≤ an ⇒ � �k
n�1an ≤ �
k
n�1an ≤ �
k
n�1an.
104. The differentiation test states that if
is an infinite series with real terms and is a real function such that for all positive integers n and exists at then
converges absolutely if and diverges otherwise. Below are some examples.
� sin 1n
, f �x� � sin x��1 � cos 1n�, f �x� � 1 � cos x
� 1n
, f �x� � x� 1n3, f �x� � x3
Divergent SeriesConvergent Series
f �0� � f�0� � 0
��
n�1Un
x � 0,d 2 fdx2f �1n� � Unf �x�
��
n�1Un
105. Using the Ratio Test,
Hence, the series converges.
�197
�1e
< 1
� limn→�
� 1�1 � �1n��n �19
7 �
� limn→�
�n � nn�1
�n � 1�n �197 �
limn→�an�1
an � limn→�
� n!�n � 1�n �19
7 �n�n � 1�!nn�1 �19
7 �n�1
298 Chapter 9 Infinite Series
106. We first prove Abel’s Summation Theorem:
If the partial sums of are bounded and if decreases to zero, then converges.
Let Let be a bound for
The series is absolutely convergent because and
converges to
Also, because bounded and Thus, converges.
Now let to finish the problem.bn �1n
��
n�1anbn � lim
n→� �
n
i�1aibibn → 0.�Sn�lim
n→�Snbn � 0
b1.��
i�1�bi � bi�1�
�Si�bi � bi�1�� ≤ M�bi � bi�1���
i�1Si�bi � bi�1�
� �n�1
i�1Si�bi � bi�1� � Snbn
� S1�b1 � b2� � S2�b2 � b3� � . . . � Sn�1�bn�1 � bn� � Snbn
� S1b1 � �S2 � S1�b2 � . . . � �Sn � Sn�1�bna1b1 � a2b2 � . . . � anbn
��Sk��.MSk � �k
i�1ai.
�anbn�bn��an
Section 9.7 Taylor Polynomials and Approximations
1.
Parabola
Matches (d)
y � �12 x2 � 1 2.
y-axis symmetry
Three relative extrema
Matches (c)
y �18 x 4 �
12 x2 � 1
3.
Linear
Matches (a)
y � e�1�2��x � 1� � 1 4.
Cubic
Matches (b)
y � e�1�2�13 �x � 1�3 � �x � 1� � 1
5.
is called the first degree Taylor polynomial for f at c.P1
−2 6
−2
10
P1
f
(1, 4)
P1�x� � �2x � 6
� 4 � ��2��x � 1�
P1�x� � f �1� � f��1��x � 1�
f��x� � �2x�3�2 f� �1� � �2
f �x� �4
x� 4x�1�2 f �1� � 4 6.
is called the first degree Taylor polynomial for f at c.P1
(8, 2)
−4
−4
14
8
P1�x� � �112
x �83
� 2 � ��1
12��x � 8�
P1�x� � f �8� � f��8��x � 8�
f��x� � �43
x�4�3 f��8� � �1
12
f �x� �43x
� 4x�1�3 f �8� � 2
Section 9.7 Taylor Polynomials and Approximations 299
7.
is called the first degree Taylor polynomial for f at c.P1
−
−1
5
P1
f
( , 2)�4
�2
4π
P1�x� � 2 � 2�x ��
4�
P1�x� � f ��
4� � f���
4��x ��
4�
f��x� � sec x tan x f���
4� � 2
f �x� � sec x f ��
4� � 2 8.
is called the first degree Taylor polynomial for f at c.P1
�2
�2
−
−3
3
P1�x� � 2x � 1 ��
2
� 1 � 2�x ��
4�
P1 � f ��
4� � f���
4��x ��
4�
f��x� � sec2 x f���
4� � 2
f �x� � tan x f ��
4� � 1
9.
� 4 � 2�x � 1� �32
�x � 1�2
P2 � f �1� � f��1��x � 1� �f��1�
2�x � 1�2
f��x� � 3x�5�2 f��1� � 3
f��x� � �2x�3�2 f��1� � �2
−2 6
−2
10
P2
f
(1, 4)
f �x� �4x
� 4x�1�2 f �1� � 4
x 0 0.8 0.9 1.0 1.1 1.2 2
Error 4.4721 4.2164 4.0 3.8139 3.6515 2.8284
7.5 4.46 4.215 4.0 3.815 3.66 3.5P2�x�
f �x�
10.
P2�x� � 2 � 2�x ��
4� �322�x �
�
4�2
P2�x� � f ��
4� � f���
4��x ��
4� �f� ���4�
2 �x ��
4�2
f��x� � sec3 x � sec x tan2 x f���
4� � 32
f��x� � sec x tan x f���
4� � 2
−30
3
4 f �x� � sec x f ��
4� � 2
x 0.585 0.685 0.885 0.985 1.785
1.1995 1.2913 1.4142 1.5791 1.8088
15.5414 1.2160 1.2936 1.4142 1.5761 1.7810 4.9475P2�x�
�4.7043�1.8270f �x�
��4�2.15
300 Chapter 9 Infinite Series
11.
(a)
3−3
−2
2
P6
P2
P4
f
P6�x� � 1 �12x2 �
124x 4 �
1720x6
P4�x� � 1 �12x2 �
124x 4
P2�x� � 1 �12x2
f �x� � cos x (b)
(c) In general, for all n.f �n��0� � Pn�n��0�
f �6��0� � �1 � P6�6��0�
f �6��x� � �cos x P�6��x� � �1
f �5��x� � �sin x P6�5��x� � �x
f �4��0� � 1 � P4�4��0�
f �4��x� � cos x P4�4��x� � 1
f����x� � sin x P4����x� � x
f ��0� � P2��0� � �1
f ��x� � �cos x P2��x� � �1
f��x� � �sin x P2��x� � �x
12.
(a)
P4�x� � x2 � x3 �12x 4
4!� x2 � x3 �
x4
2
P3�x� � x2 �6x3
3!� x2 � x3
P2�x� �2x2
2!� x2
f �4��x� � �x2 � 8x � 12�ex f �4��0� � 12
f���x� � �x2 � 6x � 6�ex f���0� � 6
f ��x� � �x2 � 4x � 2�ex f ��0� � 2
f��x� � �x2 � 2x�ex f��0� � 0
f �x� � x2ex, f �0� � 0
(b)
(c)
(d) f �n��0� � Pn�n��0�
f �4��0� � 12 � P4�4��0�
f���0� � 6 � P3���0�
f ��0� � 2 � P2��0�
−1
−3 3
f
P3
P2P4
2
13.
� 1 � x �x2
2�
x3
6
P3�x� � f �0� � f��0�x �f ��0�
2! x2 �
f����0�3!
x3
f����x� � �e�x f����0� � �1
f ��x� � e�x f ��0� � 1
f��x� � �e�x f��0� � �1
f �x� � e�x f �0� � 1 14.
�f �5��0�
5! x5 � 1 � x �
x2
2�
x3
6�
x 4
24�
x5
120
P5�x� � f �0� � f��0�x �f��0�2!
x2 �f���0�
3! x3 �
f �4��0�4!
x4
f �5��x� � �e�x f �5��0� � �1
f �4��x� � e�x f �4��0� � 1
f ���x� � �e�x f���0� � �1
f ��x� � e�x f ��0� � 1
f��x� � �e�x f��0� � �1
f �x� � e�x f �0� � 1
15.
� 1 � 2x � 2x2 �43
x3 �23
x 4
P4�x� � 1 � 2x �42!
x2 �83!
x3 �164!
x 4
f �4��x� � 162x f �4��0� � 16
f����x� � 8e2x f����0� � 8
f ��x� � 4e2x f ��0� � 4
f��x� � 2e2x f��0� � 2
f �x� � e2x f �0� � 1 16.
� 1 � 3x �92
x2 �92
x3 �278
x 4
P4�x� � 1 � 3x �92!
x2 �273!
x3 �814!
x 4
f �4��x� � 81e3x f �4��0� � 81
f���x� � 27e3x f���0� � 27
f ��x� � 9e3x f � �0� � 9
f��x� � 3e3x f��0� � 3
f �x� � e3x f �0� � 1
Section 9.7 Taylor Polynomials and Approximations 301
17.
� x �16
x3 �1
120 x5
P5�x� � 0 � �1�x �02!
x2 ��13!
x3 �04!
x 4 �15!
x5
f �5��x� � cos x f �5��0� � 1
f �4��x� � sin x f �4��0� � 0
f����x� � �cos x f����0� � �1
f ��x� � �sin x f ��0� � 0
f��x� � cos x f��0� � 1
f �x� � sin x f �0� � 0
19.
� x � x2 �12
x3 �16
x 4
P4�x� � 0 � x �22!
x2 �33!
x3 �44!
x 4
f �4��x� � xex � 4ex f �4��0� � 4
f����x� � xex � 3ex f����0� � 3
f ��x� � xex � 2ex f ��0� � 2
f��x� � xex � ex f��0� � 1
f �x� � xex f �0� � 0
18.
� � x ��3
6 x3 P3�x� � 0 � �x �
02!
x2 ��� 3
3! x3
f���x� � ��3 cos � x f���0� � ��3
f ��x� � ��2 sin � x f ��0� � 0
f��x� � � cos � x f��0� � �
f �x� � sin � x f �0� � 0
20.
� x2 � x3 �12
x 4
P4�x� � 0 � 0x �22!
x2 ��63!
x3 �124!
x 4
f �4��x� � 12e�x � 8xe�x � x2e�x f �4��0� � 12
f���x� � �6e�x � 6xe�x � x2e�x f���0� � �6
f ��x� � 2e�x � 4xe�x � x2e�x f ��0� � 2
f��x� � 2xe�x � x2e�x f��0� � 0
f �x� � x2e�x f �0� � 0
21.
� 1 � x � x2 � x3 � x 4
P4�x� � 1 � x �22!
x2 ��63!
x3 �244!
x 4
f �4��x� �24
�x � 1�5 f �4��0� � 24
f����x� ��6
�x � 1�4 f����0� � �6
f ��x� �2
�x � 1�3 f ��0� � 2
f��x� � �1
�x � 1�2 f��0� � �1
f �x� �1
x � 1 f �0� � 1
23.
P2�x� � 1 � 0x �12!
x2 � 1 �12
x2
f ��x� � sec3 x � sec x tan2 x f ��0� � 1
f��x� � sec x tan x f��0� � 0
f �x� � sec x f �0� � 1
22.
� x � x2 � x3 � x4
P4�x� � 0 � 1�x� �22
x2 �66
x3 �2424
x4
f �4��x� � �24�x � 1��5 f �4��0� � �24
f����x� � 6�x � 1��4 f����0� � 6
f��x� � �2�x � 1��3 f��0� � �2
f��x� � �x � 1��2 f��0� � 1
� 1 � �x � 1��1
f �x� �x
x � 1�
x � 1 � 1x � 1
f �0� � 0
24.
� x �13
x3 P3�x� � 0 � 1�x� � 0 �26
x3
f����x� � 4 sec2 x tan2 x � 2 sec4 x f����0� � 2
f��x� � 2 sec2 x tan x f��0� � 0
f��x� � sec2 x f��0� � 1
f �x� � tan x f �0� � 0
302 Chapter 9 Infinite Series
25.
� 1 � �x � 1� � �x � 1�2 � �x � 1�3 � �x � 1�4
�244!
�x � 1�4
P4�x� � 1 � �x � 1� �22!
�x � 1�2 ��63!
�x � 1�3
f �4��x� �24x5 f �4��1� � 24
f����x� � �6x4 f����1� � �6
f ��x� �2x3 f ��1� � 2
f��x� � �1x2 f��1� � �1
f �x� �1x f �1� � 1 26.
�5
32�x � 2�4
P4�x� �12
�12
�x � 2� �38
�x � 2�2 �14
�x � 2�3
f �4��x� � 240x�6 f �4��x� �154
f����x� � �48x�5 f����x� � �32
f��x� � 12x�4 f��2� �34
f��x� � �4x�3 f��2� � �12
f �x� � 2x�2 f �2� �12
27.
�1
16�x � 1�3 �
5128
�x � 1�4
P4�x� � 1 �12
�x � 1� �18
�x � 1�2
f �4��x� � �15
16x3x f �4��1� � �
1516
f����x� �3
8x2x f����1� �
38
f ��x� � �1
4xx f ��1� � �
14
f��x� �1
2x f��1� �
12
f �x� � x f �1� � 1
29.
�13
�x � 1�3 �14
�x � 1�4
P4�x� � 0 � �x � 1� �12
�x � 1�2
f �4��x� � �6x 4
f �4��1� � �6
f����x� �2x3 f����1� � 2
f ��x� � �1x2 f ��1� � �1
f��x� �1x f��1� � 1
f �x� � ln x f �1� � 0
28.
P3�x� � 2 �1
12�x � 8� �
1288
�x � 8�2 �5
20,736�x � 8�3
f����x� �1027
x�8�3 f����8� �1027
�128 �
53456
f� �x� � �29
x�5�3 f� �8� � �1
144
f��x� �13
x�2�3 f��8� �1
12
f �x� � x1�3 f �8� � 2
30.
P2�x� � ��2 � 2��x � �� ���2 � 2�
2�x � ��2
f ��x� � 2 cos x � 4x sin x � x2 cos x f ���� � �2 � � 2
f��x� � cos x � x 2 sin x f���� � �2�
f �x� � x2 cos x f ��� � �� 2
Section 9.7 Taylor Polynomials and Approximations 303
31.
f
P3
Q3
P5
−6
6
−�2
�2
f �5��x� � 16 sec2 x tan4 x � 88 sec4 x tan2 x � 16 sec6 x
f �4��x� � 8 sec2 x tan3 x � 16 sec4 x tan x
f����x� � 4 sec2 x tan2 x � 2 sec4 x
f ��x� � 2 sec2 x tan x
f��x� � sec2 x
f �x� � tan x (a)
(b)
� 1 � 2�x ��
4� � 2�x ��
4�2
�83�x �
�
4�3
Q3�x� � 1 � 2�x ��
4� �42!�x �
�
4�2
�163!�x �
�
4�3
n � 3, c ��
4
P3�x� � 0 � x �02!
x2 �23!
x3 � x �13
x3
n � 3, c � 0
32.
(a)
(b)
�12
�12
�x � 1� �14
�x � 1�2 �18
�x � 1�4 Q4�x� �12
� ��12��x � 1� �
1�22!
�x � 1�2 �03!
�x � 1�3 ��34!
�x � 1�4
n � 4, c � 1
P4�x� � 1 � 0x ��22!
x2 �03!
x3 �244!
x 4 � 1 � x2 � x 4
n � 4, c � 0
f �4��x� �24�5x 4 � 10x2 � 1�
�x2 � 1�5
f����x� �24x�1 � x2�
�x2 � 1�4
f ��x� �2�3x2 � 1��x2 � 1�3
f��x� ��2x
�x2 � 1�2−3 3
−2
f
Q
P4
4P2
2 f �x� �1
x2 � 1
33.
(a)
(c) As the distance increases, the accuracy decreases.
P5�x� � x �16 x3 �
1120 x5
P3�x� � x �16 x3
P1�x� � x
f �x� � sin x
x 0.00 0.25 0.50 0.75 1.00
0.0000 0.2474 0.4794 0.6816 0.8415
0.0000 0.2500 0.5000 0.7500 1.0000
0.0000 0.2474 0.4792 0.6797 0.8333
0.0000 0.2474 0.4794 0.6817 0.8417P5�x�
P3�x�
P1�x�
sin x
(b)
−3
−2
P1P3
P5
f
3
� 2�
304 Chapter 9 Infinite Series
34.
(a)
(c) As the degree increases, the accuracy increases. As the distance from to 1 increases, the accuracy decreases.x
P4�x� � �x � 1� �12�x � 1�2 �
13�x � 1�3 �
14�x � 1�4
P2�x� � �x � 1� �12�x � 1�2
P1�x� � x � 1
f �x� � ln x, c � 1
x 1.00 1.25 1.50 1.75 2.00
0 0.2231 0.4055 0.5596 0.6931
0 0.2500 0.5000 0.7500 1.0000
0 0.2188 0.375 0.4688 0.5000
0 0.2230 0.4010 0.5303 0.5833P4�x�
P2�x�
P1�x�
ln x
(b)
5
−2
−1
P1
P4P2
2
f
35.
(a)
(b)
P3�x� � x �x3
6
f �x� � arcsin x
x 0 0.25 0.50 0.75
0 0.253 0.524 0.848
0 0.253 0.521 0.820�0.253�0.521�0.820P3�x�
�0.253�0.524�0.848f �x�
�0.25�0.50�0.75
(c)
1
2
−1
−
f
x
3P
π
2π
y
36. (a)
(b)
P3�x� � x �x3
3
f �x� � arctan x (c)
x
f
1−1 12
π
π
4
4
π2
π2
−
−
P3
y
x 0 0.25 0.50 0.75
0 0.2450 0.4636 0.6435
0 0.2448 0.4583 0.6094�0.2448�0.4583�0.6094P3�x�
�0.2450�0.4636�0.6435f �x�
�0.25�0.50�0.75
37.6
6 8
4
2
−4
−6
−6x
P6 P2
P8 P4y
f (x) = cos x
f �x� � cos x
39.3
2
1
−2 2 3 4−3−4
−3
x
P6P2
P8P4
y
f (x) = ln (x2 + 1)
f �x) � ln�x2 � 1�
38.
−2
2
1
x
f x x( ) = arctan
1 3−3 −2
PP33PP11PP13P9
yPP7 PP1P5f �x� � arctan x
40.
x
2
4
−4 4
yy = 4xe(−x2/4)f �x� � 4xe�x2�4
Section 9.7 Taylor Polynomials and Approximations 305
41.
f �12� 0.6042
f �x� � e�x 1 � x �x2
2�
x3
642.
f �15� 0.0328
f �x� � x2e�x x2 � x3 �12
x4
43.
f �1.2� 0.1823
f �x� � ln x �x � 1� �12 �x � 1�2 �
13 �x � 1�3 �
14 �x � 1�4
45.
Note: you could use
Exact error: 0.000001 � 1.0 10 �6
R5�x� ≤16!
�0.3�6 � 1.0125 10�6
R5�x�: f �6��x� � �cos x, max on �0, 0.3 is 1.
R4�x� ≤15!
�0.3�5 � 2.025 10�5
f �x� � cos x; f �5��x� � �sin x ⇒ Max on �0, 0.3 is 1.
44.
f�7�
8 � �6.7954
f �x� � x2 cos x ��2 � 2��x � �� � �� 2 � 22 ��x � ��2
46.
R5�x� ≤ e1
6! �1�6 0.00378 � 3.78 10�3
f �x� � ex; f �6��x� � ex ⇒ Max on �0, 1 is e1.
47.
The exact error is Note: You could use R4.�8.5 10�4.R3�x� ≤ 7.3340
4!�0.4�4 0.00782 � 7.82 10�3.
f �x� � arcsin x; f �4��x� �x�6x2 � 9��1 � x2�7�2 ⇒ Max on �0, 0.4 is f �4��0.4� 7.3340.
48.
R3�x� ≤ 22.3672
4! �0.4�4 0.0239
⇒ Max on �0, 0.4 is f �4��0.4� 22.3672.
f �x� � arctan x; f �4��x� �24x�x2 � 1�
�1 � x2�4 49.
for all x.
By trial and error, n � 3.
