Student Note - Series

Series

Objectives: At the end of the module students are able to

⋄ Identify the difference between sequence and series

⋄ Identify the properties of some special series such as geometric series, harmonicseries and p–series

⋄ Determine whether a series converges or diverges using test of convergence

⋄ Evaluate the radius and interval of convergence of a series

§ 1 Infinite Series

Infinite series is formed by adding terms of an infinite sequence:

Sequence: {an}∞n=1 = a1 , a2 , a3 , · · · , an , · · ·

Series:∑∞

n=1 an = a1 + a2 + a3 + · · ·+ an + · · ·

Definition 1.1:

The kth partial sum Sk of the series is the sum of first k terms:

Sk =

k∑

n=1

an = a1 + a2 + · · ·+ ak

For example, the partial sum of first few terms are

S1 = a1

S2 = a1 + a2

S3 = a1 + a2 + a3

S100 = sum of first 100 terms of an

Example:

(i) Sequence:

{1

2n

}

=1

2,1

4,1

8, · · ·

Series:

∞∑

n=1

1

2n=

1

2+

1

4+

1

8+ · · ·

1

(ii) Sequence:

{(−1)k+1

k + 1

}

= −1 ,1

2,−1

3,1

4, · · ·

Series:

∞∑

k=0

(−1)k+1

k + 1= −1 +

1

2+(− 1

3

)

+1

4+ · · ·

Definition 1.2:

Given a sequence∑∞

k=1 ak = {a1 , a2 , a3 , · · · }, the nth partial sum Sn is the sum offirst n terms of the sequence, i.e., Sn = a1 + a2 + a3 + · · ·+ an =

∑nk=1 ak. The partial

sum of first n terms are

S1 = a1

S2 = a1 + a2

S3 = a1 + a2 + a3...

Sn = a1 + a2 + a3 + · · ·+ an

The column on the left (in red) forms a sequence

{Sk}nk=1 = S1 , S2 , S3 , · · · , Sn

If the sequence of the partial sums {Sn} converges to L, then the sum of the seriesconverges to L, i.e.,

limn→∞

Sn = L

S∞ =∞∑

n=1

an = L

where L is a finite number. The limit L is the sum of the infinite series∑

an and wemay write it as

L =

∞∑

n=1

an = a1 + a2 + a3 + a4 + · · ·

If the sequence of partial sum {Sn} is not convergent, then it is said to be divergent.

Example 1.1:Find the limit of the infinite series

∞∑

n=1

1

2n

2

Solution:We write down the first four partial sum of the series

S1 =1

2

S2 =1

2+

1

4=

3

4

S3 =1

2+

1

4+

1

8=

7

8

S4 =1

2+

1

4+

1

8+

1

16=

15

16

and from the pattern of the values we conclude that the partial sum is

Sn =2n − 1

2n= 1− 1

2n

So that the limit

limn→∞

Sn = limn→∞

(

1− 1

2n

)

= 1

The infinite sum of the series

S∞ =∞∑

n=1

1

2n= 1

converges to 1. �

Example 1.2:Find the limit of the infinite series

∞∑

n=1

1

n(n+ 1)

Solution:We write down the first four partial sum of the series

S1 =1

2

S2 =1

2+

1

6=

2

3

S3 =1

2+

1

6+

1

12=

3

4

S4 =1

2+

1

6+

1

12+

1

20=

4

5


Sn =n

n+ 1

3

The infinite sum of the series

S∞ = limn→∞

Sn = limn→∞

n

n + 1= lim

n→∞

1

1 + 1n

= 1

converges to 1. �

Example 1.3:Determine whether the series with the sequence

1 ,−1 , 1 ,−1 , 1 ,−1 , · · ·

is convergent/divergent.

