chapter 2 series

2-1

MAA 102 Chapter 2: Series

2.1 SERIES

Given a sequence { }na , we define a new sequence { }nS by 1 2

1

n

n k nk

S a a a a=

= = + + +

where nS is called the nth partial sum.

1 1

2 1 2

3 1 2 3

1 2 3n n

S aS a aS a a a

S a a a a

== += + +

= + + + +#

This new sequence { }nS is called the sequence of partial sums.

EXAMPLE 2.1. Derive the partial sum sequence for the following series.

(a) 11

12nn

= (b)

1

3nn

=

SOLUTION. (a)

2-2

(b)

2-3

DEFINITION 2.1. An infinite series 1 21

nn

a a a

=+ + = , with the partial sum sequence

{ }nS , converges if lim nn S exists.

If lim nn S A = , then we say that 1 nn a

= converges to A, that is Aa

nn =

=1. If lim nn S does not

exist, then the series diverges.

EXAMPLE 2.2. Determine whether the following series converges or diverges.

(a) 11

12nn

= (b)

13n

n

=

SOLUTION.

(a) 11

12nn

=

(b) 13n

n

=

2-4

EXAMPLE 2.3. Prove that the infinite series 4 8 163 23 9 27

+ + converges and find its sum.

SOLUTION.

THEOREM 2.1. (Geometric Series)

The geometric series 1 2 11

n n

nar a ar ar ar

=

= + + + + + with 0a (i) converges and has the sum

1a

r if 1r < .

(ii) diverges if 1r .

2-5

EXAMPLE 2.4. Determine whether the series 0

1 37 2

n

n

=

converges or diverges.

SOLUTION.

The converse is false, that is, if lim 0nn a = , it does not necessarily follow that the series

1n

na

= is convergent. For example, 1lim 0n n = but 1

1n n

= is divergent.

COROLLARY 2.1. (Test for Divergence)

If lim 0nn a , then the infinite series 1 nn a

= is divergent.

THEOREM 2.2. If an infinite series 1

nn

a

= is convergent, then lim 0nn a = .

2-6

EXAMPLE 2.5. Determine whether the following series converges or diverges.

(a) 1 2 3 ...3 5 7 2 1

nn

+ + + + ++ (b) ( )1 1n

n

=

SOLUTION. (a)

(b) ( )1

1 nn

= Three ways to check!

(i) Theorem 2.2

(ii) Theorem 2.1

(iii) Definition 2.1

2-7

Note that if 0 , then 1

nn

a

= and

1n

na

= will converge/diverge together.

EXAMPLE 2.6. Determine the convergence or divergence of ( )1

2 2 nnn

a =

+ if

12.75n

na

== .

SOLUTION.

THEOREM 2.3. If the convergent series 1

nn

a A

== and

1n

nb B

== , then

( )1

n nn

a b A B =

+ = + where , .

2-8

The converse of Theorem 2.3 is false, that is, if ( )1

n nn

a b =

+ is convergent, it does not necessarily follow that

1n

na

= and

1n

nb

= is convergent.

EXAMPLE 2.7. Determine the convergence or divergence of 1

1 11n n n

=

+ .

SOLUTION.

2-9

2.2 POSITIVE-TERM SERIES It is not easy to find the sum of an infinite series. We can, however, develop techniques for

using the nth term na to test a series for convergence or divergence. When applying these

tests, we shall be concerned not with the sum of the series but whether the series converges or

diverges. In this section, we shall consider only positive term series, that is, infinite series

for which every term is positive.

DEFINITION 2.2. ( ) ( )1 1

lim b

bf x dx f x dx

=

TEST 2.1. (Integral Test)

If the function ( )f x is continuous, positive valued ( )( )0f x > and decreasing for 1x , then the infinite series ( ) ( ) ( ) ( )1 2 3f f f f n+ + + + +

(i) converges if ( )1

f x dx converges.

(ii) diverges if ( )1

f x dx diverges.

2-10

EXAMPLE 2.8. Use the Integral Test to determine whether the series ( )2

231 2kk

k

= + converges or diverges.

SOLUTION.

2-11

EXAMPLE 2.9. Use the Integral Test to prove that the harmonic series 1

1n n

= diverges.

SOLUTION.

EXAMPLE 2.10. Determine whether the following series converge or diverge.

(a) 2 2 21 1 112 3 n

+ + + + + (b) 5 5 552 3 n

+ + + + +

SOLUTION. (a)

(b)

THEOREM 2.4. The p-series 1

1p

n n

=

(i) converges if 1p > . (ii) diverges if 1p .