Rn�x� ≤ 1
�n � 1�!�0.3�n�1 < 0.001
�g�n�1��x�� ≤ 1
g�x� � sin x
50.
for all and all
By trial and error, n � 2.
≤�0.1�n�1
�n � 1�! < 0.001
�Rn�x�� � � f �n�1��z�xn�1
�n � 1�! �n.x� f �n�1��x�� ≤ 1
f �x� � cos x 51.
Max on is
By trial and error, n � 5.
Rn ≤1.8221
�n � 1�! �0.6�n�1 < 0.001
e0.6 1.8221.�0, 0.6
f �n�1��x� � ex
f �x� � ex
306 Chapter 9 Infinite Series
52.
The maximum value on is
By trial and error, .n � 3
�Rn� ≤1.3499
�n � 1�! �0.3�n�1 < 0.001
e0.3 1.3499.�0, 0.3
f �n�1��x� � e x
f �x� � e x 53.
By trial and error, (See Example 9.) Using 9 terms,ln�1.5� 0.4055.
n � 9.
Rn ≤ n!
�n � 1�! �0.5�n�1 ��0.5�n�1
n � 1 < 0.0001
f �n�1��x� ���1�n n!
�x � 1�n�1 ⇒ Max on �0, 0.5 is n!.
f �x� � ln�x � 1�
55.
By trial and error, n � 16.
�Rn� ≤ ���n�1
�n � 1�! �1.3�n�1 < 0.0001
f �n�1��x� � ����n�1e��x ≤ �����n�1� on �0, 1.3
f��x� � ����e��x
f �x� � e��x, f (1.3�54.
Since this is an alternating series,
By trial and error, Using 4 terms f �0.6� 0.4257.n � 4.
Rn ≤ an�1 ��2n
�2n�! �0.6�4n < 0.0001.
f �0.6� � 1 �� 2
2! �0.6�4 �
�4
4! �0.6�8 �
� 6
6! �0.6�12 � . . .
� 1 �� 2x 4
2!�
�4x8
4!�
�6x12
6!� . . .
� 1 ���x2�2
2!�
��x2�4
4!�
��x2�6
6!� . . .
f �x� � g��x2�
g�x� � cos x � 1 �x2
2!�
x 4
4!�
x6
6!� . . .
f �x� � cos��x2�
58.
�0.3936 < x < 0.3936
�x� < 0.3936
x 4 < 0.024
�R3�x�� � �sin z4!
x4� ≤ �x4�
4! < 0.001
f �x� � sin x x �x3
3!59. fifth degree polynomial
for all and all
Note: Use a graphing utility to graphin the viewing window
to verify theanswer.��0.9467, 0.9467 ��0.001, 0.001y � cos x � �1 � x2�2 � x 4�24�
�0.9467 < x < 0.9467
�x� < 0.9467
�x�6 < 0.72
�R5�x�� ≤16!
�x�6 < 0.001
n.x� f �n�1��x�� ≤ 1
f �x� � cos x 1 �x2
2!�
x 4
4!,
56.
By trial and error, n � 7.
�Rn� ≤ 1
�n � 1�! 1n�1 ≤ 0.0001
f �n�1��x� � ��1�n�1e�x ≤ 1 on �0, 1
f��x� � �e�x
f �x� � e�x 57.
�0.3936 < x < 0
�x� < 0.3936
ez�4 < 0.3936, z < 0
�xez�4� < 0.3936
ez x4 < 0.024
R3�x� �ez
4! x 4 < 0.001
f �x� � ex 1 � x �x2
2�
x3
6, x < 0
Section 9.7 Taylor Polynomials and Approximations 307
60.
Thus
In fact, by graphing and you can verify that on ��0.19294, 0.20068�
� f �x� � y� < 0.001y � 1 � 2x � 2x2 �43 x
3,f �x� � e�2x
0 < x < 0.1970.
x < �0.0015e�2z �
1�4
0.1970e2z < 0.1970, for z < 0.
e�2zx 4 < 0.0015
R3 �x� �f 4�z�
4!�x � 0�4 �
16e�2z
24 x 4 �
23
e�2z x 4 < 0.001
f��x� � �2e�2x, f��x� � 4e�2x, f���x� � �8e�2x, f �4��x� � 16e�2x
f �x� � e�2x 1 � 2x � 2x2 �43
x3
61. The graph of the approximating polynomial P and theelementary function f both pass through the point and the slopes of P and f agree at Depending onthe degree of P, the nth derivatives of P and f agree at�c, f �c��.
�c, f �c��.�c, f �c��
62. f �c� � P2�c�, f��c� � P2��c�, and f��c� � P2��c�
63. See definition on page 650. 64. See Theorem 9.19, page 654.
65. The accuracy increases as the degree increases (for valueswithin the interval of convergence).
66.
x20−20 10
2
−2
−4
4
6
8
10 f
y
P1
P3
P2
67. (a)
(b)
(c) g�x� �sin x
x�
1x P5�x� � 1 �
x2
3!�
x 4
5!
Q6�x� � x P5�x� � x2 �x 4
3!�
x6
5!
g�x� � x sin x
P5�x� � x �x3
3!�
x5
5!
f �x� � sin x
Q5�x� � x P4�x�
Q5�x� � x � x2 �12
x3 �16
x 4 �1
24 x5
g�x� � xex
P4�x� � 1 � x �12
x2 �16
x3 �1
24 x 4
f �x� � ex 68. (a) for
This is the Maclaurin polynomial of degree 4 for
(b) for
(c)
The first four terms are the same!
R��x� � 1 � x �x2
2!�
x3
3!
R�x� � 1 � x �x2
2!�
x3
3!�
x 4
4!
Q6��x� � �x �x3
3!�
x5
5!� �P5�x�
cos xQ6�x� � 1 �x2
2�
x 4
4!�
x6
6!
g�x� � cos x.
P5��x� � 1 �x2
2!�
x 4
4!
f �x� � sin xP5�x� � x �x3
3!�
x5
5!
308 Chapter 9 Infinite Series308 Chapter 9 Infinite Series
69. (a) Q2�x� � �1 ��2�x � 2�2
32(b) R2�x� � �1 �
�2�x � 6�2
32(c) No. The polynomial will be linear.
Translations are possible atx � �2 � 8n.
70. Let f be an odd function and be the nth Maclaurin polynomial for f. Since f is odd, is even:
Similarly, is odd, is even, etc. Therefore, f, etc. are all odd functions, which implies that Hence, in the formula
all the coefficients of the even power of x are zero.Pn�x� � f �0� � f��0�x �f ��0�x2
2!� . . .
f �0� � f ��0� � . . . � 0.f �4�,f�,f��f�
f���x� � limh→0
f ��x � h� � f ��x�
h� lim
h→0
�f �x � h� � f �x�h
� limh→0
f �x � ��h�� � f �x�
�h� f��x�.
f�Pn
71. Let f be an even function and be the nth Maclaurin polynomial for f. Since f is even, is odd, is even, is odd, etc. All of the odd derivatives of f are odd and thus, all of the odd powers of x will have coefficients of zero. will only haveterms with even powers of x.
Pn
f���f�f�Pn
72. Let
For Pn�k��c� � an k! � �f �k��c�
k! �k! � f �k��c�.1 ≤ k ≤ n,
Pn�c� � a0 � f �c�
Pn �x� � a0 � a1�x � c� � a2�x � c�2 � . . . � an �x � c�n where ai �f �i��c�
i!.
73. As you move away from the Taylor Polynomial becomes less and less accurate.x � c,
Section 9.8 Power Series
1. Centered at 0 2. Centered at 0 3. Centered at 2 4. Centered at �
5.
�x� < 1 ⇒ R � 1
� limn→�
�n � 1n � 2��x� � �x�
L � limn→�
�un�1
un � � limn→�
���1�n�1xn�1
n � 2�
n � 1��1�n xn�
��
n�0��1�n
xn
n � 1
7.
2�x� < 1 ⇒ R �12
� limn→�
� 2n2x�n � 1�2� � 2�x�
L � limn→�
�un�1
un � � limn→�
� �2x�n�1
�n � 1�2 �n2
�2x�n���
n�1 �2x�n
n2
6.
2�x� < 1 ⇒ R �12
L � limn→�
�un � 1un � � lim
n→� ��2x�n�1
�2x�n � � 2�x�
��
n�0�2x�n
8.
12
�x� < 1 ⇒ R � 2
�12�x�
L � limn→�
�un�1
un � � limn→�
���1�n�1xn�1
2n�1 �2n
��1�n xn���
n�0 ��1�n xn
2n
Section 9.8 Power Series 309
9.
Thus, the series converges for all x. R � �.
� limn→�
� �2x�2
�2n � 2��2n � 1�� � 0
L � limn→�
�un�1
un � � limn→�
��2x�2n�2��2n � 2�!�2x�2n��2n�! �
��
n�0
�2x�2n
�2n�!
11.
Since the series is geometric, it converges only ifor �2 < x < 2.�x�2� < 1
��
n�0�x
2�n
10.
The series only converges at R � 0.x � 0.
� limn→�
��2n � 2��2n � 1�x2
�n � 1� � � �
L � limn→�
�un � 1un � � lim
n→� ��2n � 2�!x2n�2��n � 1�!
�2n�!x2n�n! ���
n�0
�2n�!x2n
n!
12.
is geometric, so it converges if
�x�5� < 1 ⇒ �x� < 5 ⇒ �5 < x < 5.
��
n�0�x
5�n
13.
Interval:
When the alternating series converges.
When the p-series diverges.
Therefore, the interval of convergence is �1 < x ≤ 1.
��
n�1 1n
x � �1,
��
n�1 ��1�n
n x � 1,
�1 < x < 1
� limn→�
� nxn � 1� � �x�
limn→�
�un�1
un � � limn→�
���1�n�1xn�1
n � 1�
n��1�n xn�
��
n�1 ��1�n xn
n
15.
The series converges for all x. Therefore, the interval of convergence is �� < x < �.
� limn→�
� xn � 1� � 0
limn→�
�un�1
un � � limn→�
� xn�1
�n � 1�! �n!xn�
��
n�0 xn
n!
14.
Interval:
When the series diverges.
When the series diverges.
Therefore, the interval of convergence is �1 < x < 1.
��
n�0��n � 1�x � �1,
��
n�0��1�n�1�n � 1�x � 1,
�1 < x < 1
� limn→�
��n � 2�xn � 1 � � �x�
limn→�
�un � 1un � � lim
n→� ���1�n�2�n � 2�xn�1
��1�n�n � 1�xn ���
n�0��1�n�1�n � 1�xn
16.
Therefore, the interval of convergence is �� < x < �.
� limn→�
� 3x�2n � 2��2n � 1�� � 0
limn→�
�un�1
un � � limn→�
� �3x�n�1
�2n � 1�! ��2n�!�3x�n�
��
n�0 �3x�n
�2n�!
17.
Therefore, the series converges only for x � 0.
� limn→�
��2n � 2��2n � 1� x2 � � � lim
n→� �un�1
un � � limn→�
��2n � 2�!xn�1
2n�1 �2n
�2n�!xn� ��
n�0�2n�!�x
2�n
310 Chapter 9 Infinite Series
19.
Since the series is geometric, it converges only if or �4 < x < 4.�x�4� < 1
��
n�1 ��1�n�1xn
4n
18.
Interval:
When the alternating series converges.
When the series converges by limit comparison to
Therefore, the interval of convergence is �1 ≤ x ≤ 1.
��
n�1 1n2.�
�
n�0
1�n � 1��n � 2�x � �1,
��
n�0
��1�n
�n � 1��n � 2�x � 1,
�1 < x < 1
� limn→�
��n � 1�xn � 3 � � �x� lim
n→� �un�1
un � � limn→�
� ��1�n�1xn�1
�n � 2��n � 3� ��n � 1��n � 2�
��1�n xn ���
n�0
��1�n xn
�n � 1��n � 2�
20.
Center:
Therefore, the series converges only for x � 4.
x � 4
R � 0
� limn→�
��n � 1��x � 4�3 � � � lim
n→� �un�1
un � � limn→�
���1�n�1�n � 1�!�x � 4�n�1
3n�1 �3n
��1�n n!�x � 4�n���
n�0 ��1�n n!�x � 4�n
3n
21.
Center:
Interval: or
When the p-series diverges. When the alternating series converges.
Therefore, the interval of convergence is 0 < x ≤ 10.
��
n�1
��1�n�1
nx � 10,�
�
n�1 �1n
x � 0,
0 < x < 10�5 < x � 5 < 5
x � 5
R � 5
� limn→�
�n�x � 5�5�n � 1�� �
15�x � 5� lim
n→� �un�1
un � � limn→�
���1�n�2�x � 5�n�1
�n � 1�5n�1 �n5n
��1�n�1�x � 5�n���
n�1 ��1�n�1�x � 5�n
n5n
22.
Center:
Interval: or
When the alternating series converges.
When the series diverges.
Therefore, the interval of convergence is �2 ≤ x < 6.
��
n�0
1n � 1
x � 6,
��
n�0
��1�n�1
�n � 1�x � �2,
�2 < x < 6�4 < x � 2 < 4
x � 2
R � 4
� limn→�
��x � 2��n � 1�4�n � 2� � �
14�x � 2� lim
n→� �un�1
un � � limn→�
� �x � 2�n�2
�n � 2�4n�2 ��n � 1�4n�1
�x � 2�n�1 ���
n�0
�x � 2�n�1
�n � 1�4n�1
Section 9.8 Power Series 311
23.
Center:
Interval: or
When the series diverges by the integral test.
When the alternating series converges.
Therefore, the interval of convergence is 0 < x ≤ 2.
��
n�0 ��1�n�1
n � 1x � 2,
��
n�0
1n � 1
x � 0,
0 < x < 2�1 < x � 1 < 1
x � 1
R � 1
� limn→�
��n � 1��x � 1�n � 2 � � �x � 1�lim
n→� �un�1
un � � limn→�
���1�n�2�x � 1�n�2
n � 2�
n � 1��1�n�1�x � 1�n�1�
��
n�0 ��1�n�1�x � 1�n�1
n � 1
24.
At
diverges.
At
converges.
Interval of convergence: 0 < x ≤ 4
��
n�1
��1�n�1 2n
n2n � ��
n�1
��1�n�1
n,
x � 4,
��
n�1
��1�n�1��2�n
n2n � ��
n�1
��1��2n�n2n � �
�
n�1
��1�n
,
x � 0,
�x � 22 � < 1 ⇒ �2 < x � 2 < 2 ⇒ 0 < x < 4
� limn→��x � 2
2
nn � 1� � �x � 2
2 � limn→��an�1
an � � limn→����1�n�2�x � 2�n�1
�n � 1�2n�1 ���1�n�1�x � 2�n
n2n ���
n�1
��1�n�1�x � 2�n
n2n25. is geometric. It converges if
Interval convergence: 0 < x < 6
�x � 33 � < 1 ⇒ �x � 3� < 3 ⇒ 0 < x < 6.
��
n�1�x � 3
3 �n�1
26.
Interval:
When converges.
When converges.
Therefore, the interval of convergence is �1 ≤ x ≤ 1.
x � �1, ��
n�0
��1�n�1
2n � 1
x � 1, ��
n�0
��1�n
2n � 1
�1 < x < 1
R � 1
� limn→�
��2n � 1��2n � 3�x
2� � �x2�limn→�
�un�1
un � � limn→�
���1�n�1x2n�3
�2n � 3� ��2n � 1�
��1�nx2n�1���
n�0
��1�nx2n�1
2n � 1
312 Chapter 9 Infinite Series
28.
Therefore, the interval of convergence is �� < x < �.
� limn→�
� x2
n � 1� � 0
limn→�
�un�1
un � � limn→�
���1�n�1x2n�2
�n � 1�! �n!
��1�n x2n� ��
n�0 ��1�n x2n
n!
29.
Therefore, the interval of convergence is �� < x < �.
� limn→�
� x2
�2n � 2��2n � 3�� � 0
limn→�
�un�1
un � � limn→�
� x2n�3
�2n � 3�! ��2n � 1�!
x2n�1 ���
n�0
x2n�1
�2n � 1�!30.
Therefore, the interval of convergence is �� < x < �.
� limn→�
� �n � 1�x�2n � 2��2n � 1�� � 0
limn→�
�un�1
un � � limn→�
��n � 1�!xn�1
�2n � 2�! ��2n�!n!xn �
��
n�1 n!xn
�2n�!
32.
When the series diverges by comparing it to
which diverges. Therefore, the interval of convergence is �1 < x < 1.
��
n�1
12n � 1
x � ±1,
R � 1
limn→�
�un�1
un � � limn→�
� 2 � 4 . . . �2n��2n � 2�x2n�3
3 � 5 � 7 . . . �2n � 1��2n � 3� �3 � 5 . . . �2n � 1�2 � 4 . . . �2n�x2n�1� � lim
n→� ��2n � 2�x2
�2n � 3� � � �x 2�
��
n�1
2 � 4 � 6 . . . �2n�3 � 5 � 7 . . . �2n � 1� �x
2n�1�
31.
Converges if
At diverges.
Interval of convergence: �1 < x < 1
x � ±1,
�x� < 1 ⇒ �1 < x < 1.
limn→��an�1
an � � limn→���n � 2�xn�1
�n � 1�xn � � limn→��n � 2
n � 1x� � �x�
��
n�1 2 � 3 � 4 . . . �n � 1�xn
n!� �
�
n�1�n � 1�xn
27.
Interval:
When the series diverges
by the nth Term Test.
When the alternating series diverges.
Therefore, the interval of convergence is �12
< x < 12
.
��
n�1 ��1�n�1n
n � 1x �
12
,
��
n�1
nn � 1
x � �12
,
�12
< x < 12
R �12
� limn→�
���2x��n � 1�2
n�n � 2� � � 2�x�
limn→�
�un�1
un � � limn→�
��n � 1���2x�n
n � 2�
n � 1n��2x�n�1�
��
n�1
nn � 1
��2x�n�1
Section 9.8 Power Series 313
33.
Center:
Therefore, the series converges only for x � 3.
x � 3
R � 0
� limn→�
��4n � 3��x � 3�4 � � �
limn→�
�un�1
un � � limn→�
���1�n�2 � 3 � 7 � 11 . . . �4n � 1��4n � 3��x � 3�n�1
4n�1 �4n
��1�n�1 � 3 � 7 � 11 . . . �4n � 1��x � 3�n���
n�1 ��1�n�1 3 � 7 � 11 . . . �4n � 1��x � 3�n
4n
34.
Converges if
At diverges.
At diverges.
Interval of convergence: �3 < x < 1
an �n!��2�n
1 � 3 . . . �2n � 1� � ��1�n 2 � 4 . . . 2n1 � 3 . . . �2n � 1�, x � �3,
an �n!2n
1 � 3 � 5 . . . �2n � 1� �2 � 4 � 6 . . . 2n
1 � 3 � 5 . . . �2n � 1� > 1, x � 1,
12�x � 1� < 1 ⇒ �2 < x � 1 < 2 ⇒ �3 < x < 1.
� limn→���n � 1��x � 1�
2n � 1 � �12�x � 1�
limn→��an�1
an � � limn→�� �n � 1�!�x � 1�n�1
1 � 3 � 5 . . . �2n � 1��2n � 1�� �n�!�x � 1�n
1 � 3 � 5 . . . �2n � 1����
n�1
n!�x � 1�n
1 � 3 � 5 . . . �2n � 1�
35.
Center:
Interval: or
When the series diverges.
When the series diverges.