Solution:

S1 = 1 =1

2+

1

2

S2 = 1 + (−1) = 0 =1

2− 1

2

S3 = 1 + (−1) + 1 = 1 =1

2+

1

2

S4 = 1 + (−1) + 1 + (−1) = 0 =1

2− 1

2


Sn =1

2+

1

2(−1)n+1

We may write

Sn =

{

1 if n is odd

0 if n is even

Since {Sn} oscillates between 1 and 0, it follows that limn→∞ Sn does not exist, hencethe series diverges. �

§ 2 Geometric Series

Geometric Series is a series of numbers having a common ratio r such that

Sn =n∑

k=1

ark−1 = a+ ar + ar2 + ar3 + · · ·+ arn−2 + arn−1

4

where a is the first term and r is the ratio. We note that the nth term is an = arn−1.As we multiply the series with r, we have

r · Sn =

n∑

k=1

ark = ar + ar2 + ar3 + ar4 + · · ·+ arn−1 + arn

Then we subtract the first series with the second series and find that

r · Sn − Sn = arn − a

(r − 1)Sn = a(rn − 1)

∴ Sn =a(rn − 1)

r − 1

Then, the infinite sum of the geometric series is

limn→∞

Sn = limn→∞

(a(rn − 1)

r − 1

)

= limn→∞

( arn

r − 1− a

r − 1

)

If the ratio |r| < 1, the first term on the right hand side vanishes (please refer to Theorem3 in Sequence),

∴ S∞ =a

1− r

So that the geometric series converges if |r| < 1 and diverges if |r| > 1.

Example 2.1:Determine whether the series converges/diverges,

5− 10

3+

20

9− 40

27+ · · ·

Solution:The series is also equivalent to

5 + (5)(

− 2

3

)

+ (5)(

− 2

3

)2

+ (5)(

− 2

3

)3

+ · · ·

Therefore a = 5 and r = −23. Since |r| = 2

3< 1, then the series is convergent and the

sum of the infinite series is

S =a

1− r=

5

1− (−23)= 3

�

Example 2.2:Determine whether the series conveges/diverges,

1 + 0.4 + 0.16 + 0.064 + · · ·

5

Solution:The series can be rewritten as

1 + 0.4 + (0.4)2 + (0.4)3 + · · ·

Therefore, a = 1 and r = 0.4. Since |r| < 1, the series is convergent and the sum of theinfinite series is

S =1

1− 0.4=

5

3= 1.66

�

Example 2.3:Write the number

2.317 = 2.31717171717...

as a fraction (ratio of integer).

Solution:The number can be rewritten as

2.31717171717...

= 2.3 + 0.017 + 0.00017 + 0.0000017 + · · ·= 2.3 +

17

1000+

17

100000+

17

100000+ · · ·

= 2.3 +17

(10)3+

17

(10)5+

17

(10)7+ · · ·

= 2.3 +17

(10)3+

17

(10)3· 1

(10)2+

17

(10)3· 1

(10)4+ · · ·

↑ ↑ ↑ ↑ ↑a + a × r + a × r2 + · · ·

From the second term onward are of the form of Geometric Series with a =17

(10)3and

r =1

(10)2. Thus,

∴ 2.317 = 2.3 +a

1− r

= 2.3 +

17(10)3

1− 1(10)2

=23

10+

17

990=

1147

495

�

Example 2.4:Is the series

∞∑

n=1

22n31−n

6

convergent/divergent?

Solution:

∞∑

n=1

22n31−n =∞∑

n=1

(22)n · 3 · 3−n

=

∞∑

n=1

4n · 3 · 1

3n

=

∞∑

n=1

3

(4n

3n

)

= 3

∞∑

n=1

(4

3

)n

The infinite sum is a Geometric Series with a = 43and r = 4

3. Since |r| > 1, the series

diverges. �

Exercise:

1.) Show that the series

2 +2

3+

2

32+ · · ·+ 2

3n+ · · ·

converges and find its sum.

2.) Determine whether the series converges, if so, find the infinite sum.

i)1

8− 1

4+

1

2− 1 + · · ·

ii)

∞∑

n=1

(−3)n−1

4n

§ 3 Telescoping Series

Telescoping Series is a series whose sum appears to have the situation that almost everyterm cancels with either a preceding or succeeding term. For example, the series

S∞ =

∞∑

n=1

1

n(n+ 1)=

∞∑

n=1

(1

n− 1

n+ 1

)

7

we consider the partial sums of this series up to kth term

Sk =

k∑

n=1

(1

n− 1

n + 1

)

=

(

1− 1

2

)

+

(1

2− 1

3

)

+

(1

3− 1

4

)

+

(1

4− · · ·

)

+ · · ·+(

· · · − 1

k − 1

)

+

(1

k − 1− 1

k

)

+

(1

k− 1

k + 1

)

We can see that the terms with the same colour are canceled with each other. Thus, theterms which survive are the first and the last number:

Sk = 1− 1

k + 1

Therefore, the sum of the infinite series is

S∞ = limk→∞

Sk = limk→∞

(

1− 1

k + 1

)

= 1

∴ The series converges to 1.

