tests for the convergence
Post on 03-Jun-2018
225 Views
Preview:
TRANSCRIPT
-
8/12/2019 Tests for the Convergence
1/27
Tests for the Convergence of Infinite SeriesDC-I
Semester-IILesson: Tests for the Convergence of
Infinite Series
Course Developer: Dr. Chaitanya KumarDepartment/College:
Department of Mathematics,Delhi College of Arts and Commerce (D.U.)
University of Delhi
Institute of Lifelong Learning, University of Delhi pg. 1
-
8/12/2019 Tests for the Convergence
2/27
Tests for the Convergence of Infinite Series
Table of Contents:
Chapter : Tests for the Convergence of Infinite Series
1. Learning Outcomes 2. Introduction 3. D'Alember's Ratio Test 4. Cauchy's nth Root test 5. Convergence of the infinite integral ( )
1
u x dx
o 5.1. Cauchy's Integral Test Exercises-1 6. Alternating series
o 6.1. Leibnitz testo 6.2. Absolute convergenceo 6.3. Conditional Convergence
Exercise 2 Summary References
1. Learning Outcomes
After you have read this chapter, you should be able to
Define and understand the following tests for the convergenceof an infinite sereis.
o D'Alember's Ratio Testo Cauchy's nth Root testo Cauchy's Integral Test
Define the alternating series and convergence of thealternating series
Absolute convergence Conditional Convergence
Institute of Lifelong Learning, University of Delhi pg. 2
-
8/12/2019 Tests for the Convergence
3/27
Tests for the Convergence of Infinite Series
2. IntroductionThe purpose of this chapter is to discuss the following tests
for convergence, conditional convergence and absolute convergence
of an infinite series.
3. D'Alember's Ratio TestLet
nu be a positive term series such that
n
nn 1
ulim
u += (1)
then
(i) unconverges if > 1.(ii) undiverges if < 1.(iii) Test fails if = 1.
Proof: Case (i)let > 1, we can choose > 0 such that - > 1
or > 1, = -
Using (1), there exists a positive integer m1, such that
n1
n 1
u, n m
u
+
<
n1
n 1
u, n m
u
+
< < +
Consider
( )n 1n 1
u, n m
u
+
> =
Institute of Lifelong Learning, University of Delhi pg. 3
-
8/12/2019 Tests for the Convergence
4/27
-
8/12/2019 Tests for the Convergence
5/27
Tests for the Convergence of Infinite Seriesn
nn 1
ulim 1
u +=
The series n1
un
= is convergent, but
n
n n nn 1
u n 1 1lim lim lim 1 1
u n n +
+ = = + =
The series n 21
un
= is convergent, but
( )2 2
n
2
n n nn 1
n 1u 1lim lim lim 1 1
u n n +
+ = = + =
Value Addition: Note
1. The test fails for = 1 in the sense that it fails to give anydefine information.
2. Ifn
u is a positive term series such that nn
n 1
ulim ,
u += then
nu is convergent.
Example 1: Test for convergence the seriesn 1
n 1n 1
2
3
+=
.
Solution: We have
n 1 n
n n 1n n 1
2 2u , u
3 1 3 1
+ += =
+ +
n 1 n
n
nn n n
n 1n
13
u 1 3 1 1 3lim lim lim
1u 2 3 1 21
3
+
+
+ +
= = + +
Institute of Lifelong Learning, University of Delhi pg. 5
-
8/12/2019 Tests for the Convergence
6/27
Tests for the Convergence of Infinite Series n
nn 1
u 3lim 1
u 2 += >
Hence by D'Alembert's Ratio Test the given series converges.
Example 2: Test for convergence the series2
n
2
n 1x
n 1
+
.
Solution: Here
( )
( )
22
n n 1
n n 1 22
n 1 1n 1u x , u x
n 1 n 1 1
++
+ = =
+ + +
( ) ( )
( )
22 n
n
22 n 1n nn 1
n 1 n 1 1u x 1lim lim .
u n 1 x xn 1 1 +
+
+ += =
+ +
Hence by D'Alembert's Ratio test the given series converges if
11 x 1
x> < and diverges if
11 x 1
x< > .
The test fails to give any information when x = 1,
when x = 1,2
n 2
n 1u
n 1
=
+.
2
n 2n n
n 1lim u lim 1 0
n 1
= =
+
The given series is divergent
Hence the given series converges if x < 1 and diverges if x > 1.