2-12


(a) 1

14 3nn

= + (b) 212

n

nn

=

+

SOLUTION. (a)

(b)

TEST 2.2. (Comparison Test)

Suppose 0 n na b for every positive integer n.

(i) If 1

nn

b

= converges, then

1n

na

= converges.

(ii) If 1

nn

a

= diverges, then

1n

nb

= diverges.

2-13

LIMIT COMPARISON TEST

Suppose 1

nn

a

= and

1n

nb

= are positive term series with lim nn

n

a Ab

= .

(i) If 0 A< < , then 1

nn

a

= and

1n

nb

= both converge or both diverge.

(ii) If 0A = and 1

nn

b

= converges, then

1n

na

= converges.

(iii) If A = and 1

nn

b

= diverges, then

1n

na

= diverges.

USING THE COMPARISON TEST STEP 1: Guess at whether the series na converges or diverges. STEP 2: Find a series that proves the guess to be correct. That is, if we guess that na diverges, we must find a divergent series whose

terms are smaller than the corresponding terms of na . If we guess that na converges, we must find a convergent series whose terms

are larger than the corresponding term of na . Some informal principles can be used to find this divergent or convergent series.

1. Constant terms in the denominator of na can usually be deleted without affecting the convergence or divergence of the series.

2. If a polynomial in n appears as a factor in the numerator or denominator of na , all but the leading term in the polynomial can usually be discarded without affecting the convergence or divergence of the series.

Unfortunately, it is not always so straightforward to find the series required for comparison, so the Limit Comparison Test is introduced as an alternative to the comparison test.

2-14


(a) 3 2

1

12n n

= + (b) 2121n

nn

=

++

SOLUTION.

(a)

(b)

2-15

2.3 ALTERNATING SERIES

An infinite series whose terms are alternately positive and negative is called an alternating

series. It is customary to express an alternating series in one of the forms

( ) 11 2 3 4 1 n na a a a a + + + + or

( )1 2 3 4 1 n na a a a a + + + +

If an alternating series satisfies the hypotheses of the alternating series test, and if S is the

sum of the series, then:

(a) S lies between any two successive partial sums; that is, either

1n nS S S + or 1n nS S S+ depending on which partial sum is larger.

(b) If S is approximated by Sn, then the absolute error nS S satisfies 1n nS S a + . Moreover, the sign of the error nS S is the same as that of the coefficient of 1na + .

TEST 2.3. (Alternating Series Test)

Suppose an alternating series ( ) 11

1 n nn

a +

= fulfils the following conditions:

(i) 0na > , n (ii) 1n na a+ < , n (iii) lim 0nn a = .

Then the series is convergent.

2-16

EXAMPLE 2.13. Determine whether the following alternating series converge or diverge.

(a) ( ) 11

11 nn n

+

= (b) ( ) 1 2

1

214 3

n

n

nn

=

SOLUTION.

(a)

(b)

2-17

DEFINITION 2.3. An infinite series 1

nn

a

= is absolutely convergent if the series

1 21

n nn

a a a a

== + + + +

obtained by taking the absolute value of each term is convergent. The series is said to be

absolutely divergent if the series of absolute values diverges.

Note that if 1

nn

a

= is a positive term series, then n na a= , and in this case absolute

convergence is the same as convergence.

EXAMPLE 2.14. Prove that the alternating series ( ) 121

1 n

n n

+

=

is absolutely convergent.

SOLUTION.

The converse of Theorem 2.5 is false, that is, if 1

nn

a

= is convergent, it does not necessarily

follow that 1

nn

a

= is convergent.

THEOREM 2.5. If an infinite series 1

nn

a

= is absolutely convergent, then

1n

na

= is

convergent.

2-18

DEFINITION 2.4. An infinite series 1

nn

a

= is conditionally convergent if

1n

na

= is

convergent and 1

nn

a

= is divergent.

EXAMPLE 2.15. Show that the alternating series ( )1

1 n

n n

=

is conditionally convergent.

SOLUTION.

2-19

2.4 RATIO TEST

TEST 2.4. (Ratio Test)

Let 1

nn

a

= be an infinite series of nonzero terms. If 1lim nn

n

a La+

= , then 1 nn a

=

(i) is absolutely convergent if 1L < . (ii) is divergent if 1L > or L = . (iii) may be absolutely convergent, conditionally convergent or divergent if 1L = .

2-20

EXAMPLE 2.16. Determine whether the following series are absolutely convergent,

conditionally convergent or divergent.