§ 4 Harmonic Series

The harmonic series, given by

∞∑

n=1

1

n= 1 +

1

2+

1

3+

1

4+ · · ·

is divergent infinite series in spite of the fact that the limit of the sequence when n goesto infinity is zero. The proof of the divergence of the harmonic series is shown below bycomparing the harmonic series with another divergent series:

1 +1

2+

1

3+

1

4+

1

5+

1

6+

1

7+

1

8+ · · ·

> 1 +1

2+

1

4+

1

4︸︷︷︸

+1

8+

1

8+

1

8+

1

8︸︷︷︸

+ · · ·

=1

2=

1

2

where each colour term of the harmonic series is greater than the corresponding termin the second series with the same colour, and therefore the sum of the harmonic series

8

must be greater than the sum of the second series. However, the sum of the secondseries is infinite,

1 +

(1

2

)

+

(1

4+

1

4

)

+

(1

8+

1

8+

1

8+

1

8

)

+ · · ·

= 1 +

(1

2

)

+

(1

2

)

+

(1

2

)

+ · · · = ∞

It follows that the infinite sum of the harmonic series must be divergent.

Example 4.1:Show that each of the following series are divergent.

(i)∞∑

n=1

3

n(ii)

∞∑

n=3

1

n

Solution:

(i) By the fact that constant can be factored out of the series as follows,

∞∑

n=1

3

n= 3

∞∑

n=1

1

n

Since∑∞

n=11nis divergent, any constant multiplies a divergent series it will still be

divergent and so the series has to be divergent.

(ii) The first two terms of the harmonic series is

∞∑

n=1

1

n= 1 +

1

2+

∞∑

n=3

1

n

=3

2+

∞∑

n=3

1

n

=⇒∞∑

n=3

1

n=

∞∑

n=1

1

n− 3

2

As we subtract a constant from a divergent series, the series will still be divergent.In order word, an infinity minus a constant which is still infinity. So that the seriesis divergent.

�

9

§ 5 Determining Convergence/Divergence Of Series

5.1 n–th Term Test

⋄ If limn→∞

an 6= 0 then the series

∞∑

n=1

an is divergent.

⋄ If limn→∞

an = 0, further investigation is necessary to determine whether the series∞∑

n=1

an is convergent or divergent. In other word, it is not guarantee that the series

converges.

Example 5.1:Determine whether the series

∞∑

n=1

n2

5n2 + 4

is convergent/divergent.

Solution:By using n–th term test,

limn→∞

n2

5n2 + 4= lim

n→∞

1

5 + 4n2

=1

56= 0

thus the series diverges. �

Example 5.2:Determine the following series, by using n–th term test, whether they are conver-gent/divergent?

1. Given the series:∞∑

n=1

n

2n+ 1

By using n–th term test:

limn→∞

n

2n+ 1=

1

26= 0

Conclusion: Diverges


n=1

1

n2

10


limn→∞

1

n2= 0

Conclusion: Further investigation is needed.


n=1

1√n


limn→∞

1√n= 0

Conclusion: Further investigation is needed.


n=1

en

n


limn→∞

en

n= lim

n→∞

en

1= ∞

Conclusion: Diverges �

Remark:If limn→∞ an = 0, then the series may either converges or diverges. We need to use othertest techniques to confirm.

Theorem 1:

Given a series∞∑

n=1

an = a1 + a2 + a3 + · · ·

For any positive integer k, if the sum of the same series is taken from (k + 1)–th termonwards

∞∑

n=k+1

an = ak+1 + ak+2 + ak+3 + · · ·

then both series are either convergent or divergent.

Example 5.3:Show whether the following series

∞∑

n=1

1

(n+ 2)(n+ 3)=

1

3 · 4 +1

4 · 5 +1

5 · 6 + · · ·

11

convergent/divergent.