Therefore, the interval of convergence is 0 < x < 2c.
��
n�1 1x � 2c,
��
n�1��1�n�1x � 0,
0 < x < 2c�c < x � c < c
x � c
R � c
�1c�x � c�lim
n→� �un�1
un � � limn→�
��x � c�n
cn �cn�1
�x � c�n�1���
n�1 �x � c�n�1
cn�1 36. is a positive integer.
Converges if �x�kk < 1 ⇒ R � kk.
� �x�kk
� limn→�� �n � 1�kx
�k � nk��k � 1 � nk� . . . �1 � nk�� limn→��an�1
an � � limn→����n � 1�!kxn�1
�k�n � 1�! ��n!�kxn
�kn�! ���
n�0 �n!�kxn
�kn�! , k
37.
Since the series is geometric, it converges only if or .�k < x < k.�x�k� < 1
��
n�0�x
k�n
314 Chapter 9 Infinite Series
38.
Center:
Interval: or
When the p-series diverges.
When the alternating series converges. Therefore, the interval of convergence is 0 < x ≤ 2c.��
n�1 ��1�n�1
nx � 2c,
��
n�1 �1n
x � 0,
0 < x < 2c�c < x � c < c
x � c
R � c
� limn→�
�n�x � c�c�n � 1�� �
1c�x � c�lim
n→� �un�1
un � � limn→�
���1�n�2�x � c�n�1
�n � 1�cn�1 �ncn
��1�n�1�x � c�n���
n�1 ��1�n�1�x � c�n
ncn
39.
When the series diverges and the interval of convergence is
k�k � 1� . . . �k � n � 1�1 � 2 . . . n
≥ 1��1 < x < 1.x � ±1,
R � 1
limn→�
�un�1
un � � limn→�
�k�k � 1� . . . �k � n � 1��k � n�xn�1
�n � 1�! �n!
k�k � 1� . . . �k � n � 1�xn� � limn→�
��k � n�xn � 1 � � �x�
��
n�1 k�k � 1� . . . �k � n � 1�xn
n!
40.
Interval: or
The series diverges at the endpoints. Therefore, the interval of convergence is
n!�c � 2 � c�n
1 � 3 � 5 . . . �2n � 1� �n!22
1 � 3 � 5 . . . �2n � 1�� 2 � 4 � 6 . . . �2n�
1 � 3 � 5 . . . �2n � 1� > 1�
c � 2 < x < c � 2.
c � 2 < x < c � 2�2 < x � c < 2
R � 2
limn→�
�un�1
un � � limn→�
� �n � 1�!�x � c�n�1
1 � 3 � 5 . . . �2n � 1��2n � 1� �1 � 3 � 5 . . . �2n � 1�
n!�x � c� � � limn→�
��n � 1��x � c�2n � 1 � �
12
�x � c�
��
n�1
n!�x � c�n
1 � 3 � 5 . . . �2n � 1�
41. � ��
n�1
xn�1
�n � 1�! ��
n�0 xn
n!� 1 �
x1
�x2
2� . . . 42. �
�
n�0��1�n�1�n � 1�xn � �
�
n�1��1�n�n�xn�1
43.
Replace n with n � 1.
��
n�0
x2n�1
�2n � 1�! � ��
n�1
x2n�1
�2n � 1�!44. �
�
n�0
��1�nx2n�1
2n � 1� �
�
n�1
��1�n�1x2n�1
2n � 1
45. (a) (Geometric)
(b)
(c)
(d) �2 ≤ x < 2� f �x� dx � ��
n�0
2n � 1
�x2�
n�1
,
�2 < x < 2f ��x� � ��
n�2�n
2��n � 1
2 ��x2�
n�2
,
f��x� � ��
n�1�n
2��x2�
n�1
, �2 < x < 2
f �x� � ��
n�0�x
2�n
, �2 < x < 2 46. (a)
(b)
(c)
(d) 0 ≤ x ≤ 10� f �x� dx � ��
n�1 ��1�n�1�x � 5�n�1
n�n � 1�5n ,
0 < x < 10f ��x� � ��
n�2 ��1�n�1�n � 1��x � 5�n�2
5n ,
f��x� � ��
n�1 ��1�n�1�x � 5�n�1
5n , 0 < x < 10
f �x� � ��
n�1 ��1�n�1�x � 5�n
n5n , 0 < x ≤ 10
Section 9.8 Power Series 315
48. (a)
(b)
(c)
(d) �f �x� dx � ��
n�1
��1�n�1�x � 2�n�1
n�n � 1� , 1 ≤ x ≤ 3
f��x� � ��
n�2��1�n�1�n � 1��x � 2�n�2, 1 < x < 3
f��x� � ��
n�1��1�n�1�x � 2�n�1, 1 < x < 3
f �x� � ��
n�1
��1�n�1�x � 2�n
n, 1 < x ≤ 347. (a)
(b)
(c)
(d) 0 ≤ x ≤ 2� f �x� dx � ��
n�1 ��1�n�1�x � 1�n�2
�n � 1��n � 2� ,
f ��x� � ��
n�1��1�n�1n�x � 1�n�1, 0 < x < 2
f��x� � ��
n�0 ��1�n�1�x � 1�n, 0 < x < 2
f �x� � ��
n�0 ��1�n�1�x � 1�n�1
n � 1, 0 < x ≤ 2
49.
Matches (c).S1 � 1, S2 � 1.33.
g�1� � ��
n�0�1
3�n
� 1 �13
�19
� . . . 50.
Matches (a).S1 � 1, S2 � 1.67.
g�2� � ��
n�0�2
3�n
� 1 �23
�49
� . . .
51. diverges. Matches (b).g�3.1� � ��
n�0�3.1
3 �n
52. alternating. Matches (d).g��2� � ��
n�0��
23�
n
53. converges
Matches (b).S1 � 1, S2 � 1.25, S3 � 1.3125
g�18� � �
�
n�02�1
8��n
� ��
n�0�1
4�n
� 1 �14
�1
16� . . .,
54. converges
Matches (c).S1 � 1, S2 � 0.75, S3 � 0.8125
g��18� � �
�
n�02��
18��
n
� ��
n�0��
14�
n
� 1 �14
�1
16� . . .,
55. diverges
Matches (d).S1 � 1, S2 �178
g� 916� � �
�
n�02� 9
16��n
� ��
n�0�9
8�n
, 56. converges
Matches (a).S1 � 1, S2 � 1.75
g�38� � �
�
n�02�3
8��n
� ��
n�0�3
4�n
,
57. A series of the form
is called a power series centered at c.
��
n�0an�x � c�n
58. The set of all values of x for which the power series converges is the interval of convergence. If the powerseries converges for all x, then the radius of convergenceis If the power series converges at only c, then
Otherwise, according to Theorem 8.20, thereexists a real number (radius of convergence) suchthat the series converges absolutely for anddiverges for �x � c� > R.
�x � c� < RR > 0
R � 0.R � �.
59. A single point, an interval, or the entire real line. 60. You differentiate and integrate the power series term byterm. The radius of convergence remains the same.However, the interval of convergence might change.
61. Answers will vary.
converges for . At the convergence is conditional because diverges.
converges for . At the convergence is absolute.x � ±1,�1 ≤ x ≤ 1��
n�1
xn
n2
� 1n
x � �1,�1 ≤ x < 1��
n�1
xn
n
316 Chapter 9 Infinite Series
62. Many answers possible.
(a) Geometric:
(b) converges for �1 < x ≤ 1��
n�1
��1�nxn
n
�x2� < 1 ⇒ �x� < 2��
n�1�x
2�n
(c) Geometric:
(d) converges for �2 ≤ x < 6��
n�1
�x � 2�n
n4n
�2x � 1� < 1 ⇒ �1 < x < 0��
n�1�2x � 1�n
63. (a)
(See Exercise 29.)
(b) f��x� � ��
n�0 ��1�n x2n
�2n�! � g�x�
g�x� � ��
n�0 ��1�n x2n
�2n�! , �� < x < �
f �x� � ��
n�0 ��1�n x2n�1
�2n � 1�! , �� < x < �
64. (a) (See Exercise 11)
(b) � ��
n�1
xn�1
�n � 1�! � ��
n�0 xn
n!� f �x� f��x� � �
�
n�1 nxn�1
n!
f �x� � ��
n�0 xn
n!, �� < x < � (c)
(d) f �x� � ex
f �0� � 1
f �x� � ��
n�1 xn
n!� 1 � x �
x2
2!�
x3
3!�
x 4
4!� . . .
65.
y� � y � ��
n�1 ��1�n x2n�1
�2n � 1�! � ��
n�1 ��1�n�1x2n�1
�2n � 1�! � 0
y� � ��
n�1 ��1�n �2n�x2n�1
�2n�! � ��
n�1 ��1�x2n�1
�2n � 1�!
y� � ��
n�0 ��1�n �2n � 1�x2n
�2n � 1�! � ��
n�0 ��1�nx2n
�2n�!
y � ��
n�0 ��1�n x2n�1
�2n � 1�! � ��
n�1 ��1�n�1x2n�1
�2n � 1�! 66.
y� � y � ��
n�1 ��1�n x2n�2
�2n � 2�! � ��
n�1 ��1�n�1x2n�2
�2n � 2�! � 0
y� � ��
n�1 ��1�n �2n � 1�x2n�2
�2n � 1�! � ��
n�1 ��1�nx2n�2
�2n � 2�!
y� � ��
n�1 ��1�n �2n�x2n�1
�2n�! � ��
n�1 ��1�nx2n�1
�2n � 1�!
y � ��
n�0 ��1�n x2n
�2n�! � ��
n�1 ��1�n�1x2n�2
�2n � 2�!
67.
y� � y � 0
y� � ��
n�1 �2n�x2n�1
�2n�! � ��
n�1
x2n�1
�2n � 1�! � y
y� � ��
n�0 �2n � 1�x2n
�2n � 1�! � ��
n�0
x2n
�2n�!
y � ��
n�0
x2n�1
�2n � 1�! � ��
n�1
x2n�1
�2n � 1�! 68.
y� � y � 0
y� � ��
n�1 �2n � 1�x2n�2
�2n � 1�! � ��
n�1
x2n�2
�2n � 2�! � y
y� � ��
n�1 �2n � 2�x2n�1
�2n � 2�! � ��
n�1
x2n�1
�2n � 1�!
y � ��
n�0
x2n
�2n�! � ��
n�1
x2n�2
�2n � 2�!
69.
� 0 � ��
n�0 2�n � 1�x2n ��2n � 1� � �2n � 1�
2n�1�n � 1�!
� ��
n�0�2n � 2��2n � 1�x2n
2n�1�n � 1�! ��2n � 1�x2n
2n n!�
2�n � 1�2�n � 1��
� ��
n�1 2n�2n � 1�x2n�2
2n n!� �
�
n�0 �2n � 1�x2n
2n n!
y� � xy� � y � ��
n�1 2n�2n � 1�x2n�2
2n n!� �
�
n�1 2nx2n
2n n!� �
�
n�0
x2n
2n n!
y� � ��
n�1 2n�2n � 1�x2n�2
2n n!y� � �
�
n�1 2nx2n�1
2n n!y � �
�
n�0
x2n
2n n!
(c)
(d) and g�x� � cos xf �x� � sin x
� � ��
n�0 ��1�n x2n�1
�2n � 1�! � �f �x�
g��x� � ��
n�1 ��1�n x2n�1
�2n � 1�! � ��
n�0 ��1�n�1x2n�1
�2n � 1�!
Section 9.8 Power Series 317
70.
� 0 � ��
n�1
��1�n�1 4�n � 1�x4n�2
22n�2�n � 1�! � 3 � 7 � 11 . . . �4n � 1� � ��
n�1
��1�n�1 x 4n�2
22n n! � 3 � 7 � 11 . . . �4n � 1� 22�n � 1�22�n � 1�
y� � x2y � �x2 � ��
n�2
��1�n 4nx 4n�2
22n n! � 3 � 7 � 11 . . . �4n � 5� � ��
n�1
��1�n x4n�2
22n n! � 3 � 7 � 11 . . . �4n � 1� � x2
� �x2 � ��
n�2
��1�n 4nx4n�2
22n n! � 3 � 7 � 11 . . . �4n � 5� y� � ��
n�1
��1�n 4n�4n � 1�x4n�2
22n n! � 3 � 7 � 11 . . . �4n � 1�
y� � ��
n�1
��1�n 4nx 4n�1
22n n! � 3 � 7 � 11 . . . �4n � 1�
y � 1 � ��
n�1
��1�n x4n
22n n! � 3 � 7 � 11 . . . �4n � 1�
71.
(a)
Therefore, the interval of convergence is
(b)
� ��
k�0 ��1�k x2k�2
4k �k!�2 �4k � 24k � 4
�2
4k � 4�
4k � 44k � 4� � 0
� ��
k�0 ��1�k x2k�2
4k �k!�2 ��1�2�2k � 1�4�k � 1� � ��1� 2
4�k � 1� � 1�
x2J0� � xJ0� � x2J0 � ��
k�0��1�k�1
2�2k � 1� x2k�2
4k�1�k � 1�!k!� �
�
k�0��1�k�1
2x2k�2
4k�1�k � 1�!k!� �
�
k�0��1�k
x2k�2
4k �k!�2
� ��
k�0��1�k�1
�2k � 2��2k � 1�x2k
4k�1 ��k � 1�!2 J0� � ��
k�1��1�k
2k�2k � 1�x2k�2
4k �k!�2
� ��
k�0��1�k�1
�2k � 2� x2k�1
4k�1��k � 1�!2 J0� � ��
k�1��1�k
2kx2k�1
4k �k!�2
J0 � ��
k�0��1�k
x2k
4k �k!�2
�� < x < �.
limk→�
�uk�1
uk � � limk→�
� ��1�k�1 x2k�2
22k�2 ��k � 1�!2 �22k �k!�2
��1�k x2k� � limk→�
� ��1�x2
22�k � 1�2� � 0
J0�x� � ��
k�0 ��1�k x2k
22k �k!�2
(d)
(integral is approximately 0.9197304101)
� 1 �1
12�
1320
0.92
� ��
k�0
��1�k
4k�k!�2�2k � 1�
� ��
k�0
��1�k x2k�1
4k�k!�2�2k � 1��1
0
�1
0 J0dx � �1
0 �
�
k�0 ��1�k x2k
4k �k!�2 dx
72.
(a)
Therefore, the interval of convergence is
—CONTINUED—
�� < x < �.
limk→�
�uk�1
uk � � limk→�
� ��1�k�1 x2k�3
22k�3�k � 1�!�k � 2�! �22k�1k!�k � 1�!
��1�k x2k�1 � � limk→�
� ��1�x2
22 �k � 2��k � 1�� � 0
J1�x� � x ��
k�0
��1�k x2k
22k�1k!�k � 1�! � ��
k�0
��1�k x2k�1
22k�1k!�k � 1�!
(c)
−6 6
−5
3
P6�x� � 1 �x2
4�
x4
64�
x6
2304
318 Chapter 9 Infinite Series
72. —CONTINUED
(b)
(c)
(d)
Note: J0��x� � �J1�x� �J1�x� � � ��
k�0
��1�kx2k�1
22k�1k!�k � 1�! � ��
k�0
��1�k�1x2k�1
22k�1k!�k � 1�!
J0��x� � ��
k�0 ��1�k�12�k � 1�x2k�1
22k�2�k � 1�!�k � 1�! � ��
k�0
��1�k�1x2k�1
22k�1k!�k � 1�! −6 6
−4
4P7�x� �x2
�1
16 x3 �
1384
x5 �1
18,432 x7
� ��
k�0 ��1�kx2k�3���1� � 1
22k�1k!�k � 1�! � 0
� ��
k�0
��1�k�1x2k�3
22k�1k!�k � 1�! � ��
k�0
��1�kx2k�3
22k�1k!�k � 1�!
� ��
k�1
��1�kx2k�1
22k�1�k � 1�!k!� �
�
k�0
��1�k x2k�3
22k�1k!�k � 1�!
� ��
k�1 ��1�kx2k�14k�k � 1�
22k�1k!�k � 1�! � ��
k�0
��1�kx2k�3
22k�1k!�k � 1�!
� ��
k�1 ��1�kx2k�1��2k � 1��2k� � �2k � 1� � 1
22k�1k!�k � 1�! � ��
k�0
��1�kx2k�3
22k�1k!�k � 1�!
�x2
� ��
k�1
��1�kx2k�1
22k�1k!�k � 1�!� � ��
k�0
��1�kx2k�3
22k�1k!�k � 1�!
� ��
k�1 ��1�k�2k � 1��2k�x2k�1
22k�1k!�k � 1�! �x2
� ��
k�1 ��1�k�2k � 1�x2k�1
22k�1k!�k � 1�!
� ��
k�0
��1�kx2k�3
22k�1k!�k � 1�! � ��
k�0
��1�k x2k�1
22k�1k!�k � 1�!
x2J1� � xJ1� � �x2 � 1�J1 � ��
k�1 ��1�k �2k � 1��2k�x2k�1
22k�1k!�k � 1�! � ��
k�0 ��1�k�2k � 1�x2k�1
22k�1k!�k � 1�!
J1��x� � ��
k�1 ��1�k �2k � 1��2k�x2k�1
22k�1k!�k � 1�!
J1��x� � ��
k�0 ��1�k �2k � 1�x2k
22k�1k!�k � 1�!
J1�x� � ��
k�0
��1�k x2k�1
22k�1k!�k � 1!�
73.
−2
2
−2� 2�
f �x� � ��
n�0��1�n
x2n
�2n�! � cos x 74.
� sin x−
−2
� �
2 f �x� � ��
n�0��1�n
x2n�1
�2n � 1�!
75. Geometric
−1 10
3
�1
1 � ��x� �1
1 � x for �1 < x < 1
f �x� � ��
n�0��1�nxn � �
�
n�0��x�n 76.
−2.5 2.5
− �2
�2
f �x� � ��
n�0��1�n
x2n�1
2n � 1� arctan x, �1 ≤ x ≤ 1
Section 9.8 Power Series 319
77.
(a)
(c) The alternating series converges more rapidly. The partial sums of the series of positive terms approach the sum from below. The partial sums of the alternating series alternate sides of the horizontal line representing the sum.
−1 60
1.80
�1
1 � �3�8� �85
� 1.6 ��
n�0�3�4
2 �n
� ��
n�0�3
8�n
��
n�0�x
2�n
(b)
(d) �N
n�0�3
2�n
> M
0.60−1 6
1.10
�1
1 � ��3�8� �8
11 0.7272
��
n�0��3�4
2 �n
� ��
n�0��
38�
n
M 10 100 1000 10,000
N 4 9 15 21
78. converges on
(a)
(c) The alternating series converges more rapidly. The partial sums in (a) approach the sum 2 from below. The partial sums in (b) alternate sides of the horizontal line y �
23.
3
60
−1
��
n�0�3�1
6��n
� ��
n�0�1
2�n
�1
1 � �1�2� � 2x �16
:
��13
, 13�.�
�
n�0�3x�n
(b)
(d) �N
n�0�3 �
23�
n
� �N
n�02n > M
1.5
6
−0.5
−1
�1
1 � �1�2� �23
��
n�0�3��
16��
n
� ��
n�0��
12�
n
x � �16
:
M 10 100 1000 10,000
N 3 6 9 13
79. False;
converges for but diverges for x � �2.x � 2
��
n�0 ��1�nxn
n2n
81. True; the radius of convergence is for both series.R � 1
80. True; if
converges for then we know that it must convergeon ��2, 2.
x � 2,
��
n�0 an x
n
82. True
� ��
n�0
an
n � 1 �1
0 f �x� dx � �1
0 ��
�
n�0 an x
n� dx � ��
n�0 an x
n�1
n � 1 �1
0
320 Chapter 9 Infinite Series
83.