(a) 1

1!n n

= (b) ( )

1

!1 n nn

ne

=

(c) 1

1n n

=

SOLUTION. (a)

(b)

(c)

2-21

2.5 ROOT TEST

EXAMPLE 2.17. Determine the convergence or divergence of the following series.

(a) 11 2nn

n

= (b)

1

23

nn

nn

=

SOLUTION.

(a)

(b)

TEST 2.5. (Root Test)

Let 1

nn

a

= be an infinite series. If Lan nn =lim , then 1 nn a

=

(i) is absolutely convergent if 1L < . (ii) is divergent if 1L > or L = . (iii) may be absolutely convergent, conditionally convergent or divergent if 1L = .

2-22

SUMMARY

To determine the convergence of an infinite series1

nn

a

= :

STEP 1 : Find lim nn a .

(i) If lim 0nn a , then 1 nn a

= is divergent (Collorary 2.1).

(ii) If lim 0nn a = , then no conclusion can be made. Go to Step 2.

STEP 2 : Determine whether 1

nn

a

= is positive term series or alternating series.

(i) If positive term series, use

integral test comparison test ratio test root test

(ii) If alternating series, use

alternating series test ratio test root test

Ratio Test and Root Test can be used for any series, namely positive term series and

alternating series.

Ratio Test : Series with factorials (!) and (expression)n.

Root Test : Series with (expression)n

2-23

STRATEGY FOR SERIES

STEP 1: With a quick glance does it look like the series terms dont converge to zero in the

limit, i.e. does lim 0nn a ? If so, use the Divergence Test. Note that you should only do the divergence test if a quick glance suggests that the series terms may

not converge to zero in the limit.

STEP 2: Is the series a p-series ( 1pn ) or a geometric series ( 1 nn ar

= or 1

1

n

nar

= )? If so

use the fact that p-series will only converge if 1p > and a geometric series will only converge if 1r < . Remember as well that often some algebraic manipulation is required to get a geometric series into the correct form.

STEP 3: Is the series similar to a p-series or a geometric series? If so, try the Comparison

Test.

STEP 4: Is the series a rational expression involving only polynomials or polynomials under

radicals (i.e. a fraction involving only polynomials or polynomials under radicals)?

If so, try the Comparison Test and/or the Limit Comparison Test. Remember

however, that in order to use the Comparison Test and the Limit Comparison Test

the series terms all need to be positive.

STEP 5: Does the series contain factorials or constants raised to powers involving n? If so,

then the Ratio Test may work. Note that if the series term contains a factorial then

the only test that weve got that will work is the Ratio Test.

STEP 6: Can the series terms be written in the form ( )1 nn na b= or ( ) 11 nn na b+= ? If so, then the Alternating Series Test may work.

STEP 7: Can the series terms be written in the form ( )nn na b= ? If so, then the Root Test may work.

STEP 8: If ( )na f n= for some positive, decreasing function and ( )a f x dx is easy to evaluate then the Integral Test may work.

2-24

EXTRA NOTES

Divergence Test

If lim 0nn a then na will diverge

Integral Test

Suppose that ( )f x is a positive, decreasing function on the interval [ ),k and that ( ) nf n a= then,

1. If ( ) k

f x dx is convergent so is n

n ka

= .

2. If ( ) k

f x dx is divergent so is n

n ka

= .

Comparison Test

Suppose that we have two series na and nb with , 0n na b for all n and n na b for all n. Then,

1. If nb is convergent then so is na . 2. If na is divergent then so is nb .

Limit Comparison Test

Suppose that we have two series na and nb with , 0n na b for all n. Define, lim n

nn

acb

= .

If c is positive (i.e. 0c > ) and is finite (i.e. c < ) then either both series converge or both series diverge.

2-25

Alternating Series Test

Suppose that we have a series na and either ( )1 nn na b= or ( ) 11 nn na b+= where 0nb for all n. Then if,

1. lim 0nn b = and, 2. { }nb is eventually a decreasing sequence,

Then the series is convergent.

Ratio Test

Suppose we have the series na . Define, 1lim n

nn

aLa+

=

Then,

1. if 1L < the series is absolutely convergent (and hence convergent). 2. if 1L > the series is divergent. 3. if 1L = the series may be divergent, conditionally convergent, or absolutely

convergent.

Root Test

Suppose that we have the series na . Define,

1

lim limn nn nn nL a a = = Then,

1. if 1L < the series is absolutely convergent (and hence convergent). 2. if 1L > the series is divergent. 3. if 1L = the series may be divergent, conditionally convergent, or absolutely

convergent.

chapter 2 series

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