Solution:Recall the series we have considered earlier (Telescoping series):

∞∑

n=1

1

n(n + 1)=

1

1 · 2 +1

2 · 3 +1

3 · 4 +1

4 · 5 +1

5 · 6 + · · ·

=1

1 · 2 +1

2 · 3 +

∞∑

n=1

1

(n+ 2)(n+ 3)

The series in the question is the sum of this series started from third term onwards, thismeans that

∞∑

n=1

1

(n+ 2)(n+ 3)=

∞∑

n=3

1

n(n+ 1)

As we found before the series∑∞

n=11

n(n+1)converges to 1, and thus by the above Theorem,

the series in the question is convergent and converges to

S∞ =∞∑

n=1

1

(n+ 2)(n+ 3)= 1− 1

1 · 2 − 1

2 · 3 =1

3

�

Theorem 2:

If∑

an and∑

bn are convergent series with sums A and B, respectively, then

i)∑

(an + bn) converges with sum A+B

ii)∑

(c× an) converges with sum c× A, for every c 6= 0

iii)∑

(an − bn) converges with sum A− B

Example 5.4:Find the sum of series

∞∑

n=1

(3

n(n+ 1)+

1

2n

)

Solution:By using Theorem 2,

∞∑

n=1

(3

n(n + 1)+

1

2n

)

= 3

∞∑

n=1

1

n(n+ 1)+

∞∑

n=1

1

2n

From the previous results where the first series is a telescoping series and the secondseries has been considered in Example 1.1, the sum of the series given in the question is

12

convergent and converges to 3× 1 + 1 = 4 �

Example 5.5:Find the sum of series

∞∑

n=1

(1

5n+

1

n

)

Solution:By using the theorem,

∞∑

n=1

(1

5n+

1

n

)

=

∞∑

n=1

1

5n+

∞∑

n=1

1

n

Since the first series is a geometric series with r = 15< 1, this series is convergent. The

second series is a harmonic series, it is divergent. The sum of these two series is thusdivergent. �

5.2 p–Series

The p–series is a series of the following form∞∑

n=1

1

np=

1

1p+

1

2p+

1

3p+

1

4p+ · · ·

Theorem 3:

Given a p–series,∞∑

n=1

1

np

⋄ If p > 1, the series is convergent.

⋄ If p ≤ 1, the series is divergent.

e.g.,The series

∞∑

n=1

1

n4=

1

14+

1

24+

1

34+

1

44+ · · ·

is convergent because it is a p–series with p = 4 > 1.

The series∞∑

n=1

1

n1/2=

1

11/2+

1

21/2+

1

31/2+

1

41/2+ · · ·

is divergent because it is a p–series with p = 1/2 < 1.

13

Exercise:Determine whether the following series is convergent or divergent:

i) 1 +1

8+

1

27+

1

64+

1

125+ · · ·

ii)

∞∑

n=1

n−1.4 + 3n−1.2

iii)∞∑

n=1

(n + 1)2

n(n+ 2)

5.3 Comparison Test

Assume 0 ≤ an ≤ bn for all n.

⋄ Suppose that the sum of series∑

bn is convergent, then the series∑

an is alsoconvergent. In other word, a series of positive terms is convergent if its termsare less than the corresponding terms of a positive series which is known to beconvergent.

⋄ Suppose that the sum of series∑

an is divergent, then the series∑

bn is alsodivergent. In other word, a series is divergent if its terms are greater than thecorresponding terms of a series which is known to be divergent.

Example 5.6:Determine whether the series

∞∑

n=1

5

2n2 + 4n+ 3

converges/diverges.

Solution:Since n ≥ 1, thus 2n2 + 4n + 3 > 2n2, then

5

2n2 + 4n+ 3<

5

2n2

The series∞∑

n=1

5

2n2=

5

2

∞∑

n=1

1

n2

is a p–series with p = 2 > 1, by Theorem 3 the series is convergent. Therefore, byComparison Test the series

∞∑

n=1

5

2n2 + 4n+ 3is convergent .

14

�

Example 5.7:Determine whether

∞∑

n=3

lnn

n

converges/diverges.

Solution:Since ln(e) = ln(2.7183) = 1, therefore, lnn > 1 for n ≥ 3. Thus,

lnn

n>

1

nfor n ≥ 3 .