Thus, the series converges for all x : R � �.
limn→�
�an�1
an � � limn→�
� �n � 1 � p�!�n � 1�!�n � 1 � q�! x
n�1� �n � p�!n!�n � q�! x
n� � limn→�
� �n � 1 � p�x�n � 1��n � 1 � q�� � 0
84. (a)
Since each series is geometric,
(b) For g�x� �1
1 � x2 � 2x1
1 � x2 �1 � 2x1 � x2.�x� < 1,
R � 1.
limn→�
S2n � ��
n�0x2n � 2x �
�
n�0x2n
� �1 � x2 � x 4 � . . . � x2n� � 2x�1 � x2 � x 4 � . . . � x2n�
S2n � 1 � 2x � x2 � 2x3 � x 4 � . . . � x2n � 2x2n�1
g�x� � 1 � 2x � x2 � 2x3 � x 4 � . . .
85. (a)
Each series is geometric, and the interval of convergence is
(b) For f �x� � c01
1 � x3 � c1x1
1 � x3 � c2x2 11 � x3 �
c0 � c1x � c2x2
1 � x3 .�x� < 1,
��1, 1�.R � 1,
limn→�
S3n � c0 ��
n�0x3n � c1x �
�
n�0x3n � c2x2 �
�
n�0x3n
S3n � c0�1 � x3 � . . . � x3n� � c1x�1 � x3 � . . . � x3n� � c2x2�1 � x3 � . . . � x3n�
� c0 � c1x � c2x2 � c0x
3 � c1x4 � c2x
5 � c0x6 � . . .
f �x� � ��
n�0cnxn, cn�3 � cn
86. For the series
For the series
limn→�
�bn�1
bn � � limn→�
�cn�1x2n�2
cn x2n � � limn→�
�cn�1
cn x
2� < 1 ⇒ �x2� < � cn
cn�1� � R ⇒ �x� < �R.
�cn x2n,
limn→�
�an�1
an � � limn→�
�cn�1xn�1
cnxn � � limn→�
�cn�1
cn
x� < 1 ⇒ �x� < � cn
cn�1� � R
�cn xn,
87. At the series is which converges.
At the series is which diverges.
Hence, at we have converges, diverges.
The series converges conditionally. ⇒
��
n�0�cn��R�n� � �
�
n�0cnRn�
�
n�0cn��R�nx0 � R,
��
n�0cnRn,x0 � R,
��
n�0cn�x0 � x0 � R�n � �
�
n�0cn��R�nx0 � R,
Section 9.9 Representation of Functions by Power Series
Section 9.9 Representation of Functions by Power Series 321
1. (a)
This series converges on
(b)
�
x3
8�
x 4
16
x3
8
x2
4�
x3
8
x2
4
x2
�x2
4
x2
1 �x2
2 � x ) 1
12
�x4
�x2
8�
x3
16� . . .
��2, 2�.
� ��
n�0 12
�x2�
n
� ��
n�0
xn
2n�1
1
2 � x�
1�21 � �x�2� �
a1 � r
2. (a)
This series converges on
(b)
�
4x3
125�
4x 4
625
4x3
125
425
x2 �4x3
125
425
x2
45
x �4
25x2
45
x
4 �45
x
5 � x ) 4
45
�4
25x �
4125
x2 �4x3
625� . . .
��5, 5�.
� ��
n�0 45
�x5�
n
� ��
n�0
4xn
5n�1
f �x� �4
5 � x�
4�51 � x�5
�a
1 � r
3. (a)
This series converges on
(b)
�
�x3
8�
x 4
16
�x3
8
x2
4�
x3
8
x2
4
�x2
�x2
4
�x2
1 �x2
2 � x ) 1
12
�x4
�x2
8�
x3
16� . . .
��2, 2�.
� ��
n�0 12
��x2�
n
� ��
n�0 ��1�n xn
2n�1
1
2 � x�
1�21 � ��x�2� �
a1 � r
4. (a)
This series converges on
(b)
��x3 � x 4�x3
x2 � x3x2
�x � x2�x
1 � x1 � x ) 1
1 � x � x2 � x3 � . . .��1, 1�.
� ��
n�0��x�n � �
�
n�0��1�n xn
1
1 � x�
11 � ��x� �
a1 � r
322 Chapter 9 Infinite Series
5. Writing in the form we have
which implies that and Therefore, the power series for is given by
� ��
n�0 �x � 5�n
��3�n�1, �x � 5� < 3 or 2 < x < 8.
1
2 � x� �
�
n�0 arn � �
�
n�0 �
13��
13
�x � 5�n
f �x�r � ��1�3��x � 5�.a � �1�3
12 � x
�1
�3 � �x � 5� ��1�3
1 � �1�3��x � 5�
a��1 � r�,f �x�
7. Writing in the form we have
which implies that and Therefore, thepower series for is given by
� �3 ��
n�0�2x�n, �2x� < 1 or �
12
< x < 12
.
3
2x � 1� �
�
n�0 arn � �
�
n�0 ��3��2x�n
f �x�r � 2x.a � �3
32x � 1
��3
1 � 2x�
a1 � r
a��1 � r�,f �x�
6. Writing in the form we have
Therefore, the power series for is given by
or �9 < x < 5�x � 2� < 7
� ��
n�0
4�x � 2�n
7n�1 .
4
5 � x� �
�
n�0arn � �
�
n�0
47�
17
�x � 2��n
f �x�
45 � x
�4
7 � �x � 2� �4�7
1 � 1�7�x � 2� �a
1 � r.
a��1 � r�,f �x�
8. Writing in the form we have
which implies that and Therefore, the power series for is given by
or 12
< x < 72
.�x � 2� < 32
� ��
n�0 ��2�n�x � 2�n
3n ,
3
2x � 1� �
�
n�0 arn � �
�
n�0��
23
�x � 2�n
,
f �x�r � ��2�3��x � 2�.a � 1
32x � 1
�3
3 � 2�x � 2� �1
1 � �2�3��x � 2� �a
1 � r
a��1 � r�,f �x�
9. Writing in the form we have
which implies that and Therefore, the power series for is given by
or �172
< x < 52
.�x � 3� < 112
� � ��
n�0 2n�x � 3�n
11n�1 ,
1
2x � 5� �
�
n�0 arn � �
�
n�0��
111��
211
�x � 3�n
f �x�r � �2�11��x � 3�.a � �1�11
��1�11
1 � �2�11��x � 3� �a
1 � r
1
2x � 5�
�111 � 2�x � 3�
a��1 � r�,f �x� 10. Writing in the form we have
which implies that and Therefore,the power series for is given by
or �52
< x < 52
.�x� < 52
12x � 5
� ��
n�0 ar n � �
�
n�0��
15��
25
x�n
� � ��
n�0 2n xn
5n�1,
f �x�r � �2�5�x.a � �1�5
12x � 5
�1
�5 � 2x�
�1�51 � �2�5�x �
a1 � r
a��1 � r�,f �x�
11. Writing in the form we have
which implies that and Therefore,the power series for is given by
or �2 < x < 2.�x� < 2
� 3 ��
n�0 ��1�n xn
2n�1 �32
��
n�0 ��
x2�
n
,
3
x � 2� �
�
n�0 arn � �
�
n�0 32
��12
x�n
f �x�r � ��1�2�x.a � 3�2
3x � 2
�3
2 � x�
3�21 � �1�2�x �
a1 � r
a��1 � r�,f �x� 12. Writing in the form we have
which implies that and Therefore, the power series for is given by
or �23
< x < 143
.�x � 2� < 83
�12
��
n�0 ��3�n �x � 2�n
8n ,
4
3x � 2� �
�
n�0 arn � �
�
n�0 12��
38
�x � 2�n
f �x�r � ��3�8��x � 2�.a � 1�2
43x � 2
�4
8 � 3�x � 2� �1�2
1 � �3�8��x � 2� �a
1 � r
a��1 � r�,f �x�
Section 9.9 Representation of Functions by Power Series 323
13.
Writing as a sum of two geometric series, we have
The interval of convergence is since
� limn→�
��1 � ��2�n�1�x�2 � ��2�n�1 � � �x�.lim
n→� �un�1
un � � limn→�
��1 � ��2�n�1�xn�1
��2�n�1 ���2�n
�1 � ��2�n� xn��1 < x < 1
� ��
n�0 � 1
��2�n � 1xn. 3x
x2 � x � 2� �
�
n�0��
12
x�n
� ��
n�0��1��x�n
f �x�
�1
1 � �1�2�x ��1
1 � x
3xx2 � x � 2
�2
x � 2�
1x � 1
�2
2 � x�
1�1 � x
14.
Writing as a sum of two geometric series, we have
or �12
< x < 12
.�x� < 12
4x � 72x2 � 3x � 2
� ��
n�0�3
2���12
x�n
� ��
n�0 2�2x�n � �
�
n�0�3��1�n
2n�1 � 2n�1xn,
f �x�
4x � 72x2 � 3x � 2
�3
x � 2�
22x � 1
�3
2 � x�
2�1 � 2x
�3�2
1 � �1�2�x �2
1 � 2x
16. First finding the power series for we have
Now replace x with
The interval of convergence is or since
limn→�
�un�1
un � � limn→�
���1�n�1x2n�2
4n�1 �4n
��1�n x2n� � ��x2
4 � � �x2�4
.
�2 < x < 2�x2� < 4
4
4 � x2 � ��
n�0 ��1�n x2n
4n .
x2.
1
1 � �1�4�x � ��
n�0��
14
x�n
� ��
n�0 ��1�n xn
4n
4��4 � x�,
15.
Writing as a sum of two geometric series, we have
The interval of convergence is or since limn→�
�un�1
un � � limn→�
�2x2n�2
2x2n � � �x2�.�1 < x < 1�x2� < 1
21 � x2 � �
�
n�0 xn � �
�
n�0��x�n � �
�
n�0�1 � ��1�n�xn � �
�
n�0 2x2n.
f �x�
21 � x2 �
11 � x
�1
1 � x
17.
(See Exercise 15.) � 2 � 0x � 2x2 � 0x3 � 2x4 � 0x5 � 2x6 � . . . � ��
n�0 2x2n, �1 < x < 1
h�x� ��2
x2 � 1�
11 � x
�1
1 � x� �
�
n�0��1�n xn � �
�
n�0xn � �
�
n�0��1�n � 1�xn
11 � x
� ��
n�0��1�n ��x�n � �
�
n�0��1�2n xn � �
�
n�0 xn
11 � x
� ��
n�0��1�n xn
324 Chapter 9 Infinite Series
18.
� � ��
n�0x2n�1, 1 < x < 1 �
12 �
�
n�0��2�x2n�1
�12
0 � 2x � 0x2 � 2x3 � 0x4 � 2x5 � . . .� �12 �
�
n�0��1�n � 1�xn
�12 �
�
n�0��1�nxn �
12 �
�
n�0xn h�x� �
xx2 � 1
�1
2�1 � x� �1
2�1 � x�
19. By taking the first derivative, we have Therefore,
� ��
n�0��1�n�1�n � 1�xn, �1 < x < 1.
�1
�x � 1�2 �ddx� �
�
n�0��1�n xn � �
�
n�1��1�n nxn�1
ddx�
1x � 1 �
�1�x � 1�2.
20. By taking the second derivative, we have Therefore,
�ddx
� ��
n�1��1�nnxn�1 � �
�
n�2��1�n n�n � 1�xn�2 � �
�
n�0��1�n �n � 2��n � 1�xn, �1 < x < 1.
2
�x � 1�3 �d 2
dx2 � ��
n�0��1�n xn
d 2
dx2 � 1x � 1 �
2�x � 1�3.
21. By integrating, we have Therefore,
To solve for C, let and conclude that Therefore,
ln�x � 1� � ��
n�0 ��1�n xn�1
n � 1, �1 < x ≤ 1.
C � 0.x � 0
�1 < x ≤ 1.ln�x � 1� � �� ��
n�0��1�n xn dx � C � �
�
n�0 ��1�n xn�1
n � 1,
� 1
x � 1 dx � ln�x � 1�.
22. By integrating, we have
and
Therefore,
To solve for C, let and conclude that Therefore,
ln�1 � x2� � � ��
n�0 x2n�2
n � 1, �1 < x < 1.
C � 0.x � 0
� C � ��
n�0 ��1�n � 1�xn�1
n � 1� C � �
�
n�0 �2x2n�2
2n � 2� C � �
�
n�0 ��1�x2n�2
n � 1
� � � ��
n�0��1�n xn dx � � � �
�
n�0 xn dx � �C1 � �
�
n�0 ��1�n xn�1
n � 1 � �C2 � ��
n�0
xn�1
n � 1
ln�1 � x2� � � 1
1 � x dx � �
11 � x
dx
f �x� � ln�1 � x2� � ln�1 � x� � �ln�1 � x��.
� 1
1 � x dx � �ln�1 � x� � C2.�
11 � x
dx � ln�1 � x� � C1
23.1
x2 � 1� �
�
n�0��1�n�x2�n � �
�
n�0��1�n x2n, �1 < x < 1
Section 9.9 Representation of Functions by Power Series 325
24. (See Exercise 23.)
Since we have
To solve for C, let and conclude that Therefore,
�1 ≤ x ≤ 1.ln�x2 � 1� � ��
n�0 ��1�n x2n�2
n � 1,
C � 0.x � 0
�1 ≤ x ≤ 1.ln�x2 � 1� � � � ��
n�0��1�n 2x2n�1 dx � C � �
�
n�0 ��1�n x2n�2
n � 1,
ddx
�ln�x2 � 1�� �2x
x2 � 1,
� ��
n�0��1�n 2x2n�1
2x
x 2 � 1� 2x �
�
n�0��1�n x2n
25. Since, we have �12
< x < 12
.1
4x2 � 1� �
�
n�0��1�n�4x2�n � �
�
n�0��1�n 4n x2n � �
�
n�0��1�n �2x�2n,
1x � 1
� ��
n�0��1�n xn,
26. Since we can use the result of Exercise 25 to obtain
To solve for C, let and conclude that Therefore,
�12
< x ≤ 12
.arctan�2x� � 2 ��
n�0 ��1�n 4nx 2n�1
2n � 1,
C � 0.x � 0
�12
< x ≤ 12
.arctan�2x� � 2� 1
4x2 � 1 dx � 2�� �
�
n�0��1�n 4n x2n dx � C � 2 �
�
n�0 ��1�n 4n x2n�1
2n � 1,
� 1
4x2 � 1 dx �
12
arctan�2x�,
27.
−4 8
−3
S3f
S2
5
x �x2
2 ≤ ln�x � 1� ≤ x �
x2
2�
x3
3x 0.0 0.2 0.4 0.6 0.8 1.0
0.000 0.180 0.320 0.420 0.480 0.500
0.000 0.182 0.336 0.470 0.588 0.693
0.000 0.183 0.341 0.492 0.651 0.833x �x2
2�
x3
3
ln�x � 1�
x �x2
2
28.
≤ x �x2
2�
x3
3�
x 4
4�
x5
5−4 8
−3
f
S5
S4
5
x �x2
2�
x3
3�
x 4
4 ≤ ln�x � 1�
x 0.0 0.2 0.4 0.6 0.8 1.0
0.0 0.18227 0.33493 0.45960 0.54827 0.58333
0.0 0.18232 0.33647 0.47000 0.58779 0.69315
0.0 0.18233 0.33698 0.47515 0.61380 0.78333x �x 2
2�
x3
3�
x 4
4�
x5
5
ln�x � 1�
x �x 2
2�
x3
3�
x 4
4
326 Chapter 9 Infinite Series
29.
(a)
(b) From Example 4,
� ln x, 0 < x ≤ 2, R � 1.
��
n�1
��1�n�1�x � 1�n
n� �
�
n�0
��1�n�x � 1�n�1
n � 1
3
−3
0 4
n = 1n = 3
n = 6n = 2
��
n�1
��1�n�1�x � 1�n
n�
�x � 1�1
��x � 1�2
2�
�x � 1�3
3� . . .
30.
(a)
(b) ��
n�0
��1�nx2n�1
�2n � 1�! � sin x, R � �
4
6
−4
−6
��
n�0
��1�nx2n�1
�2n � 1�! � x �x3
3!�
x5
5!� . . .
(c)
(d) This is an approximation of
The error is approximately 0.
sin�12�.
��
n�0
��1�n�1�2�2n�1
�2n � 1�! 0.4794255386
31. line
Matches (c)
g�x� � x 32. cubic with 3 zeros.
Matches (d)
g�x� � x �x3
3,
33.
Matches (a)
g�x� � x �x3
3�
x5
5 34.
Matches (b)
g�x� � x �x3
3�
x5
5�
x7
7,
In Exercises 35-38, arctan x � ��
n�0��1�n
x2n�1
2n � 1.
35.
Since we can approximate the series by its first two terms: arctan 14 14 �
1192 0.245.1
5120 < 0.001,
�14
�1
192�
15120
� . . . arctan 14
� ��
n�0��1�n
�1�4�2n�1
2n � 1� �
�
n�0
��1�n
�2n � 1�42n�1
(c)
(d) This is an approximation of The error is approxi-mately 0. The error is less than the first omitted term,1��51 � 251� 8.7 � 10�18.�
ln�1
2�.��
n�1
��1�n�1��1�2�n
n� �
�
n�1
��1�2�n
n �0.693147
x � 0.5:
36.
Since we can approximate the series by its first two terms: 0.134.177,147�230,686,720 < 0.001,
�27192
�2187
344,064�
177,147230,686,720
�3�4
0 arctan x2 dx � �
�
n�0��1�n
�3�4�4n�3
�4n � 3��2n � 1� � ��
n�0��1�n 34n�3
�4n � 3��2n � 1�44n�3
� arctan x2 dx � ��
n�0��1�n
x 4n�3
�4n � 3��2n � 1� � C, C � 0
arctan x2 � ��
n�0��1�n
x4n�2
2n � 1
Section 9.9 Representation of Functions by Power Series 327
37.
Since we can approximate the series by its first term: �1�2
0 arctan x2
x dx 0.125.
11152
< 0.001,
�1�2
0 arctan x2
x dx � �
�
n�0��1�n
1�4n � 2��2n � 1�24n�2 �
18
�1
1152� . . .
� arctan x2
x dx � �
�
n�0��1�n
x 4n�2
�4n � 2��2n � 1� � C �Note: C � 0�
arctan x2
x�
1x �
�
n�0��1�n
�x2�2n�1
2n � 1� �
�
n�0��1�2
x 4n�1
2n � 1
38.
Since we can approximate the series by its first term: �1�2
0x2 arctan x dx 0.016.
11152
< 0.001,
�1�2
0 x2 arctan x dx � �
�
n�0��1�n
1�2n � 4��2n � 1�22n�4 �
164
�1
1152� . . .
� x2 arctan x dx � ��
n�0��1�n
x2n�4
�2n � 4��2n � 1�
x2 arctan x � ��
n�0��1�n
x2n�3
2n � 1
In Exercises 39–43, use 1
1 � x� �
�
n�0 xn, �x� < 1.