Example 3: Test for convergence the series2 3 4
1 1 1 1...
2 2.2 3.2 4.2+ + + +
Solution: Here,
Institute of Lifelong Learning, University of Delhi pg. 6
-
8/12/2019 Tests for the Convergence
7/27
Tests for the Convergence of Infinite Series( )n n 12 n 1
1 1u ,u
n.2 n 1 2+ +
= =+
( ) ( )n 1
n
nn n n
n 1
n 1 2 n 1 .2ulim lim lim
u n.2 n
+
+
+ += =
=n
1lim 2 1 2 1
n
+ = >
By D'Alembert's Ratio test the given series converges.
Example 4: Test for convergence the series
( )
2 3 4x x x x
... x 01.3 2.4 3.5 4.6+ + + + > .
Solution: Here
( ) ( ) ( )
n n 1
n n 1
x xu ,u
n n 2 n 1 x 3
+
+= =+ + +
( ) ( )
( )
n
n
n 1n nn 1
x n 1 x 3ulim lim
u n n 2 .x +
+
+ +=
+
( ) ( )( )n
n 1 n 3 1lim .
n n 2 x
+ +=
+
2
n2
1 3n 1 1
1 1n nlim .
2 x xn 1
n
+ +
= =
+
By D'Alembert's Ratio test,n
u converges if1
1x
> i.e. x < 1 and
diverges if1
1x
< i.e. x > 1 and test fails for x = 1.
For x = 1,( )n 2
1 1u
2n n 2n 1
n
= =+
+
Institute of Lifelong Learning, University of Delhi pg. 7
-
8/12/2019 Tests for the Convergence
8/27
Tests for the Convergence of Infinite SeriesLet n 2
1v
n= which is convergent since p = 2 > 1.
2
n
n n2n
u nlim lim 1
2v
n 1 1n
= =
+
So by comparison testn
u and nv both converge and diverge
together.
Since n 21
vn
= converges consequently nu converges for x=1.
Hence the given series converges for x < 1 and diverges for x > 1.
Example 5: Show that the series2 3 4
1 2! 3! 4!...
5 5 5 5+ + + + is divergent.
Solution: We have
n n
n!u
5= and
( )n 1 n 1
n 1 !u
5+ +
+=
( )
n 1
n
nn nn 1
5u n!lim lim
u n 1 !5
+
+
=+
=n
5lim 0 1
n 1= .
Solution: Here
p
n
nu
n!= and
( )( )
p
n 1
n 1u
n 1 !+
+=
+
Institute of Lifelong Learning, University of Delhi pg. 8
-
8/12/2019 Tests for the Convergence
9/27
Tests for the Convergence of Infinite Series( )
( )( )
pp
n
pn n n
n 1
n n 1 !u nlim lim lim n 1
u n 1n! n 1 +
+ = = + + +
p
n
n 1lim
1nn
+=
+
Hence by Ratio test, the given series is convergent.
4. Cauchy's nth Root testIf
nu is a positive term series such that
( )1
nn
nlim u
=
Then (i)n
u converges if < 1 (ii) nu diverges if > 1 (iii) the
fails if = 1.
Proof Case I. < 1.
Let us select a positive number such that + < 1.
Let + = < 1
Since ( )1
nn
nlim u
= therefore there exists a positive integer m such
that
( )1
nnu , n m <
( )1
nnu , x m < < +
( ) ( )n n n
nu , n m < < + =
nnu , n m<
Institute of Lifelong Learning, University of Delhi pg. 9
-
8/12/2019 Tests for the Convergence
10/27
Tests for the Convergence of Infinite SeriesBut since n is a convergent geometric series with common ratio
< 1, therefore by comparison test the seriesn
u converges.
Case II. > 1
Let us select a positive number such that
- > 1
Let - = > 1
Since ( )1
nn
nlim u ,
= therefore there exists a positive integer m, such
that
( )1
nn 1
u , n m < < +
( ) ( )n n
n 1u , n m < < +
( )n n
n 1u , n m > =
But since n is a divergent geometric series with common ratio
> 1, therefore by comparison test, the seriesn
u diverges.
Value addition:NoteThe test fails to give any definite information for = 1.
Consider the two series1
n and 2
1
n . The series
1
n diverges,
while
1
n
2n
1lim
n
=
and the series2
1
n , converges, while
1
n
2n
1lim
n
=
.