We know that the series∑∞

n=11nis a harmonic series and it is divergent. By Theorem 1

the series∑∞

n=31nis also divergent, and by Comparison Test,

∞∑

n=3

lnn

ndiverges .

�

Exercise:Determine, by using Comparison Test, whether the series converges or diverges:

i)∞∑

n=1

2

n4 + 2ii)

∞∑

n=1

4 + 3n

2n

5.4 Limit Comparison Test

Given the positive terms series∞∑

n=1

an and∞∑

n=1

bn

We require that all an and bn are positive. Suppose

limn→∞

anbn

= c

where c > 0 is a finite number.

⋄ If the limit of anbn

is positive,

limn→∞

anbn

= c > 0 ,

then the sums∑∞

n=1 an and∑∞

n=1 bn are either both convergent or both divergent.In other word, if the limit is positive, the terms of these two series are growing atthe same rate, so either both series converge or diverge together.

15


is zero, and the sum∑

bn converges, then the sum∑

an converges.In other word, if the limit is zero, the bottom terms are growing more quickly thanthe upper terms. So that if the bottom series converges, the upper series which isgrowing more slowly must also converge.


is infinite, and the sum∑

bn diverges, then the sum∑

an diverges.In other word, if the limit is infinite, the bottom series is growing more slowly. Ifthe bottom series diverges, the upper series must also diverge.

Remark:The Limit Comparison Test only applies to series with non-negative terms, i.e., an , bn >0 for all n.

Example 5.8:By using Limit Comparison Test, determine whether the series

∞∑

n=1

1

2n − 1

converges/diverges.

Solution:

Let

an =1

2n − 1

We choose

bn =1

2n

Then,

limn→∞

anbn

=1

2n−112n

= limn→∞

1

1− 12n

= 1 > 0

Since the series

∞∑

n=1

1

2nis a convergent geometric series (please see Example 1.1), by

Limit Comparison Test, the series∞∑

n=1

1

2n − 1converges. �

Example 5.9:By using Limit Comparison Test, determine whether the series

∞∑

k=1

2k2 + 3k√5 + k5

16

converges/diverges.

Solution:The highest power of the numerator is 2k2 and the highest power of the denominatorproportional to k5/2. We let

ak =2k2 + 3k√5 + k5

and choose

bk =2k2

√k5

=2

k1/2

Then the limit

limk→∞

anbn

= limk→∞

2k2+3k√5+k5

2k1/2

= limk→∞

2k2 + 3k√5 + k5

× k1/2

2

= limk→∞

k2(2 + 3k)

k5/2

√5k5

+ 1× k1/2

2

= limk→∞

2 + 3k

2√

5k5

+ 1= 1 > 0

Since∑∞

n=1 bk is p–series with p = 12, it is divergent. By Limit Comparison Test,

∞∑

k=1

2k2 + 3k√5 + k5

diverges. �

Exercise:By Limit Comparison Test, determine whether the following series converges or diverges.

i)

∞∑

k=1

4k2 − 2k + 6

8k7 + k − 8ii)

∞∑

n=1

5

3n + 1

17

5.5 The Ratio Test

Given a series∞∑

n=1

an.

i) If limn→∞

∣∣∣∣

an+1

an

∣∣∣∣= L < 1, then the series

∞∑

n=1

an is convergent.

ii) If limn→∞

∣∣∣∣

an+1

an

∣∣∣∣= L > 1, then the series

∞∑

n=1

an is divergent.

iii) If limn→∞

∣∣∣∣

an+1

an

∣∣∣∣= L = 1, then the series

∞∑

n=1

an may be convergent or convergent.

Example 5.10:Use the ratio test, determine whether the given series is convergent/divergent:

i)

∞∑

n=1

1

n!ii)

∞∑

k=1

(2k)!

4kiii)

∞∑

k=1

1

2k − 1

Solution:

i) Let an = 1n!. By ratio test,

limn=∞

∣∣∣∣

an+1

an

∣∣∣∣

= limn=∞

∣∣∣∣∣

1(n+1)!

1n!

∣∣∣∣∣

= limn=∞

∣∣∣∣

n!

(n+ 1)!