39.1
�1 � x�2 �ddx�
11 � x �
ddx� �
�
n�0xn � �
�
n�1nxn�1, �x� < 1 40.
x�1 � x�2 � x �
�
n�1nxn�1 � �
�
n�1nxn, �x� < 1
41.
� ��
n�0�2n � 1�xn, �x� < 1
� ��
n�1n�xn�1 � xn�, �x� < 1
1 � x
�1 � x�2 �1
�1 � x�2 �x
�1 � x�2 42.
(See Exercise 41.)
x�1 � x��1 � x�2 � x �
�
n�0�2n � 1�xn � �
�
n�0�2n � 1�xn�1, �x� < 1
43.
Since the probability of obtaining a head on a single toss is it is expected that, on average, a head will be obtained in two tosses.
12 ,
�12
1
1 � �1�2��2 � 2
E�n� � ��
n�1 nP�n� � �
�
n�1 n�1
2�n
�12
��
n�1 n�1
2�n�1
P�n� � �12�
n
44. (a)
(b) �9
100�
11 � �9�10��2 � 9
110
��
n�1 n� 9
10�n
�9
100 �
�
n�1 n� 9
10�n�1
13
��
n�1 n�2
3�n
�29
��
n�1 n�2
3�n�1
�29
1
1 � �2�3��2 � 2
In Exercise 44,1
1 � x� �
�
n�0 xn, �x�1.
328 Chapter 9 Infinite Series
45. Replace x with ��x�. 46. Replace x with x2. 47. Replace x with andmultiply the series by 5.
��x� 48. Integrate the series andmultiply by ��1�.
50. (a) From Exercise 49, we have
(b)
(see part (a).)4 arctan 15
� arctan 1
239� arctan
120119
� arctan 1
239�
�
4
4 arctan 15
� 2 arctan 15
� 2 arctan 15
� arctan 5
12� arctan
512
� arctan� 2�5�12�1 � �5�12�2 � arctan
120119
2 arctan 15
� arctan 15
� arctan 15
� arctan � 2�1�5�1 � �1�5�2 � arctan
1024
� arctan 5
12
� arctan � �120�119� � ��1�239�1 � �120�119���1�239� � arctan �28,561
28,561� � arctan 1 ��
4
arctan 120119
� arctan 1
239� arctan
120119
� arctan ��1
239�
51. (a)
(b) � � 8 arctan 12
� 4 arctan 17
8�12
��0.5�3
3�
�0.5�5
5�
�0.5�7
7 � 4�17
��1�7�3
3�
�1�7�5
5�
�1�7�7
7 3.14
2 arctan 12
� arctan 17
� arctan 43
� arctan ��17� � arctan� �4�3� � �1�7�
1 � �4�3��1�7� � arctan 2525
� arctan 1 ��
4
2 arctan 12
� arctan 12
� arctan 12
� arctan�12 �
12
1 � �1�2�2 � arctan 43
49. Let Then,
Therefore, for xy � 1.arctan x � arctan y � arctan� x � y1 � xy� arctan� x � y
1 � xy� � .
x � y
1 � xy� tan
tan�arctan x� � tan�arctan y�
1 � tan�arctan x� tan�arctan y� � tan
tan�arctan x � arctan y� � tan
arctan x � arctan y � .
52. (a)
(b)
� 4�12
��1�2�3
3�
�1�2�5
5�
�1�2�7
7 � 4�13
��1�3�3
3�
�1�3�5
5�
�1�3�7
7 4�0.4635� � 4�0.3217� 3.14
� � 4�arctan 12
� arctan 13
arctan 12
� arctan 13
� arctan� �1�2� � �1�3�1 � �1�2��1�3� � arctan�5�6
5�6� ��
4
54. From Exercise 53, we have
� ln�13
� 1� � ln 43
0.2877.
��
n�1��1�n�1
13n n
� ��
n�1 ��1�n�1�1�3�n
n
53. From Exercise 21, we have
Thus,
� ln�12
� 1� � ln 32
0.4055.
��
n�1��1�n�1
12n n
� ��
n�1 ��1�n�1�1�2�n
n
� ��
n�1 ��1�n�1xn
n.
ln�x � 1� � ��
n�0 ��1�n xn�1
n � 1� �
�
n�1 ��1�n�1xn
n
Section 9.10 Taylor and Maclaurin Series 329
56. From Example 5, we have
� arctan 1 ��
4� 0.7854
��
n�0��1�n
12n � 1
� ��
n�0��1�n
�1�2n�1
2n � 1
arctan x � ��
n�0��1�n
x2n�1
2n � 1.55. From Exercise 53, we have
� ln�25
� 1� � ln 75
� 0.3365.
��
n�1��1�n�1
2n
5n n� �
�
n�1 ��1�n�1�2�5�n
n
57. From Exercise 56, we have
� arctan 12
� 0.4636.
��
n�0��1�n
122n�1�2n � 1� � �
�
n�0��1�n
�1�2�2n�1
2n � 1
58. From Exercise 56, we have
� arctan 13
� 0.3218.
� ��
n�0��1�n �1�3�2n�1
2n � 1
��
n�1��1�n�1
132n�1�2n � 1� � �
�
n�0��1�n
132n�1�2n � 1�
59. is an odd function (symmetric to the origin).f �x� � arctan x 60. The approximations of degree have relative extrema.�4n � 1, n � 1, 2, . . .�
3, 7, 11, . . . ,
61. The series in Exercise 56 converges to its sum at a slowerrate because its terms approach 0 at a much slower rate.
62. Because the radius of
convergence is the same, 3.
ddx� �
�
n�0anxn � �
�
n�1nanxn�1,
64. Using a graphing utility, you obtain the following partialsums for the left hand side. Note that
n � 1: S1 � 0.3183098862
n � 0: S0 � 0.3183098784
1�� � 0.3183098862.
65.
From Example 5 we have
� 3 ��
6� � 0.9068997
� 3 arctan � 13�
� 3 ��
n�0
��1�n�1�3�2n�1
2n � 1
��
n�0
��1�n
3n�2n � 1� � ��
n�0
��1�n
�3�2n�2n � 1� 33
arctan x � ��
n�0��1�n
x2n�1
2n � 1.
��
n�0
��1�n
3n�2n � 1�66.
� sin��
3� �32
� 0.866025
��
n�0
��1�n� 2n�1
32n�1�2n � 1�! � ��
n�0��1�n
���3�2n�1
�2n � 1�!
63. Because the first series is the derivative of the secondseries, the second series converges for andperhaps at the endpoints, x � 3 and x � �5.�
�x � 1� < 4 �
Section 9.10 Taylor and Maclaurin Series
1. For we have:
e2x � 1 � 2x �4x2
2!�
8x3
3!�
16x 4
4!� . . . � �
�
n�0 �2x�n
n!.
f �n��x� � 2n e2x ⇒ f �n��0� � 2n
f �x� � e2x
c � 0,
4. For we have:
and so on. Therefore we have:
�22
���
n�0 ��1�n�n�1��2 x � ���4��n�1
�n � 1�! � 1�.
�22
�1 � �x ��
4� � x � ���4��2
2!�
x � ���4��3
3!�
x � ���4��4
4!� . . .
sin x � ��
n�0 f �n����4� x � ���4��n
n!
f �4��x� � sin x f �4���
4� �22
f���x� � �cos x f����
4� � �22
f ��x� � �sin x f ���
4� � �22
f��x� � cos x f���
4� �22
f �x� � sin x f��
4� �22
c � ��4,
330 Chapter 9 Infinite Series
2. For we have:
e3x � 1 � 3x �9x2
2!�
27x3
3!� . . . � �
�
n�0
�3x�n
n!.
f �n��x� � 3ne3x ⇒ f �n��0� � 3n
f �x� � e3x
c � 0,
3. For we have:
and so on. Therefore, we have:
[Note: 1, 1, 1, . . .]�1,�1,�1,�1,��1�n�n�1��2 � 1,
�22
��
n�0 ��1�n�n�1��2 x � ���4��n
n!.
�22
�1 � �x ��
4� � x � ���4��2
2!�
x � ���4��3
3!�
x � ���4��4
4!� . . .
cos x � ��
n�0 f �n����4� x � ���4��n
n!
f �4��x� � cos�x� f �4���
4� �22
f����x� � sin�x� f�����
4� �22
f � �x� � �cos�x� f ���
4� � �22
f��x� � �sin�x� f���
4� � �22
f �x� � cos�x� f ��
4� �22
c � ��4,
Section 9.10 Taylor and Maclaurin Series 331
5. For we have,
and so on. Therefore, we have:
� ��
n�0 ��1�n
�x � 1�n�1
n � 1.
� �x � 1� ��x � 1�2
2�
�x � 1�3
3�
�x � 1�4
4�
�x � 1�5
5� . . .
� 0 � �x � 1� ��x � 1�2
2!�
2�x � 1�3
3!�
6�x � 1�4
4!�
24�x � 1�5
5!� . . .
ln x � ��
n�0 f �n��1��x � 1�n
n!
f �5��x� �24x5 f �5��1� � 24
f �4��x� � �6x 4
f �4��1� � �6
f����x� �2x3 f����1� � 2
f ��x� � �1x2 f ��1� � �1
f��x� �1x f��1� � 1
f �x� � ln x f �1� � 0
c � 1,
6. For we have:
ex � ��
n�0 f �n��1��x � 1�n
n!� e �1 � �x � 1� �
�x � 1�2
2!�
�x � 1�3
3!�
�x � 1�4
4!� . . . � e �
�
n�0 �x � 1�n
n!.
f �n��x� � ex ⇒ f �n��1� � e
f �x� � ex
c � 1,
7. For we have:
and so on. Therefore, we have:
� 2x �8x3
3!�
32x5
5!�
128x7
7!� . . . � �
�
n�0 ��1�n�2x�2n�1
�2n � 1�! .
� 0 � 2x �0x2
2!�
8x3
3!�
0x 4
4!�
32x5
5!�
0x6
6!�
128x7
7!� . . . sin 2x � �
�
n�0 f �n��0�xn
n!
f �7��x� � �128 cos 2x f �7��0� � �128
f �6��x� � �64 sin 2x f �6��0� � 0
f �5��x� � 32 cos 2x f �5��0� � 32
f �4��x� � 16 sin 2x f �4��0� � 0
f���x� � �8 cos 2x f���0� � �8
f ��x� � �4 sin 2x f ��0� � 0
f��x� � 2 cos 2x f��0� � 2
f �x� � sin 2x f �0� � 0
c � 0,
332 Chapter 9 Infinite Series
10. For we have;
tan�x� � ��
n�0 f �n��0�xn
n!� x �
2x3
3!�
16x5
5!� . . . � x �
x3
3�
215
x5 � . . ..
f �5��x� � 8 2 sec6�x� � 11 sec4�x� tan2�x� � 2 sec2�x� tan4�x�� f �5��0� � 16
f �4��x� � 8 sec4�x� tan�x� � sec2�x� tan3�x�� f �4��0� � 0
f���x� � 2 sec4�x� � 2 sec2�x� tan2�x�� f���0� � 2
f ��x� � 2 sec2�x� tan�x� f ��0� � 0
f��x� � sec2�x� f��0� � 1
f �x� � tan�x� f �0� � 0
c � 0,
8. For we have:
and so on. Therefore, we have:
� x2 �x 4
2�
x6
3� . . . � �
�
n�0 ��1�n x2n�2
n � 1.
� 0 � 0x �2x2
2!�
0x3
3!�
12x 4
4!�
0x5
5!�
240x6
6!� . . . ln�x2 � 1� � �
�
n�0 f �n��0�xn
n!
f �6��x� ��240�5x6 � 15x 4 � 15x2 � 1�
�x2 � 1�6 f �6��0� � 240
f �5��x� �48x�x 4 � 10x2 � 5�
�x2 � 1�5 f �5��0� � 0
f �4��x� �12��x 4 � 6x2 � 1�
�x2 � 1�4 f �4��0� � �12
f���x� �4x�x2 � 3��x2 � 1�3 f���0� � 0
f ��x� �2 � 2x2
�x2 � 1�2 f ��0� � 2
f��x� �2x
x2 � 1 f��0� � 0
f �x� � ln�x2 � 1� f �0� � 0
c � 0,
9. For we have:
sec�x� � ��
n�0 f �n��0�xn
n!� 1 �
x2
2!�
5x 4
4!� . . ..
f �4��x� � 5 sec5�x� � 18 sec3�x� tan2�x� � sec�x� tan4�x� f �4��0� � 5
f���x� � 5 sec3�x� tan�x� � sec�x� tan3�x� f���0� � 0
f ��x� � sec3�x� � sec�x� tan2�x� f ��0� � 1
f��x� � sec�x� tan�x� f��0� � 0
f �x� � sec�x� f �0� � 1
c � 0,
Section 9.10 Taylor and Maclaurin Series 333
11. The Maclaurin series for is
Because or we have for all z. Hence by Taylor’s Theorem,
Since it follows that as Hence, the Maclaurin series for converges to for all x.cos xcos xn →�.Rn�x� → 0limn→�
�x�n�1
�n � 1�! � 0,
0 ≤ �Rn�x�� � �f �n�1��z��n � 1�!x
n�1� ≤ �x�n�1
�n � 1�!.
� f �n�1��z�� ≤ 1±cos x,f �n�1��x� � ±sin x
��
n�0
��1�n x2n
�2n�! .f �x� � cos x
12. The Maclaurin series for is
Hence, by Taylor’s Theorem,
Since it follows that as
Hence, the Maclaurin Series for converges to for all x.e�2xe�2x
n →�.Rn�x� → 0limn→�
���2�n�1xn�1
�n � 1�! � � limn→�
� �2x�n�1
�n � 1�!� � 0,
0 ≤ �Rn�x�� � �f �n�1��z��n � 1�!x
n�1� � ���2�n�1e�2z
�n � 1�! xn�1�.f �n�1��x� � ��2�n�1e�2x.
��
n�0
��2x�n
n!.f �x� � e�2x
13. The Maclaurin series for is
For fixed ,
The argument is the same if . Hence, the Maclaurin series for converges to for all x.sinh xsinh xf �n�1��x� � cosh x��
0 ≤ �Rn�x�� � � f �n�1��z��n � 1�! xn�1� � � sinh�z�
�n � 1�! xn�1� → 0 as n →�.
xf �n�1��x� � sinh x �or cosh x�.
��
n�0
x2n�1
�2n � 1�!.f �x� � sinh x
14. The Maclaurin series for is
For fixed ,
The argument is the same if . Hence, the Maclaurin series for converges to for all x.cosh xcosh xf �n�1��x� � cosh x��
0 ≤ �Rn�x�� � � f �n�1��z��n � 1�! xn�1� � � sinh�z�
�n � 1�! xn�1� → 0 as n →�.
xf �n�1��x� � sinh x �or cosh x�.
��
n�0
x2n
�2n�!.f �x� � cosh x
15. Since we have
� ��
n�0��1�n �n � 1�xn.
�1 � x��2 � 1 � 2x �2�3�x2
2!�
2�3��4�x3
3!�
2�3��4��5�x4
4!� . . . � 1 � 2x � 3x2 � 4x3 � 5x 4 � . . .
�1 � x��k � 1 � kx �k�k � 1�x2
2!�
k�k � 1��k � 2�x3
3!� . . .,
16. Since we have
� 1 � ��
n�1 1 � 3 � 5 . . . �2n � 1�xn
2n n!.
� 1 �x2
��1��3�x2
222!�
�1��3��5�x3
233!� . . .
�1 � ��x��1�2
� 1 � �12�x �
�1�2��3�2�x2
2!�
�1�2��3�2��5�2�x3
3!� . . .
�1 � x��k � 1 � kx �k�k � 1�x2
2!�
k�k � 1��k � 2�x3
3!� . . .,
334 Chapter 9 Infinite Series
23.
sin 3x � ��
n�0
��1�n�3x�2n�1
�2n � 1�!
sin x � ��
n�0
��1�nx2n�1
�2n � 1�! 24.
� 1 �16x2
2!�
256x4
4!� . . .
cos 4x � ��
n�0
��1�n�4x�2n
�2n�! � ��
n�0
��1�n42nx2n
�2n�!
cos x � ��
n�0
��1�nx2n
�2n�! � 1 �x2
2!�
x4
4!�
x6
6!� . . .
26.
� 2x3 �2x9
3!�
2x15
5!� . . .
� 2�x3 �x9
3!�
x15
5!� . . .
2 sin x3 � 2 ��
n�0
��1�n�x3�2n�1
�2n � 1�!
sin x � ��
n�0
��1�nx2n�1
�2n � 1�!25.
� 1 �x3
2!�
x6
4!� . . .
� ��
n�0 ��1�n x3n
�2n�!
cos x3�2 � ��
n�0 ��1�n �x3�2�2n
�2n�!
cos x � ��
n�0 ��1�n x2n
�2n�! � 1 �x2
2!�
x 4
4!� . . .
17. and since we have
14 � x2
�12
�1 � ��
n�1 ��1�n 1 � 3 � 5 . . . �2n � 1��x�2�2n
2n n! �12
� ��
n�1 ��1�n 1 � 3 � 5 . . . �2n � 1�x2n
23n�1n!.
�1 � x��1�2 � 1 � ��
n�1 ��1�n 1 � 3 � 5 . . . �2n � 1�xn
2nn!,
14 � x2
� �12��1 � �x
2�2
�1�2
19. Since
we have �1 � x2�1�2 � 1 �x2
2� �
�
n�2 ��1�n�1 1 � 3 � 5 . . . �2n � 3�x2n
2n n!.
�1 � x�1�2 � 1 �x2
� ��
n�2 ��1�n�1 1 � 3 � 5 . . . �2n � 3�xn
2n n!
21.
ex2�2 � ��
n�0 �x2�2�n
n!� �
�
n�0
x2n
2n n!� 1 �
x 2
2�
x 4
222!�
x6
233!�
x8
244!� . . .
ex � ��
n�0 xn
n!� 1 � x �
x2
2!�
x3
3!�
x 4
4!�
x5
5!� . . .
18.
� 1 �14
x � ��
n�2
��1�n�1 3 � 7 � 11 . . . �4n � 5�4nn!
xn
� 1 �14
x �3
422!x2 �
3 � 7433!
x3 �3 � 7 � 11
444!x 4 � . . .
�1 � x�1�4 � 1 �14
x ��1�4���3�4�
2!x2 �
�1�4���3�4���7�4�3!
x3 � . . .
20. Since
we have �1 � x3�1�2 � 1 �x3
2� �
�
n�2 ��1�n�1 1 � 3 � 5 . . . �2n � 3�x3n
2n n!.
�1 � x�1�2 � 1 �x2
� ��
n�2 ��1�n�1 1 � 3 � 5 . . . �2n � 3�xn
2n n!
22.
e�3x � ��
n�0 ��3x�n
n!� �
�
n�0 ��1�n 3n xn
n!� 1 � 3x �
9x2
2!�
27x3
3!�
81x 4
4!�
243x5
5!� . . .
ex � ��
n�0 xn
n!� 1 � x �
x2
2!�
x3
3!�
x 4
4!�
x5
5!� . . .
Section 9.10 Taylor and Maclaurin Series 335
27.
� x �x3
3!�
x5
5!�
x7
7!� . . . � �
�
n�0
x2n�1
�2n � 1�!
sinh�x� �12
�ex � e�x�
ex � e�x � 2x �2x3
3!�
2x5
5!�
2x7
7!� . . .
e�x � 1 � x �x2
2!�
x3
3!�
x 4
4!�
x5
5!� . . .
ex � 1 � x �x2
2!�
x3
3!�
x 4
4!�
x5
5!� . . . 28.