Institute of Lifelong Learning, University of Delhi pg. 10
-
8/12/2019 Tests for the Convergence
11/27
Tests for the Convergence of Infinite Series
Example 7: Test for convergence the series( )
nn 2
1
logn
= .
Solution: Let( )
n n
1u
logn=
( )( )
1
n1
nn n
n n
1lim u lim
logn
=
=n
1lim 0 1
logn= <
Hence by Cauchy's nth root test the given series is convergent.
Example 8: Test for convergence the series whose general term is
32n
11
n
+
.
Solution: Let3
2n
n
1u
11
n
=
+
, then
( ) 32
1
n
1
nn
n n n
1lim u lim
1
1 n
=
+
nn
1 1lim 1
e11
n
= = 1 and divergent if p 1
and divergent if 0 < p < 1.
Solution: Let ( )( )
( )p1
u x , p 0x logx
= >
Clearly, for x > 2, u(x) is non negative, monotonically decreasing
and integrable function. Also u(n) = un, n N
Consider ( )( )
t t
p
2 2
1u x dx dx
x logx=
( )
( )( )
t1 p
2
t
2
1log x , if p 1
1 p
log log x , if p 1
=
=
or ( ) ( ) ( )
( ) ( )
1 p 1 pt
2
1log t log 2 ,if p 1
1 pu x dx
log log t log log 2 , if p 1
= =
Institute of Lifelong Learning, University of Delhi pg. 16
-
8/12/2019 Tests for the Convergence
17/27
Tests for the Convergence of Infinite Series
( ) ( )t
1 p
t2
, if p 1
1lim u x dx log 2 , if p 1
p 1
, if p 1
=
Thus ( )2
u x dx
is convergent if p > 1 and divergent if 0 < p < 1.
Hence, by Cauchy's integral test the given series is convergent if p
> 1 and divergent if 0 < p 2 and
diverges if p
Institute of Lifelong Learning, University of Delhi pg. 17
-
8/12/2019 Tests for the Convergence
18/27
-
8/12/2019 Tests for the Convergence
19/27
Tests for the Convergence of Infinite Series6.1. Leibnitz test: If the alternating series
( )n 1
n
n 1
1 2 3 4 nu u u u1 , u , nu 0
=
+ + > = is such that
(i) un+1
< un,
v nand
(ii)n
nlim u 0
=
then the series converges.
Proof: Let Sn= u1 u2+ u3- u4+ + (-1)nun
Now for all n,
2n 2 2n 2n 1 2n 2S S u u 0+ + + =
2n 2 2nS S+
The sequence is a monotonic increasing sequence.
Again 2n 1 2 3 2n 1 2nS u u u ... u u= + +
( ) ( ) ( )1 2 3 4 5 2n 2 2n 1 2nu u u u u ... u u u =
But since n 1 nu u+ for all n, therefore each bracket on the right is
positive and hence
2n 1S u , v n<
Thus the monotonic increasing sequence is bounded above
and is consequently convergent.
Let2n
nlimS S
=
We shall now show that the sequence also converges to
the same limit S.
Institute of Lifelong Learning, University of Delhi pg. 19
-
8/12/2019 Tests for the Convergence
20/27
Tests for the Convergence of Infinite SeriesNow 2n 1 2n 2n 1S S u+ += +
2n 1 2n 2n 1
n n nlimS limS lim u+ +
= +
But by condition (ii)
2n 1nlim u 0+
=
2n 1 2n
n nlimS limS S+
= =
Thus the sequences and both converge to the same
limits.
We shall now show that the sequence also converges to S
Let > 0 be given.
Since the sequences and both converge to S,
therefore there exists positive integers m1, m2such that
2n 1S S , n m < (1)
and 2n 1 2S S , n m+ < (2)
Thus from (1) and (2), we have
( )n 1 2S S , n max m ,m <
converges to S
The series ( )n 1
n1 u
converges.
Example 14: Show that the seriesp p p p
1 1 1 1...
1 2 3 4 + + converges for
p > 0.
Institute of Lifelong Learning, University of Delhi pg. 20
-
8/12/2019 Tests for the Convergence
21/27
Tests for the Convergence of Infinite SeriesSolution: Let n n 1p p 1
1 1u u
n n+ +
= =
Here n 1 nu u , v n+ <
and n pn n
1lim u lim 0
n = =
Hence by Leibnitg test the alternating series( )
n 1
p
1
n
converges.