∣∣∣∣

= limn=∞

∣∣∣∣

1

n+ 1

∣∣∣∣= 0 < 1

Therefore, the series is convergent.

ii) Let ak = (2k)!4k

. By ratio test,

limk=∞

∣∣∣∣

ak+1

ak

∣∣∣∣

= limk=∞

∣∣∣∣∣∣

(2(k+1)

)!

4k+1

(2k)!4k

∣∣∣∣∣∣

= limk=∞

∣∣∣∣

(2k + 2)!

(2k)!× 1

4

∣∣∣∣

=1

4limk=∞

∣∣∣∣

(2k)!(2k + 1)(2k + 2)

(2k)!

∣∣∣∣

=1

4limk=∞

∣∣(2k + 1)(2k + 2)

∣∣ = +∞

18

Therefore, the series is divergent.

iii) Let ak = 12k−1

. By ratio test,

limk=∞

∣∣∣∣

ak+1

ak

∣∣∣∣

= limk=∞

∣∣∣∣∣

12(k+1)−1

12k−1

∣∣∣∣∣

= limk=∞

∣∣∣∣

2k − 1

2k + 1

∣∣∣∣

= limk=∞

∣∣∣∣

2− 1k

2 + 1k

∣∣∣∣= 1

Therefore, the series may be converges or diverges. �

Exercise:Use ratio test to determine whether the following series are convergent/divergent.

i)

∞∑

k=1

3k

k!ii)

∞∑

n=1

1

5n

5.6 The Root Test

Consider the sum∞∑

n=1

an.

1. If limn→∞

|an|1

n = L < 1, then the series

∞∑

n=1

an converges absolutely.

2. If limn→∞

|an|1

n = L > 1, then the series

∞∑

n=1

an diverges.

3. If limn→∞

|an|1

n = 1, then we may conclude nothing from this (inconclusive).

Example 5.11:Use the Root Test to determine whether the following series

i)

∞∑

k=1

(4k − 5

2k + 1

)k

ii)∞∑

n=1

1(ln(n+ 1)

)n

converge or diverge.

19

Solution:

i) By using Root Test,

limk→∞

((4k − 5

2k + 1

)k) 1

k

= limk→∞

4k − 5

2k + 1

= limk→∞

4− 5k

2 + 1k

= 2 > 1

Therefore, the series diverges.

ii) By using Root Test,

limn→∞

(1

(ln(n+ 1)

)n

) 1

n

= limn→∞

((1

ln(n + 1)

)n) 1

n

= limn→∞

1

ln(n+ 1)= 0 < 1

Therefore, the series converges.

�

Exercise:Determine, by using Root Test, whether the following series

1)

∞∑

n=1

(3n+ 1

2n− 1

)n

2)

∞∑

k=1

(

1− e−k)k

3)

∞∑

k=1

nn

2× (16n)

4)

∞∑

k=1

n1

n

converges/diverges.

5.7 The Alternating Series Test (AST)

An alternating series is a series whose terms are alternately positive and negative, e.g.,

∞∑

n=1

(−1)n−1

n= 1− 1

2+

1

3− 1

4+

1

5− 1

6+ · · ·

20

The alternating series test is used when the terms of the underlying sequence alternate.Suppose that we have a series

∞∑

n=1

(−1)n−1an = a1 − a2 + a3 − a4 + · · ·

where an ≥ 0 for all n ∈ N. Then if

i) {an} is a decreasing sequence, i.e., a1 ≥ a2 ≥ a3 ≥ · · · , and

ii) limn→∞

an = 0,

then the series is convergent.

Example 5.12:Use AST determine whether the following series

1)

∞∑

n=1

(−1)n−1

n2)

∞∑

k=1

(−1)k+1(k + 3)

k(k + 1)

converges/diverges.

Solution:

1) The series is given by

∞∑

n=1

(−1)n−1

n= 1− 1

2+

1

3− 1

4+

1

5− 1

6+ · · ·

i) we observe that an > an+1 because 1n> 1

n+1, and

ii) limn→∞ an = limn→∞1n= 0.

Thus, both conditions of AST are satisfied and it may conclude that the series isconvergent.