2 cos h�x� � ex � e�x � ��
n�02
x2n
�2n�!
ex � e�x � 2 �2x2
2!�
2x4
4!� . . .
e�x � 1 � x �x2
2!�
x3
3!� . . .
ex � 1 � x �x2
2!�
x3
3!� . . .
29.
�12�1 � �
�
n�0 ��1�n�2x�2n
�2n�! �12�1 � 1 �
�2x�2
2!�
�2x�4
4!�
�2x�6
6!� . . .
cos2�x� �12
1 � cos�2x��
30. The formula for the binomial series gives which implies that
� x �x3
2 � 3�
1 � 3x5
2 � 4 � 5�
1 � 3 � 5x7
2 � 4 � 6 � 7� . . . .
� x � ��
n�1 ��1�n1 � 3 � 5 . . . �2n � 1�x2n�1
2n�2n � 1�n!
ln�x � x2 � 1� � � 1
x2 � 1 dx
�1 � x2��1�2 � 1 � ��
n�1 ��1�n 1 � 3 � 5 . . . �2n � 1�x2n
2n n!
�1 � x��1�2 � 1 � ��
n�1 ��1�n 1 � 3 � 5 . . . �2n � 1�xn
2n n!,
31.
� ��
n�0
��1�nx2n�2
�2n � 1�!
� x2 �x4
3!�
x6
5!� . . .
x sin x � x�x �x3
3!�
x5
5!� . . .� 32.
� ��
n�0
��1�nx2n�1
�2n�!
� x �x3
2!�
x5
4!� . . .
x cos x � x�1 �x2
2!�
x4
4!� . . .�
33.
� ��
n�0
��1�nx2n
�2n � 1�!, x 0 � 1 �x2
2!�
x4
4!� . . .
sin x
x�
x � �x3�3!� � �x5�5!� � . . .
x34.
� ��
n�0
�2n�!x2n
�2nn!�2�2n � 1�, x 0
arcsin x
x� �
�
n�0
�2n�!x2n�1
�2nn!�2�2n � 1� �1x
35.
eix � e�ix
2i� x �
x3
3!�
x5
5!�
x7
7!� . . . � �
�
n�0 ��1�n x2n�1
�2n � 1�! � sin�x�
eix � e�ix � 2ix �2ix3
3!�
2ix5
5!�
2ix7
7!� . . .
e�ix � 1 � ix ���ix�2
2!�
��ix�3
3!�
��ix�4
4!� . . . � 1 � ix �
x2
2!�
ix3
3!�
x 4
4!�
ix5
5!�
x6
6!� . . .
eix � 1 � ix ��ix�2
2!�
�ix�3
3!�
�ix�4
4!� . . . � 1 � ix �
x2
2!�
ix3
3!�
x 4
4!�
ix5
5!�
x6
6!� . . .
336 Chapter 9 Infinite Series
40.
−3 3
−3
f P5
3
� x �x2
2�
x3
3�
3x5
40� . . .
� x � �x2 �x2
2 � � �x3
3�
x3
2�
x3
2 � � ��x 4
4�
x 4
3�
x 4
4�
x 4
6 � � �x5
5�
x5
4�
x5
6�
x5
12�
x5
24� � . . .
� �1 � x �x2
2�
x3
6�
x 4
24� . . .��x �
x2
2�
x3
3�
x 4
4�
x5
5� . . .�
f �x� � ex ln�1 � x�
37.
� x � x2 �x3
3�
x5
30� . . .
� x � x2 � �x3
2�
x3
6 � � �x 4
6�
x 4
6 � � � x5
120�
x5
12�
x5
24� � . . .
� �1 � x �x2
2�
x3
6�
x4
24� . . .��x �
x3
6�
x5
120� . . .�
−6 6
−2
P5
f
14 f �x� � ex sin x
36. (See Exercise 35.)
eix � e�ix
2� 1 �
x2
2!�
x 4
4!�
x6
6!� . . . � �
�
n�0 ��1�n x2n
�2n�! � cos�x�
eix � e�ix � 2 �2x2
2!�
2x 4
4!�
2x6
6!� . . .
38.
� 1 � x �x3
3�
x 4
6� . . .
� 1 � x � �x2
2�
x2
2 � � �x3
6�
x3
2 � � �x 4
24�
x 4
4�
x 4
24� � . . .
� �1 � x �x2
2�
x 4
6�
x 4
24� . . .��1 �
x2
2�
x 4
24� . . .�
−6 6
−40
g
P4
8 g�x� � ex cos x
39.
� x �x2
2�
x3
6�
3x5
40� . . .
� x �x2
2� �x3
3�
x3
2 � � �x 4
4�
x 4
4 � � �x5
5�
x5
6�
x5
24� � . . .
� �1 �x2
2�
x 4
24� . . .��x �
x2
2�
x3
3�
x 4
4�
x5
5� . . .� −3 9
−4
h
P5
4 h�x� � cos x ln�1 � x�
Section 9.10 Taylor and Maclaurin Series 337
41. Divide the series for sin x by
�
�5x4
6�
5x5
6
�5x4
6�
x5
120
5x3
6�
5x4
6
5x3
6� 0x4
�x2 � x3
�x2 � x3
6
x � x2
1 � x� x � 0x2 � x3
6� 0x4 �
x5
120� . . .
x � x2 �5x3
6�
5x4
6�
�1 � x�.g�x� �sin x
1 � x.
−6 6
−4
g
P4
4
g�x� � x � x2 �5x3
6�
5x 4
6� . . .
42. Divide the series for by
�
3x4
8�
3x5
8
3x4
8�
x5
120
�x3
3�
x4
3
�x3
3�
x4
24
x2
2�
x3
2
0 �x2
2�
x3
6
1 � x
1 � x� 1 � x �x2
2�
x3
6�
x4
24�
x5
120� . . .
1 �x2
2�
x3
3�
3x4
8� . . .
�1 � x�.exf �x� �ex
1 � x.
−6 6
−6
fP4
6
f �x� � 1 �x2
2�
x3
3�
3x4
8� . . .
43.
Matches (c)
f �x� � x sin x
y � x2 �x 4
3!� x�x �
x3
3!�
45.
Matches (a)
f �x� � xex
y � x � x2 �x3
2!� x�1 � x �
x2
2!�
44.
Matches (d)
f �x� � x cos x
y � x �x3
2!�
x5
4!� x�1 �
x2
2!�
x 4
4!�
46.
Matches (b)
f �x� � x 2 1
1 � x
y � x2 � x3 � x 4 � x2�1 � x � x2�
338 Chapter 9 Infinite Series
53. Since
we have limx→0
1 � cos xx
� limx→0
��
n�0 ��1�x2n�1
�2n � 2�! � 0.
1 � cos
x�
x2!
�x3
4!�
x5
6!�
x7
8!� . . . � �
�
n�0
��1�nx2n�1
�2n � 2�!
1 � cos x �x2
2!�
x 4
4!�
x6
6!�
x8
8!� . . . � �
�
n�0
��1�nx2n�2
�2n � 2�!
cos x � ��
n�0
��1�nx2n
�2n�! � 1 �x2
2!�
x 4
4!�
x6
6!�
x8
8!� . . .
47.
� �x
0 � �
�
n�0 ��1�n�1t2n�2
�n � 1�! dt � � ��
n�0
��1�n�1 t2n�3
�2n � 3��n � 1�!x
0� �
�
n�0
��1�n�1x2n�3
�2n � 3��n � 1�!
�x
0 �e�t 2 � 1� dt � �x
0 ���
�
n�0 ��1�n t2n
n! � � 1 dt
49. Since
we have (10,001 terms)ln 2 � 1 �12
�13
�14
� . . . � ��
n�1��1�n�1
1n
� 0.6931.
�0 < x ≤ 2�ln x � ��
n�0 ��1�n�x � 1�n�1
n � 1� �x � 1� �
�x � 1�2
2�
�x � 1�3
3�
�x � 1�4
4� . . .,
50. Since we have
(4 terms)sin�1� � ��
n�0
��1�n
�2n � 1�! � 1 �13!
�15!
�17!
� . . . � 0.8415.
sin�x� � ��
n�0 ��1�nx2n�1
�2n � 1�! � x �x3
3!�
x5
5!�
x7
7!� . . . ,
52. Since we have
and (6 terms)e � 1
e� 1 � e�1 � 1 �
12!
�13!
�14!
�15!
�17!
� . . . � ��
n�1 ��1�n�1
n!� 0.6321.
e�1 � 1 � 1 �12!
�13!
�14!
�15!
� . . .ex � ��
n�0 xn
n!� 1 � x �
x2
2!�
x3
3!�
x 4
4!�
x5
5!� . . . ,
48.
� x �x 4
8� �
�
n�2 ��1�n�11 � 3 � 5 . . . �2n � 3�x3n�1
�3n � 1�2nn!
� �t �t4
8� �
�
n�2 ��1�n�11 � 3 � 5 . . . �2n � 3�t3n�1
�3n � 1�2n n! x
0
�x
0 1 � t3 dt � �x
0 �1 �
t3
2� �
�
n�2 ��1�n�11 � 3 � 5 . . . �2n � 3�t3n
2n n! dt
51. Since
we have (12 terms)e2 � 1 � 2 �22
2!�
23
3!� . . . � �
�
n�0 2n
n!� 7.3891.
ex � ��
n�0 xn
n!� 1 � x �
x2
2!�
x3
3!� . . . ,
54. Since
we have limx→0
sin x
x� lim
x→0 �
�
n�0 ��1�nx2n
�2n � 1�! � 1.
sin x
x� 1 �
x2
3!�
x 4
5!�
x6
7!� . . . � �
�
n�0 ��1�nx2n
�2n � 1�!
sin x � ��
n�0 ��1�n x2n�1
�2n � 1�! � x �x3
3!�
x5
5!�
x7
7!� . . .
Section 9.10 Taylor and Maclaurin Series 339
55.
Since we need three terms:
three nonzero
Note: We are using limx→0�
sin xx
� 1.
terms��using �1
0 sin x
x dx � 1 �
13 � 3!
�1
5 � 5!� . . . � 0.9461.
1��7 � 7!� < 0.0001,
�1
0 sin x
x dx � �1
0 � �
�
n�0 ��1�nx2n
�2n � 1�! dx � � ��
n�0
��1�nx2n�1
�2n � 1��2n � 1�!1
0� �
�
n�0
��1�n
�2n � 1��2n � 1�!
56.
Since we have
Note: We are using limx→0�
arctan x
x� 1.
�1�2
0 arctan x
x dx � �1
2�
13223 �
15225 �
17227 �
19229� � 0.4872.
1��9229� < 0.0001,
�1�2
0 arctan x
x dx � �1�2
0 �1 �
x2
3�
x 4
5�
x6
7� . . .� dx � �x �
x3
32 �x5
52 �x7
72 � . . .1�2
0
57.
Since we need two terms.
�0.3
0.1 1 � x3 dx � ��0.3 � 0.1� �
18
�0.34 � 0.14� � 0.201.
156 �0.37 � 0.17� < 0.0001,
�0.3
0.1 1 � x3 dx � �0.3
0.1 �1 �
x3
2�
x6
8�
x9
16�
5x12
128� . . .� dx � �x �
x 4
8�
x7
56�
x10
160�
5x13
1664� . . .
0.3
0.1
58.
Since �1�4
0x ln�x � 1� dx �
�1�4�3
3�
�1�4�4
8� 0.00472.
�1�4�5
15< 0.0001,
� �x3
3�
x4
4 � 2�
x5
5 � 3�
x6
6 � 4� . . .
1�4
0 �14
0x ln�x � 1� dx � �1�4
0�x2 �
x3
2�
x4
3�
x5
4� . . .� dx
59.
Since we need five terms.
�1
0 x cos x dx � 2����2�3�2
3�
���2�7�2
14�
���2�11�2
264�
���2�15�2
10,800�
���2�19�2
766,080 � 0.7040.
2���2�23�2��23 � 10!� < 0.0001,
� � ��
n�0 ��1�nx�4n�3��2
�4n � 32 ��2n�!
��2
0
� � ��
n�0 ��1�n2x�4n�3��2
�4n � 3��2n�! ��2
0���2
0 x cos x dx � ���2
0 � �
�
n�0 ��1�nx�4n�1��2
�2n�! dx
60.
Since we have
�1
0.5 cos x dx � ��1 � 0.5� �
14
�1 � 0.52� �1
72 �1 � 0.53� �
12880
�1 � 0.54� �1
201,600 �1 � 0.5�5 � 0.3243.
1201,600 �1 � 0.55� < 0.0001,
�1
0.5 cos x dx � �1
0.5 �1 �
x2!
�x2
4!�
x3
6!�
x4
8!� . . .� dx � �x �
x2
2�2!� �x3
3�4!� �x 4
4�6!� �x5
5�8!� � . . .1
0.5
61. From Exercise 21, we have
�1
2� �1 �
12 � 1 � 3
�1
22 � 2! � 5�
123 � 3! � 7 � 0.3414.
�1
2� �
�
n�0
��1�n
2nn!�2n � 1� 1
2� �1
0 e�x2�2 dx �
12�
�1
0 �
�
n�0 ��1�nx2n
2nn! dx �
12�
� ��
n�0
��1�nx2n�1
2nn!�2n � 1�1
0
340 Chapter 9 Infinite Series
66.
The polynomial is a reasonable approximation on the interval �0.48, 1.75�.
� 1.3025 ��x � 1�4
4! � 5.5913 ��x � 1�5
5!
P5�x� � 0.7854 � 0.7618�x � 1� � 0.3412 ��x � 1�2
2! � 0.0424 ��x � 1�3
3! −2 4
−1
f
P5
3 f �x� � 3x � arctan x, c � 1
67. See Guidelines, page 680. 68. (odd coefficients are zero)a2n�1 � 0
69. (a) Replace x with (c) Multiply series by x.
(b) Replace x with (d) Replace x with then replace x with and add thetwo together.
�2x,2x,3x.
��x�.
70. The binomial series is The radius of convergence is R � 1.�1 � x�k � 1 � kx �k�k � 1�
2!x2 �
k�k � 1��k � 2�3!
x3 � . . . .
62. From Exercise 21, we have
�8191
26 � 6! � 13�
32,76727 � 7! � 15
�131,071
28 � 8! � 17�
524,28729 � 9! � 19 � 0.1359.
�1
2� �1 �
72 � 1 � 3
�31
22 � 2! � 5�
12723 � 3! � 7
�511
24 � 4! � 9�
204725 � 5! � 11
�1
2� �
�
n�0 ��1�n�2n�1 � 1�
2nn!�2n � 1�
1
2� �2
1 e�x2�2dx �
12�
�2
1 �
�
n�0 ��1�nx2n
2nn! dx �
12�
� ��
n�0
��1�nx2n�1
2nn!�2n � 1�2
1
63.
The polynomial is a reasonable approximation on the
interval �34 , 34�.
−3 3
−2
f
P5
2
P5�x� � x � 2x3 �2x5
3
f �x� � x cos 2x � ��
n�0 ��1�n4nx2n�1
�2n�!
65.
The polynomial is a reasonable approximation on the interval 14 , 2�.
P5�x� � �x � 1� ��x � 1�3
24�
�x � 1�4
24�
71�x � 1�5
1920 −2 4
−2
gP5
3 f �x� � x ln x, c � 1
64.
The polynomial is a reasonable approximation on theinterval ��0.60, 0.73�.
−4 8
−4
f
P5
4
P5�x� �x2
2�
x3
4�
7x 4
48�
11x5
96
f �x� � sin x2
ln�1 � x�
Section 9.10 Taylor and Maclaurin Series 341
71.
� �tan ��x �gx2
2v02 cos2 �
�kgx3
3v03 cos3 �
�k2gx 4
4v04 cos4 �
� . . .
� �tan ��x �gx
kv0 cos ��
gxkv0 cos �
�gx2
2v02 cos2 �
�gkx3
3v03 cos3 �
�gk2x4
4v04 cos4 �
� . . .
� �tan ��x �gx
kv0 cos ��
gk2��
kxv0 cos �
�12�
kxv0 cos ��
2
�13�
kxv0 cos ��
3
�14�
kxv0 cos ��
4
� . . .
y � �tan � �g
kv0 cos ��x �gk2 ln�1 �
kxv0 cos ��
72.
� 3x � 32 ��
n�2
2n xn
n�64�n �16�n�2 � 3x � 32 ��
n�2
xn
n�32�n �16�n�2
� 3x � 32� 22x2
2�64�2 �23x3
3�64�316�
24x 4
4�64�4�16�2 � . . .
y � 3x �32x2
2�64�2�1�2�2 ��1�16��32�x3
3�64�3�1�2�3 ��1�16�2�32�x 4
4�64�4�1�2�4 � . . .
� � 60�, v0 � 64, k �1
16, g � �32
73.
(a) (b)
Let Then
Thus, and we have
(c) This series converges to f at only.x � 0��
n�0 f �n��0�
n! xn � f �0� �
f��0�x1!
�f ��0�x2
2!� . . . � 0 f �x�
f��0� � 0.y � e�� � 0
ln y � limx→0
ln�e�1�x2
x � � limx→0�
��1x2 � ln x � lim
x→0� ��1 � x2 ln x
x2 � ��.
y � limx→0
e�1�x2
x.
f��0� � limx→0
f �x� � f �0�
x � 0� lim
x→0 e�1�x2
� 0x
21 3
2
1
−1−2−3x
y
f �x� � �e�1�x2,
0, x 0 x � 0
74. (a) (b)
From Exercise 8, you obtain:
(c)
(d) The curves are nearly identical for Hence, the integrals nearly agree on that interval.0 < x < 1.
G�x� � �x
0P8�t� dt
F�x� � �x
0 ln�t2 � 1�
t 2 dt
P8 � 1 �x2
2�
x4
3�
x6
4�
x8
5.
P �1x2 �
�
n�0 ��1�nx2n�2
n � 1� �
�
n�0 ��1�nx2n
n � 1
0 2
−0.5
1.5f �x� �ln�x2 � 1�
x2 .
x 0.25 0.50 0.75 1.00 1.50 2.00
0.2475 0.4810 0.6920 0.8776 1.1798 1.4096
0.2475 0.4810 0.6924 0.8865 1.6878 9.6063G�x�
F�x�
342 Chapter 9 Infinite Series
82. Assume is rational. Let and form the following.
Set a positive integer. But,
a contradiction. �1
N � 1 �1 �1
N � 1�
1�N � 1�2 � . . . �
1N � 1� 1
1 � � 1N � 1� �
1N
,
a � N!� 1�N � 1�! �
1�N � 2�! � . . . �
1N � 1
�1
�N � 1��N � 2� � . . . < 1
N � 1�
1�N � 1�2 � . . .
a � N!�e � �1 � 1 � . . . �1N!�,
e � �1 � 1 �12!
� . . . �1
N! �1
�N � 1�! �1
�N � 2�! � . . .
N > qe � p�q
83.
Equating coefficients,
etc.
In general, The coefficients are the Fibonacci numbers.an � an�1 � an�2.
a4 � a3 � a2 � 3,
a3 � a2 � a1 � 0 ⇒ a3 � 2
a2 � a1 � a0 � 0 ⇒ a2 � 1
a1 � a0 � 1 ⇒ a1 � 1
a0 � 0
x � a0 � �a1 � a0�x � �a2 � a1 � a0�x2 � �a3 � a2 � a1�x3 � . . .
x � �1 � x � x2��a0 � a1x � a2x2 � . . .�
g�x� �x
1 � x � x2 � a0 � a1x � a2x2 � . . .