6.2. Absolute convergence: A series unis said to be absolutely
convergent if the seriesn
u is convergent.
6.3. Conditional Convergence: A seriesn
u is said to be
conditionally convergent, if
(i) unis convergent and
(ii) unis not absolutely convergent.
Value Addition: Illustrations
1. The series n 2 31 1 1u 1 ...2 2 2
= + + is absolutely convergent,
since n 2 31 1 1
u 1 ...2 2 2
= + + + + , being a geometric series with
common ratio1
12
< , is convergent.
2. The series n 1 1 1u 1 ...2 3 4
= + + is not absolutely convergent,
since n1 1 1
u 1 ... ...2 3 n
= + + + + + is not convergent.
Theorem 2: Every absolutely convergent series is convergent.
Institute of Lifelong Learning, University of Delhi pg. 21
-
8/12/2019 Tests for the Convergence
22/27
Tests for the Convergence of Infinite SeriesProof: Let
nu be absolutely convergent, so that nu is
convergent.
Hence for any > 0, by Cauchy's General principle of convergence,
there exists a positive number m such that
n 1 n 2 n pu u ... u , n m+ + ++ + + < and p > 1
Also for n and p > 1,
n 1 n 2 n p n 1 n 2 n pu u ... u u u ... u , n m+ + + + + ++ + + + + + < and p >1.
Hence by Cauchy's General principle of convergence the series un
converges.
Value Addition: Remark
The divergence of |un| does not imply the divergence of un.
For example, if( )
n 1
n
1u
n
= , we have seen above that |un| is
divergent, whereas unis convergent.
Example 15: Test for convergence and absolute convergence
the series.
1 1 1 1 ...1 3 5 7
+ +
Solution: We have
n
1u
2n 1=
n 1
1u
2n 1+ =
+
Institute of Lifelong Learning, University of Delhi pg. 22
-
8/12/2019 Tests for the Convergence
23/27
Tests for the Convergence of Infinite Series n 1 nu u ; v n.+ <
Now,n
n n
1lim u lim 0
2n 1 = =
Hence, by leibnitg's test, the series unis convergent. Now we show
that unis not absolutely convergent.
We haven
1u
2n 1=
Letn
1v
n=
nn n
n
u n 1lim lim 0
v 2n 1 2 = =
Since vn is divergent, so |un| is divergent. Hence un is not
absolutely convergent.
Example 16: Show that for any fixed value of x the series
2n 1
sin nx
n
=
is convergent.
Solution: Let n 2sin nx
un
= so thatn 2
sin nxu
n=
Nown 2 2
sin nx 1u , v n
n n=
and2
1
n converges.
Hence by comparison test, the series2
sin nx
n converges.
2
sin nx
n is absolutely convergent.
Institute of Lifelong Learning, University of Delhi pg. 23
-
8/12/2019 Tests for the Convergence
24/27
Tests for the Convergence of Infinite SeriesSince every absolutely convergent series is convergent, therefore
2
sin nx
n is convergent.
Example 17: Show that the series
2 3x xx ...2! 3!+ + + converges
absolutely for all values of x.
Solution: Letn
n
xu
n!= and
( )
n 1
n 1
xu
n 1 !
+
+ = +
Now nn n
n 1
u n 1lim lim
u x +
+= , except, when x = 0. Hence by Ratio
test the series converges absolutely for all x except possibly zero.
But for x = 0, the series evidently converges absolutely. Hence the
series converges absolutely for all values of x.
Note: Since for a convergent series un, nnlim u 0
=
n
n
xlim 0n!
= , is a useful result.
Example 18: Test for convergence and absolute convergence the
series( )
( )
n 1
n 1
1 1 1 1...
log n 1 log 2 log 3 log 4
+
=
= +
+
Solution: We have
( )n
1u , n N
log n 1
=
+
Clearly( )nn n1
lim u lim 0log n 1
= =+
Since log x is an increasing function for all x > 0.
( ) ( ) ( )log n 2 log n 1 n 2 n 1+ > + + > +
Institute of Lifelong Learning, University of Delhi pg. 24
-
8/12/2019 Tests for the Convergence
25/27
Tests for the Convergence of Infinite Series
( ) ( )1 1
, v nlog n 2 log n 1
top related