2) i) To show ak > ak+1,

ak − ak+1 =(k + 3)

k(k + 1)− k + 4

(k + 1)(k + 2)

=(k + 2)(k + 3)− k(k + 4)

k(k + 1)(k + 2)

=k2 + 5k + 6− k2 − 4k

k(k + 1)(k + 2)

=k + 6

k(k + 1)(k + 2)> 0

21

for all k ≥ 1. We found that ak − ak+1 > 0 ⇒ ak > ak+1 which satisfies the firstcondition of AST.

ii) The limit

limk→∞

k + 3

k(k + 1)= lim

k→∞

1 + 3k

k + 1= 0

Thus, the second condition of AST is satisfied and the series converges.

�

Exercise:Use AST determine whether the following series

1)∞∑

n=1

(−1)n+1

n22)

∞∑

k=1

(−1)k+1e−k

converges/diverges.

§ 6 Types of Convergence

A series∑

an is called absolutely convergent if the series of absolute value of∑

|an| isconvergent. E.g., the series

∞∑

n=1

(−1)n−1

n2= 1− 1

22+

1

32− · · ·

is absolutely convergent because

∞∑

n=1

∣∣∣∣

(−1)n−1

n2

∣∣∣∣=

∞∑

n=1

1

n2= 1 +

1

22+

1

32+ · · ·

is convergent because it is a p–series with p = 2 > 1. If∑

an is convergent and∑

|an|is divergent, we call the series conditionally convergent.

Theorem 4:If a series

∑an is called absolutely convergent, then the series is convergent.

Example 6.1:The series

∞∑

n=1

(−1)n−1

n= 1− 1

2+

1

3− · · ·

22

is convergent (please see Example 5.12), however the series

∞∑

n=1

∣∣∣∣

(−1)n−1

n

∣∣∣∣=

∞∑

n=1

1

n= 1 +

1

2+

1

3+ · · ·

is divergent (because it is Harmonic series). Therefore, the series is conditionally con-vergent. �

6.1 The Ratio Test for Absolute Convergence

Given an alternating series∑∞

n=1 an.

⋄ If limn→∞

∣∣∣∣

an+1

an

∣∣∣∣= L < 1, then the series

∑∞n=1 an is absolutely convergent (and

therefore converges).

⋄ If limn→∞

∣∣∣∣

an+1

an

∣∣∣∣= L > 1, then the series

∑∞n=1 an diverges.

⋄ If limn→∞

∣∣∣∣

an+1

an

∣∣∣∣= 1, then the series

∑∞n=1 an may either be convergent or divergent

(inconclusive).

Example 6.2:Determine the type of convergence of the following series

∞∑

n=1

(−1)nn3

3n

Solution:Let

an = (−1)nn3

3n, then an+1 = (−1)n+1 (n+ 1)3

3n+1

By using ratio test

limn→∞

∣∣∣∣

an+1

an

∣∣∣∣

= limn→∞

∣∣∣∣

(−1)n+1(n+ 1)3/3n+1

(−1)nn3/3n

∣∣∣∣

= limn→∞

(n + 1)3

3n+2× 3n

n3

= limn→∞

1

3

(n+ 1

n

)3

= limn→∞

1

3

(1 + 1

n

1

)3

=1

3

Therefore, by ratio test, the series converges absolutely. �

23

§ 7 Power Series

Objective: At the end of the module, you should be able to

⋄ Identify properties of the power series,

⋄ determine radius of convergence,

⋄ determine interval of convergence, represent some functions using power series.

7.1 What is Power Series?

It is a series of the form

∞∑

n=0

cnxn = c0 + c1x+ c2x

2 + c3x3 + · · ·

where x is variable and cn’s are constant coefficient of the series.E.g.,

i)

∞∑

n=0

xn = 1 + x+ x2 + x3 + · · ·

ii)

∞∑

n=0

(−1)nx2n+1

2n + 1= x− x3

3+

x5

5− · · ·

For each fixed x value, the series becomes as what we had learned earlier (no variable) andwe can use the appropriate techniques to test whether the series converges or diverges.E.g.,

∞∑

n=0

xn = 1 + x+ x2 + x3 + · · ·

If x = 2, the series becomes

∞∑

n=0

2n = 1 + 2 + 22 + 23 + · · ·

which is divergent.