75. By the Ratio Test: which shows that converges for all x.��
n�0 xn
n!lim
n →� � xn�1
�n � 1�! �n!xn� � lim
n →� �x�n � 1
� 0
76.
�ln 3 � 1.098612�
� 1 �1
12�
180
�1
448� 1.098065 ln 3 � ln�1 � 1�2
1 � 1�2� � 2�12��1 �
�1�2�2
3�
�1�2�4
5�
�1�2�6
7
� 2x ��
n�0
x2n
2n � 1, R � 1 � 2x � 2
x3
3� 2
x5
5� . . . �
� �x �x2
2�
x3
3� . . . � � ��x �
x2
2�
x3
3� . . . �
ln�1 � x1 � x� � ln�1 � x� � ln�1 � x�
80.
��91729
� �0.12483
��1�35 � �
��1�3���4�3���7�3���10�3���13�3�5!
81.
Example: �1 � x�2 � ��
n�0�2n�xn � 1 � 2x � x2
�1 � x�k � ��
n�0�kn�xn
77. �53� �
5 � 4 � 33!
�606
� 10 78. ��22 � �
��2���3�2!
� 3 79.
� �0.0390625 � �5
128
�0.54 � �
�0.5���0.5���1.5���2.5�4!
Review Exercises for Chapter 9 343
Review Exercises for Chapter 9
1. an �1n!
2. an �n
n2 � 13. 6, 5, 4.67, . . .
Matches (a)
an � 4 �2n
:
4. 3.5, 3, . . .
Matches (c)
an � 4 �n2
: 5. 10, 3, . . .
Matches (d)
an � 10�0.3�n�1:6. . . .
Matches (b)
6, �4,an � 6��23�
n�1
:
7.
The sequence seems to converge to 5.
� limn→�
�5 �2n� � 5
limn→�
an � limn→�
5n � 2
n120
0
8an �5n � 2
n 8.
The sequence seems to diverge(oscillates).
1, 0, , 0, 1, 0, . . .�1sin n�
2:
0 12
−2
2an � sin n�
2
9.
Converges
limn→�
n � 1
n2 � 0 10.
Converges
limn→�
1
�n� 0 11. lim
n→�
n3
n2 � 1� � 12.
Diverges
� �
limn→�
n
ln�n� � limn→�
1
1�n
13. Converges� limn→�
1
�n � 1 � �n� 0lim
n→� ��n � 1 � �n� � lim
n→� ��n � 1 � �n��n � 1 � �n
�n � 1 � �n
14.
Converges; k � 2n
limn→�
�1 �1
2n�n
� limk→�
��1 �1k�
k
�1�2
� e1�2 15.
Converges
limn→�
sin �n�n
� 0
84. Assume the interval is Let
Hence,
Since and
Thus,
Note: Let Then and f� �1� � 2. f � �x� � 1 f��x� ≤ 1,f �x� �12�x � 1�2 � 1.
f��x� ≤ 2.
� 3 � x2 ≤ 4.
≤ 1 � 1 �12�1 � x2� �
12�1 � x�2
2 f��x� ≤ f �1� � f ��1� �12�1 � x�2 f ��c� �
12�1 � x�2 f ��d�
f � �x� ≤ 1, f �x� ≤ 1
2 f��x� � f �1� � f ��1� �12�1 � x�2f ��c� �
12�1 � x�2f ��d�.
f �1� � f ��1� � 2 f��x� �12�1 � x�2 f ��c� �
12�1 � x�2 f ��d�
f ��1� � f �x� � ��1 � x� f�(x) �12��1 � x�2 f �(d), d � ��1, x�.
f �1� � f �x� � �1 � x� f�(x) �12�1 � x�2 f �(c), c � �x, 1�
x � �1, 1�,�1, 1�.
22. (a)
The series converges, by the Limit Comparison Test with � 1n2.
21. (a)
The series converges by the Alternating Series Test.
20. (a)
The series converges by the Alternating Series Test.
19. (a)
The series diverges geometric r �32 > 1�.�
344 Chapter 9 Infinite Series
k 5 10 15 20 25
13.2 113.3 873.8 6448.5 50,500.3Sk
(b)
120
−10
120
k 5 10 15 20 25
0.4597 0.4597 0.4597 0.4597 0.4597Sk
(b)
120
−0.1
1
23. Converges. Geometric series, r � 0.82, r < 1.
25. Diverges. nth-Term Test. limn→�
an � 0.
k 5 10 15 20 25
0.3917 0.3228 0.3627 0.3344 0.3564Sk
(b)
0 120
1
k 5 10 15 20 25
0.8333 0.9091 0.9375 0.9524 0.9615Sk
(b)
0 120
1
24. Diverges. Geometric series, r � 1.82 > 1.
26. Diverges. nth-Term Test, limn→�
an �23
.
27. Geometric series with and
S �a
1 � r�
11 � �2�3� �
11�3
� 3
r �23
.a � 1��
n�0�2
3�n
28.
See Exercise 27.
��
n�0
2n�2
3n � 4 ��
n�0�2
3�n
� 4�3� � 12
16. Let
Assume and note that the terms
converge as Hence converges.ann →�.
bn ln b � cn ln cbn � cn �
bn ln bbn � cn �
cn ln cbn � cn
b ≥ c
limn→�
ln y � limn→�
1
bn � cn�bn ln b � cn ln c�.
ln y �ln�bn � cn�
n
y � �bn � cn�1�n
18. (a)
(b) V5 � 120,000�0.70�5 � $20,168.40
n � 1, 2, 3, 4, 5Vn � 120,000�0.70�n,
17.
(a)
(b) A40 8218.10
A4 5254.73 A8 5522.43
A3 5189.85 A7 5454.25
A2 5125.78 A6 5386.92
A1 � 5062.50 A5 5320.41
n � 1, 2, 3
An � 5000�1 �0.05
4 �n
� 5000�1.0125�n
Review Exercises for Chapter 9 345
32. (a)
(b) 0.923076 �0.923076
1 � 0.000001�
923,076999,999
�12�76,923�13�76,923� �
1213
0.923076 � 0.9230761 � 0.000001 � �0.000001�2 � . . .� � ��
n�0 �0.923076��0.000001�n
33.
meters � �8 � ��
n�016�0.7�n � �8 �
161 � 0.7
� 4513
D � 8 � 16�0.7� � 16�0.7�2 � . . . � 16�0.7�n � . . .�
D2 � 0.7�8� � 0.7�8� � 16�0.7�
D1 � 8
35. See Exercise 110 in Section 9.2.
$5087.14
�200�e�0.06��2� � 1�
e0.06�12 � 1
A �P�ert � 1�er�12 � 1
34.
$4,371,379.65
S � �39
n�032,000�1.055�n �
32,000�1 � 1.05540�1 � 1.055
36. See Exercise 110 in Section 9.2.
$14,343.25
� 100� 120.035���1 �
0.03512 �
120
� 1�
A � P�12r ���1 �
r12�
12t
� 1�
37.
By the Integral Test, the series converges.
��
1x�4 ln�x� dx � lim
b→� ��
ln x3x3 �
19x3�
b
1� 0 �
19
�19
39.
Since the second series is a divergent p-series while the first series is a convergent p-series, the differencediverges.
��
n�1� 1
n2 �1n� � �
�
n�1 1n2 � �
�
n�1 1n
38.
Divergent p-series, p �34 < 1
��
n�1
14�n3
� ��
n�1
1n3�4
40.
The first series is a convergent p-series and the secondseries is a convergent geometric series. Therefore, theirdifference converges.
��
n�1� 1
n2 �12n� � �
�
n�1 1n2 � �
�
n�1 12n
41.
By a limit comparison test with the convergent p-series
the series converges.��
n�1
1n3�2,
limn→�
1��n3 � 2n
1��n3�2� � limn→�
n3�2
�n3 � 2n� 1
��
n�1
1
�n3 � 2n42.
By a limit comparison test with the series diverges.��
n�1 1n
,
limn→�
�n � 1��n�n � 2�
1�n� lim
n→� n � 1n � 2
� 1
��
n�1
n � 1n�n � 2�
29. �1
1 � �1�2� �1
1 � �1�3� � 2 �32
�12�
�
n�0� 1
2n �13n� � �
�
n�0�1
2�n
� ��
n�0�1
3�n
30.
�1
1 � �2�3� � ��1 �12� � �1
2�
13� � �1
3�
14� � . . .� � 3 � 1 � 2
��
n�0��2
3�n
�1
�n � 1��n � 2�� � ��
n�0�2
3�n
� ��
n�0� 1
n � 1�
1n � 2�
31. (a)
(b) 0.09 �0.09
1 � 0.01�
111
0.09 � 0.09 � 0.0009 � 0.000009 � . . . � 0.09�1 � 0.01 � 0.0001 � . . .� � ��
n�0 �0.09��0.01�n
346 Chapter 9 Infinite Series
45. Converges by the Alternating Series Test. (Conditional convergence)
47. Diverges by the nth-Term Test.
46.
By the Alternating Series Test, the series converges.
limn→�
�n
n � 1� 0
an�1 ��n � 1n � 2
≤ �n
n � 1� an
��
n�1
��1�n�nn � 1
48. Converges by the Alternating Series Test.
an�1 �3 ln�n � 1�
n � 1<
3 ln nn
� an, limn→�
3 ln n
n� 0
49.
By the Ratio Test, the series converges.
� �0��1� � 0 < 1
� limn→�
� 1e2n�1��n � 1
n �
� limn→�
en2�n � 1�en2�2n�1n
limn→�
an�1
an � limn→�
n � 1e�n�1�2
en2
n ��
n�1
nen2
51.
Therefore, by the Ratio Test, the series diverges.
� limn→�
2n3
�n � 1�3 � 2limn→�
an�1
an � limn→�
2n�1
�n � 1�3 n3
2n��
n�1 2n
n3
50.
By the Ratio Test, the series diverges.
� limn→�
n � 1
e� �
limn→�
an�1
an � limn→�
�n � 1�!en�1
en
n!��
n�1 n!en
52.
By the Ratio Test, the series converges.
limn→�
an�1
an � limn→�
1 3 . . . �2n � 1��2n � 1�2 5 . . . �3n � 1��3n � 2�
2 5 . . . �3n � 1�1 3 . . . �2n � 1� � lim
n→� 2n � 13n � 2
�23
��
n�1 1 3 5 . . . �2n � 1�2 5 8 . . . �3n � 1�
43.
Since diverges (harmonic series),
so does the original series.
��
n�1
12n
�12
��
n�1 1n
� �32
54
. . . 2n � 12n � 2�
12n
>1
2n
an �1 3 5 . . . �2n � 1�
2 4 6 . . . �2n�
��
n�1 1 3 5 . . . �2n � 1�
2 4 6 . . . �2n�44. Since converges, converges by the
Limit Comparison Test.
��
n�1
13n � 5�
�
n�1
13n
54. (a) The series converges by the Alternating Series Test.
(b) (c)
(d) The sum is approximately 0.0714. 0 120
0.3
Review Exercises for Chapter 9 347
x 5 10 15 20 25
0.0871 0.0669 0.0734 0.0702 0.0721Sn
55. (a) ��
N
1x2 dx � ��
1x�
�
N�
1N
N 5 10 20 30 40
1.4636 1.5498 1.5962 1.6122 1.6202
0.2000 0.1000 0.0500 0.0333 0.0250��
N
1x2 dx
�N
n�1 1n2
(b)
The series in part (b) converges more rapidly. The integral values represent the remainders of the partial sums.
��
N
1x5 dx � ��
14x4�
�
N�
14N4
N 5 10 20 30 40
1.0367 1.0369 1.0369 1.0369 1.0369
0.0004 0.0000 0.0000 0.0000 0.0000��
N
1x5 dx
�N
n�1
1n5
56. No. Let then The series is a convergent p-series.��
n�1
3937.5n2a75 � 0.7.an �
3937.5n2 ,
57.
f����x� � �18
e�x�2 f����0� � �18
� 1 �12
x �18
x2 �1
48x3 f��x� �
14
e�x�2 f��0� �14
� 1 �12
x �14
x2
2!�
18
x3
3! f��x� � �
12
e�x�2 f��0� � �12
P3�x� � f �0� � f��0�x � f��0�x2
2!� f����0�x3
3! f �x� � e�x�2 f �0� � 1
58.
f����x� � 4 sec2 x tan2 x � 2 sec4 x f������
4� � 16
f��x� � 2 sec2 x tan x f����
4� � �4
f��x� � sec2 x f����
4� � 2
P3�x� � �1 � 2�x ��
4� � 2�x ��
4�2
�83�x �
�
4�3
f �x� � tan x f ���
4� � �1
53. (a) Converges
(b) (c)
(d) The sum is approximately 3.75.−1
0 12
4
� limn→�
�n � 1n ��3
5� �35
< 1,Ratio Test: limn→�
an�1
an � limn→�
�n � 1��3�5�n�1
n�3�5�n
x 5 10 15 20 25
2.8752 3.6366 3.7377 3.7488 3.7499Sn
348 Chapter 9 Infinite Series
60. cos�0.75� 1 ��0.75�2
2!�
�0.75�4
4!�
�0.75�6
6! 0.7317
61. ln�1.75� �0.75� ��0.75�2
2�
�0.75�3
3�
�0.75�4
4�
�0.75�5
5�
�0.75�6
6� . . . �
�0.75�14
14 0.559062
62. e�0.25 1 � 0.25 ��0.25�2
2!�
�0.25�3
3!�
�0.25�4
4! 0.779
63.
(a)
This inequality is true for
(b)
This inequality is true for
(c)
This inequality is true for
(d)
This inequality is true for n � 10.
Rn�x� ≤ 2n�1
�n � 1�! < 0.0001
n � 5.
Rn�x� ≤ �0.5�n�1
�n � 1�! < 0.0001
n � 6.
Rn�x� ≤ �1�n�1
�n � 1�! < 0.001
n � 4.
Rn�x� ≤ �0.5�n�1
�n � 1�! < 0.001
f �n�1��z� ≤ 1 ⇒ Rn�x� ≤ xn�1
�n � 1�!
Rn�x� �f �n�1��z��n � 1�!x
n�1
c � 0f �x� � cos x, 64.
−10 10
−10
f
P4
P10P6
10
P10�x� � 1 �x2
2!�
x4
4!�
x6
6!�
x8
8!�
x10
10!
P6�x� � 1 �x2
2!�
x4
4!�
x6
6!
P4�x� � 1 �x2
2!�
x4
4!
f �x� � cos x
65.
Geometric series which converges only if or �10 < x < 10.
x�10 < 1
��
n�0� x
10�n
66.
Geometric series which converges only if or�
12 < x < 12 .
2x < 1
��
n�0�2x�n
67.
Center: 2
Since the series converges when and when ,the interval of convergence is 1 ≤ x ≤ 3.
x � 3x � 1
R � 1
� x � 2
limn→�
un�1
un � limn→�
��1�n�1�x � 2�n�1
�n � 2�2 �n � 1�2
��1�n�x � 2�n ��
n�0
��1�n�x � 2�n
�n � 1�2 68.
Center: 2
Since the series converges at and diverges at , theinterval of convergence is 53 ≤ x < 73 .
73
53
R �13
� 3x � 2
limn→�
un�1
un � limn→�
3n�1�x � 2�n�1
n � 1
n3n�x � 2�n
��
n�1
3n�x � 2�n
n
59. Since
sin 95 � sin�95�
180� 95�
180�
�95��3
18033!�
�95��5
18055!�
�95��7
18077! 0.99594
�95��9
1809 9!< 0.001, use four terms.
Review Exercises for Chapter 9 349
69.
which implies that the series converges only at the centerx � 2.
limn→�
un�1
un � limn→�
�n � 1�!�x � 2�n�1
n!�x � 2�n � �
��
n�0n!�x � 2�n 70.
Geometric series which converges only if
or 0 < x < 4.x � 22 < 1
��
n�0 �x � 2�n
2n � ��
n�0�x � 2
2 �n
71.
� ��
n�0���1�n�11
4n�n!�2 � ��1�n 1
4n�n!�2�x2n�2 � 0
� ��
n�0���1�n�14�n � 1�2
4n�1�n � 1�!�2 � ��1�n 1
4n�n!�2�x2n�2
� ��
n�0���1�n�1�2n � 2��2n � 1 � 1�
4n�1�n � 1�!�2 � ��1�n 1
4n�n!�2�x2n�2
� ��
n�0���1�n�1
�2n � 2��2n � 1�4n�1�n � 1�!�2 �
��1�n�1�2n � 2�4n�1�n � 1�!�2 �
��1�n
4n�n!�2�x2n�2
x2y� � xy� � x2y � ��
n�0 ��1�n�1�2n � 2��2n � 1�x2n�2
4n�1�n � 1�!�2 � ��
n�0
��1�n�1�2n � 2�x2n�2
4n�1�n � 1�!�2 � ��
n�0��1�n
x2n�2
4n�n!�2
y� � ��
n�0 ��1�n�1�2n � 2��2n � 1�x2n
4n�1�n � 1�!�2
y� � ��
n�1 ��1�n�2n�x2n�1
4n�n!�2 � ��
n�0 ��1�n�1�2n � 2�x2n�1
4n�1�n � 1�!�2
y � ��
n�0��1�n
x2n
4n�n!�2
72.
� ��
n�1 ��1�n3n�1x2n
2nn!�2n � 2n� � 0
� ��
n�1 ��1�n�13n�1x2n
2nn!�2n� � �
�
n�1 ��1�n3n�1x2n
2n�1�n � 1�! 2n2n
� ��
n�0 ��1�n3n�1x2n
2nn!��2n� � �
�
n�0 ��1�n�13n�2x2n�2
2nn!
� ��
n�0 ��1�n3n�1x2n
2nn!��2n � 1� � 1� � �
�
n�0 ��1�n�13n�2x2n�2
2nn!
� ��
n�0 ��1�n�13n�1�2n � 2�x2n
2nn!� �
�
n�0 ��1�n�13n�2x2n�2
2nn!� �
�
n�0 ��1�n3n�1x2n
2nn!
y� � 3xy� � 3y � ��
n�0 ��3�n�1�2n � 2��2n � 1�x2n
2n�1�n � 1�! � ��
n�0
��1�n�13n�2�2n � 2�x2n�2
2n�1�n � 1�! � ��
n�0 ��1�n3n�1x2n
2nn!
y� � ��
n�0 ��3�n�1�2n � 2��2n � 1�x2n
2n�1�n � 1�!
y� � ��
n�1 ��3�n�2n�x2n�1
2nn!� �
�
n�0 ��3�n�1�2n � 2�x2n�1
2n�1�n � 1�!
y � ��
n�0 ��3�nx2n
2nn!
73.
��
n�0 23 �
x3�
n
� ��
n�0
2xn
3n�1
23 � x
�2�3
1 � �x�3� �a
1 � r74.
��
n�0
32��
x2�
n
� ��
n�0
��1�n3xn
2n�1
32 � x
�3�2
1 � �x�2� �3�2
1 � ��x�2� �a
1 � r
350 Chapter 9 Infinite Series
75. Power series
Derivative:
� ��
n�0 29
�n � 1��x3�
n
��
n�1 23
n�x3�
n�1
�13� � �
�
n�1 29
n�x3�
n�1
��
n�0 23�
x3�
n
g�x� �2
3 � x. 76. Integral: �
�
n�0
��1�n3xn�1
�n � 1�2n�1
77. �32
< x < 32
1 �23
x �49
x2 �8
27x3 � . . . � �
�
n�0�2x
3 �n
�1
1 � �2x�3� �3
3 � 2x,
79.
��22 �
�
n�0 ��1�n�n�1��2x � �3��4��n
n! �
�22
��22 �x �
3�
4 � ��2
2 2!�x �3�
4 �2
� . . .
sin�x� � ��
n�0 f �n��x�x � �3��4��n
n!
f���x� � �cos x, . . .
f��x� � �sin x
f��x� � cos x
f �x� � sin x
78.
�1 < x < 7 �32
4 � �x � 3� �32
1 � x,
8 � 2�x � 3� �12
�x � 3�2 �18
�x � 3�3 � . . . � ��
n�0 8���x � 3�
4 �n
�8
1 � ��x � 3��4�
80.
��22 �1 � �x �
�
4� � ��
n�1 ��1�n�n�1���2x � ���4��n�1
�n � 1�! �
��22
��22 �x �
�
4� ��2
2 2!�x ��
4�2
��2
2 3!�x ��
4�3
��2
2 4!�x ��
4�4
� . . .cos x � ��
n�0 f �n�����4�x � ���4��n
n!
f���x� � sin x
f��x� � �cos x
f��x� � �sin x
f �x� � cos x
81. and since we have � 1 � x ln 3 �x2 ln 3�2
2!�
x3 ln 3�3
3!�
x4 ln 3�4
4!� . . . . 3x � �
�
n�0 �x ln 3�n
n!ex � �
�
n�0 xn
n!,3x � �eln�3��x � ex ln�3�
82.
� 1 �12!�x �
�
2�2
�54!�x �
�
2�4
� . . .csc�x� � ��
n�0 f �n����2�x � ���2��n
n!
f �4��x� � 5 csc5�x� � 15 csc3�x� cot2�x� � csc�x� cot4�x�
f���x� � �5 csc3�x� cot�x� � csc�x� cot3�x�
f��x� � csc3�x� � csc�x� cot2�x�
f��x� � �csc�x� cot�x�
f �x� � csc�x�
Review Exercises for Chapter 9 351
83.
� ��
n�0 �n!�x � 1�n
n!� � �
�
n�0�x � 1�n, �2 < x < 0
1x
� ��
n�0
f �n���1��x � 1�n
n!
f����x� � �6x4, . . .
f��x� �2x3
f��x� � �1x2
f �x� �1x
84.
� 2 ��x � 4�
22 � ��
n�2
��1�n�11 3 5 . . . �2n � 3��x � 4�n
23n�1n!
� 2 ��x � 4�
22 ��x � 4�2
252!�
1 3�x � 4�3
283!�
1 3 5�x � 4�4
2114!� . . . �x � �
�
n�0 f �n��4��x � 4�n
n!
f �4��x� � ��12��
12��
32��
52�x�7�2, . . .
f���x� � �12��
12��
32�x�5�2
f��x� � ��12��
12�x�3�2
f��x� �12
x�1�2
f �x� � x1�2
85.
� 1 �x5
�2
25x2 �
6125
x3 � . . .
� 1 �x5
� ��
n�2
��1�n�14 9 14 . . . �5n � 6�xn
5nn!
� 1 �15
x �1 4x2
522!�
1 4 9x3
533!� . . .
�1 � x�1�5 � 1 �x5
��1�5���4�5�x2
2!�
1�5��4�5���9�5�x3
3!� . . .
�1 � x�k � 1 � kx �k�k � 1�x2
2!�
k�k � 1��k � 2�x3
3!� . . .
86.
� ��
n�0 ��1�n�n � 2�!xn
2n!� �
�
n�0 ��1�n�n � 2��n � 1�xn
2
1�1 � x�3 � 1 � 3x �
12x2
2!�
60x3
3!�
360x4
4!�
2520x5
5!� . . .
h�5��x� � �2520�1 � x��8
h�4��x� � 360�1 � x��7
h���x� � �60�1 � x��6
h��x� � 12�1 � x��5
h��x� � �3�1 � x��4
h�x� � �1 � x��3
352 Chapter 9 Infinite Series
87. 0 < x ≤ 2
� ��
n�1��1�n�1
14nn
0.2231
ln�54� � �
�
n�1��1�n�1��5�4� � 1
n �n
ln x � ��
n�1��1�n�1
�x � 1�n
n,
89.
e1�2 � ��
n�0 �1�2�n
n!� �
�
n�0
12nn!
1.6487
�� < x < �ex � ��
n�0 xn
n!,
88. 0 < x ≤ 2
� ��
n�1��1�n�1
15nn
0.1823
ln�65� � �
�
n�1��1�n�1��6�5� � 1
n �n
ln x � ��
n�1��1�n�1
�x � 1�n
n,
90.
� ��
n�0
2n
3nn! 1.9477e2�3 � �
�
n�0 �2�3�n
n!
�� < x < �ex � ��
n�0 xn
n!,
91.
cos�23� � �
�
n�0��1�n
22n
32n�2n�! 0.7859
�� < x < � cos x � ��
n�0��1�n
x2n
�2n�!,
93. The series for Exercise 41 converges very slowly because the terms approach 0 at a slow rate.
95. (a)
� 1 � 2x � 2x2 �43
x3P�x� � 1 � 2x �4x2
2!�
8x3
3!
f ���x� � 8e2x f ���0� � 8
f��x� � 4e2x f ��0� � 4
f��x� � 2e2x f��0� � 2
f �x� � e2x f �0� � 1 (b)
(c)
P � 1 � 2x � 2x 2 �43
x 3
e x e x � �1 � x �x2
2!� . . .��1 � x �
x2
2!� . . .�
P�x� � 1 � 2x � 2x2 �43
x3
e x � ��
n�0 xn
n!, e2x � �
�
n�0 �2x�n
n!
92.
sin�13� � �
�
n�0��1�n
132n�1�2n � 1�! 0.3272
�� < x < � sin x � ��
n�0��1�n
x2n�1
�2n � 1�!,
94.
� 1 �x 3
2� �
�
n�2
��1�n1 3 . . . �2n � 1�2nn!
x 3n
� 1 �12
�x 3� ���1�2���3�2�
2!x6 �
��1�2���3�2���5�2�3!
x9 � . . .
1
�1 � x 3� �1 � x 3��1�2, k � �
12
96. (a)
—CONTINUED—
� 2x �43
x3 �4
15x5 �
8315
x7 � . . .sin 2x � 0 � 2x �0x2
2!�
8x3
3!�
0x4
4!�
32x5
5!�
0x6
6!�
128x7
7!� . . .
f �7��x� � �128 cos 2x f �7��0� � �128
f �6��x� � �64 sin 2x f �6��0� � 0
f �5��x� � 32 cos 2x f �5��0� � 32
f �4��x� � 16 sin 2x f �4��0� � 0
f���x� � �8 cos 2x f���0� � �8
f��x� � �4 sin 2x f��0� � 0
f��x� � 2 cos 2x f��0� � 2
f �x� � sin 2x f �0� � 0
Review Exercises for Chapter 9 353
97.
� ��
n�0
��1�nx2n�1
�2n � 1��2n � 1�!
�x
0 sin t
t dt � � �
�
n�0
��1�nt2n�1
�2n � 1��2n � 1�!�x
0
sin t
t� �
�
n�0 ��1�nt2n
�2n � 1�!
sin t � ��
n�0 ��1�nt2n�1
�2n � 1�!
99.
� ��
n�0 ��1�nxn�1
�n � 1�2�x
0 ln�t � 1�
t dt � � �
�
n�0 ��1�nt n�1
�n � 1�2 �x
0
ln�t � 1�
t� �
�
n�0 ��1�nt n
n � 1
ln�1 � t� � � 11 � t
dt � ��
n�0 ��1�ntn�1
n � 1
1
1 � t� �
�
n�0��1�ntn
96. —CONTINUED—
(b)
(c)
� 2�x �2x3
3�
2x5
15�
4x7
315� . . .� � 2x �
43
x3 �4
15x5 �
8315
x7 � . . .
� 2�x � ��x3
2�
x3
6 � � � x5
24�
x5
12�
x5
120� � ��x7
720�
x7
144�
x7
240�
x7
5040� � . . .�
� 2�x �x3
6�
x5
120�
x7
5040� . . .��1 �
x2
2�
x4
24�
x6
720� . . .�
sin 2x � 2 sin x cos x
� 2x �8x3
6�
32x5
120�
128x7
5040� . . . � 2x �
43
x3 �4
15x5 �
8315
x7 � . . .
sin 2x � ��
n�0 ��1�n�2x�2n�1
�2n � 1�! � 2x ��2x�3
3!�
�2x�5
5!�
�2x�7
7!� . . .
sin x � ��
n�0 ��1�nx2n�1
�2n � 1�!
98.
� ��
n�0
��1�nx n�1
22n�2n�!�n � 1�
�x
0cos
�t2
dt � � ��
n�0
��1�n t n�1
22n�2n�!�n � 1��x
0
cos �t2
� ��
n�0 ��1�n t n
22n�2n�!
cos t � ��
n�0 ��1�nt 2n
�2n�!
100.
�x
0 et � 1
t dt � � �
�
n�1
tn
n n!�x
0� �
�
n�1
xn
n n!
et � 1
t� �
�
n�1 tn�1
n!
et � 1 � ��
n�1 tn
n!
et � ��
n�0 tn
n!
101.
By L’Hôpital’s Rule, limx→0�
arctan x
�x� lim
x→0� � 1
1 � x2�� 1
2�x�� lim
x→0�
2�x1 � x2 � 0.
limx→0�
arctan x
�x� 0
arctan x
�x� �x �
x5�2
3�
x 9�2
5�
x13�2
7�
x17�2
9� . . .
arctan x � x �x3
3�
x5
5�
x7
7�
x9
9� . . .
354 Chapter 9 Infinite Series
102.
By L’Hôpital’s Rule, limx→0
arcsin x
x� lim
x→0 � 1�1 � x2�
1� 1.
limx→0
arcsin x
x� 1
arcsin x
x� 1 �
x2
2 � 3�
1 � 3x4
2 � 4 � 5�
1 � 3 � 5x6
2 � 4 � 6 � 7� . . .
arcsin x � x �x3
2 � 3�
1 � 3x5
2 � 4 � 5�
1 � 3 � 5x7
2 � 4 � 6 � 7� . . .
Problem Solving for Chapter 9
1. (a)
(b)
(c) limn→�
Cn � 1 � ��
n�0 13 �
23�
n
� 1 � 1 � 0
0, 13
, 23
, 1, etc.
�1�3
1 � �2�3�� 1
1�1
3� � 2�1
9� � 4� 1
27� � . . . � ��
n�0
1
3 �2
3�n
2. Let
Thus, S ��2
6�
14
�2
6�
�2
6 �34� �
�2
8.
� S �122 ��2
6 �. � S �122�1 �
122 �
132 � . . .
� S �122 �
142 � . . .
Then �2
6�
112 �
122 �
132 �
142 � . . .
S � ��
n�1
1
�2n � 1�2�
1
12�
1
32�
1
52� . . ..
3. If there are n rows, then
For one circle,
and
For three circles,
and
For six circles,
and
Continuing this pattern,
limn→�
An ��
2�
14
��
8
An ��
2
n�n � 1�2�3 � 2�n � 1��2
Total Area � ��rn2�an � �� 1
2�3 � 2�n � 1��2
n�n � 1�
2
rn �1
2�3 � 2�n � 1�.
r3 �1
2�3 � 4.
1 � 2�3r3 � 4r3a3 � 6
1 3 r3
r3r32
r2 �1
2 � 2�3.
1 � 2�3r2 � 2r2a2 � 3
1 3 r2
r2r22
r1 �1
3��3
2 � ��3
6�
1
2�3.a1 � 1
1 32
12
r1
an �n�n � 1�
2.
Problem Solving for Chapter 9 355
4. (a) Position the three blocks as indicated in the figure. The bottom block extends overthe edge of the table, the middle block extends over the edge of the bottom block,and the top block extends over the edge of the middle block.
The centers of gravity are located at
bottom block:
middle block:
top block:16
�14
�12
�12
�5
12.
16
�14
�12
� �1
12
16
�12
� �13
1�21�4
1�6
16
512
1112
14
16
12
0
The center of gravity of the top 2 blocks is
which lies over the bottom block. The center of gravity of the 3 blocks is
which lies over the table. Hence, the far edge of the top block lies
beyond the edge of the table.
(b) Yes. If there are n blocks, then the edge of the top block lies from the
edge of the table. Using 4 blocks,
which shows that the top block extends beyond the table.
(c) The blocks can extend any distance beyond the table because the series diverges:
��
i�1 12i
�12�
�
i�1 1i
� �.
�4
i�1 12i
�12
�14
�16
�18
�2524
�n
i�1 12i
16
�14
�12
�1112
��13
�1
12�
512��3 � 0
��1
12�
512��2 �
16
,
5. (a)
because each series in the second line has
(b)
�Assume all an > 0.� R � 1
� �a0 � a1x � . . . � ap�1x p�1� 11 � x p
� �a0 � a1x � . . . � ap�1xp�1��1 � xp � . . .�
� a0�1 � xp � . . .� � a1x�1 � xp � . . .� � . . . � ap�1xp�1�1 � xp � . . .�
�anxn � �a0 � a1x � . . . � ap�1xp�1� � �a0x p � a1x
p�1 � . . .� � . . .
R � 1.R � 1
� �1 � 2x � 3x2� 11 � x3
� �1 � x3 � x6 � . . .�1 � 2x � 3x2�
� �1 � x3 � x6 � . . .� � 2�x � x4 � x7 � . . .� � 3�x2 � x5 � x8 � . . .�
�anxn � 1 � 2x � 3x2 � x3 � 2x4 � 3x5 � . . .
356 Chapter 9 Infinite Series
6.
If converges conditionally.
If diverges.
No values of a and b give absolute convergence. implies conditional convergence.a � b
a � b, ��
n�1
��1�n�1�a � b�2n
� ��
n�1
a � b2n
a � b, ��
n�1
��1�n�1�2a�2n
� a ��
n�1
��1�n�1
n
a �b2
�a3
�b4
� . . . � ��
n�1
��1�n�1�a � b� � �a � b�2n
7. (a)
Letting you have Letting
Thus,
(b) Differentiating,
Letting
2e � ��
n�0 n � 1
n!� 5.4366.
x � 1,
xex � ex � ��
n�0
�n � 1�xn n!
.
��
n�1
1�n � 2�n!
�12
.
e � e � 1 � ��
n�0
1�n � 2�n!
�12
� ��
n�1
1�n � 2�n!
.
x � 1,C � 1.x � 0,
xex dx � xex � ex � C � ��
n�0
xn�2
�n � 2�n!
xex � ��
n�0 xn�1
n!
ex � 1 � x �x2
2!� . . . � �
�
n�0 xn
n!8.
f �12��0�
12!�
16!
⇒ f �12��0� �12!6!
� 665,280
ex2� 1 � x2 �
x4
2!� . . . �
x12
6!� . . .
ex � 1 � x �x2
2!� . . .
9. Let
Then,
Since and this series converges.an�1 < an,limn→�
an � 0
�
0 sin x
x dx � a1 � a2 � a3 � a4 � . . . .
a1 � �
0 sin x
x dx, a2 � � 2�
�
sin x
x dx, a3 � 3�
2�
sin x
x dx, etc.
10. (a) If diverges.
If converges.
If diverges.
(b) diverges by part (a).��
n�4
1n ln�n2� �
12 �
�
n�4
1n ln n
p < 1,
p > 1, �
2
1x�ln x�p dx � lim
b→� ��ln b�1�p
1 � p�
�ln 2�1�p
1 � p
p � 1, �
2
1x ln x
dx � ln ln x�
2
Problem Solving for Chapter 9 357
11. (a)
[See part (b) for proof.]
(b) Use mathematical induction to show the sequence is increasing. Clearly,
Now assume Then
Use mathematical induction to show that the sequence is bounded above by a. Clearly,
Now assume Then and implies
Hence, the sequence converges to some number L. To find L, assume
Hence, L �1 � �1 � 4a
2.
L �1 ± �1 � 4a
2.
L � �a � L ⇒ L2 � a � L ⇒ L2 � L � a � 0
an�1 � an � L:
a > �an � a � an�1.
a2 > an � a
a2 � a > an
a�a � 1� > an�1�
a � 1 > 1a > anan < a.
a1 � �a < a.
an�1 > an.
�an � a > �an�1 � a
an � a > an�1 � a
an > an�1.
a2 � �a � a1 � �a�a > �a � a1.
limn→�
an �1 � �13
2
a6 � 2.30146
a5 � 2.29672
a4 � 2.27493
a3 � 2.17533
a2 � 1.73205
a1 � 3.0
12. Let
as
By the Root Test, converges converges.⇒ �anr n�bn
Lr <1rr � 1.
n →�. �bn�1�n � �anr n�1�n � an1�n � r → Lr
bn � anrn.
13. (a)
—CONTINUED—
S5 �4532
�1
16�
4732
S4 �118
�1
32�
4532
S3 �98
�14
�118
S1 � 1 �18
�98
S1 �120 � 1
��
n�1
12n���1�n �
121�1 �
122�1 �
123�1 �
124�1 �
125�1 � . . .
358 Chapter 9 Infinite Series
14. (a)
(b)
� 1.0204081632 . . .
� 1 � 0.02 � 0.0004 � . . .
� 1 � 0.02 � �0.02�2 � . . .
1
0.98�
11 � 0.02
� ��
n�0�0.02�n
� 1.010101 . . .
� 1 � 0.01 � �0.01�2 � . . .
1
0.99�
11 � 0.01
� ��
n�0�0.01�n
13. —CONTINUED—
(b)
This sequence is which diverges.
(c)
converges because and and n�2 →1.n�1�2 →1�2��1�n� �12, 2, 12, 2, . . . �
1
2 � n�2��1�n → 1
2< 1
n� 12n���1�n � � 1
2n � 2��1�n�1�n
18, 2, 18, 2, . . .
an�1
an
�2n���1�n
2�n�1����1�n�1 �2��1�n
21���1�n�1
15.
S10 � 1490 � 810 � 440 � 2740
S9 � 810 � 440 � 240 � 1490
S8 � 440 � 240 � 130 � 810
S7 � 240 � 130 � 70 � 440
S6 � 130 � 70 � 40 � 240
16. (a)
(b)
(c)
converges. �43
� ��
n�1
1n3�2
W �43
��1 �1
23�2 �1
33�2 � . . .
S � 4��1 �12
�13
� . . . � 4� ��
n�1
1n
� �
� 2 ��
n�1
1n1�2 � � �p-series, p �
12
< 1�
Height � 2�1 �1�2
�1�3
� . . .
17. (a) diverges (harmonic)
(b) Let By the Mean Value Theorem,
where is between and . Thus,
Because converges, the Comparison Theorem
tells us that converges.��
n�1�sin� 1
2n� � sin� 12n � 1�
��
n�1
12n�2n � 1�
�1
2n�2n � 1�
≤ � 12n
�1
2n � 1� 0 ≤ �sin� 1
2n� � sin� 12n � 1��
yxc
� f �x� � f �y�� � f��c��x � y� � cos�c��x � y� ≤ �x � y�,f �x� � sin x.
��
n�1
12n
�12
��
n�1 1n