If x =1

2, the series becomes

∞∑

n=0

(1

2

)n

= 1 +1

2+

(1

2

)2

+

(1

2

)3

+ · · ·

which is convergent (geometric series with r = 1/2).

24

What we need to do here is to find what is/are the possible value of x that will makethe series converges. There is a number R such that the power series will converge for|x| < R and will diverge for |x| > R. The number R is called the radius of convergenceof the series. This can be written in interval form as

|x| < R ⇒ −R < x < R

This is called interval of convergence. Note that the series may or may not converge if|x| = R. There are four possibilities of interval of convergence:

i) −R < x < R , (−R,R)

ii) −R ≤ x < R , [−R,R)

iii) −R < x ≤ R , (−R,R]

iv) −R ≤ x ≤ R , [−R,R]

Therefore, we must check whether the end point values should be included or not sothat the power series will converge for these values. To completely identify the intervalof convergence all that we have to do is determine if the power series will converge forx = −R and x = R.E.g.,

∞∑

n=0

xn = 1 + x+ x2 + x3 + · · ·

This is a geometric series, we know for this series to converge when

|x| < 1 or − 1 < x < 1

Here, the radius of convergence R = 1. The end point values −1 and 1 are not inclusive.

Example 7.1:For what values of x is the series

∞∑

n=0

n!xn

converges?

Solution:By Ratio Test,

limn→∞

∣∣∣∣

an+1

an

∣∣∣∣

= limn→∞

∣∣∣∣

(n+ 1)!xn+1

n!xn

∣∣∣∣

= limn→∞

∣∣∣∣

n!(n+ 1)xn · xn!xn

∣∣∣∣

= limn→∞

|(n+ 1)x|= lim

n→∞(n+ 1)|x| = ∞

25

The series diverges for all values of x except when x = 0, thus, the radius of convergenceR = 0. �

Example 7.2:Find the radius of convergence and the interval of convergence of the series

∞∑

n=0

(−1)nxn

n+ 1


limn→∞

∣∣∣∣

(−1)n+1xn+1/(n+ 2)

(−1)nxn/(n + 1)

∣∣∣∣

= limn→∞

∣∣∣∣

xn+1

n + 2× n+ 1

xn

∣∣∣∣

= limn→∞

|x|∣∣∣∣

n+ 1

n+ 2

∣∣∣∣

= limn→∞

|x|∣∣∣∣

1 + 1n

1 + 2n

∣∣∣∣= |x|

For the series to converge, |x| < 1. Thus, the radius of convergence R = 1, and theinterval of convergence is −1 < x < 1. To test the inequality signs let x = −1, then

∞∑

n=0

(−1)n(−1)n

n + 1=

∞∑

n=0

(−1)2n

n + 1=

∞∑

n=0

1

n+ 1=

∞∑

n=1

1

n

This is a Harmonic series and thus divergent. So that x = −1 is not included in theinterval of convergence. Now let x = 1,

∞∑

n=0

(−1)n(1)n

n+ 1=

∞∑

n=0

(−1)n

n + 1=

∞∑

n=1

(−1)n−1

n

By AST, the series converges (See Example 5.12). So that x = 1 is included in theinterval of convergence. Therefore the interval of convergence is −1 < x ≤ 1. �


∞∑

n=1

(−1)n−1xn

n3

26


limn→∞

∣∣∣∣

(−1)nxn+1/(n+ 1)3

(−1)n−1xn/n3

∣∣∣∣

= limn→∞

∣∣∣∣

xn+1

(n+ 1)3× n3

xn

∣∣∣∣

= limn→∞

∣∣∣∣

(n

n+ 1

)3∣∣∣∣|x|

= limn→∞

∣∣∣∣

(1

1 + 1n

)3∣∣∣∣|x|

= |x|

The series converges if |x| < 1. Therefore, the radius of convergence R = 1. For find theinterval of convergence, we consider first at x = −1,

∞∑

n=1

(−1)n−1(−1)n

n3=

∞∑

n=1

(−1)2n−1

n3= −

∞∑

n=1

1

n3

where 2n− 1 is an odd number for all integer n and thus (−1)2n−1 = −1. Since the lastseries is a p–series with p = 3, thus the series converges. At x = 1,

∞∑

n=1

(−1)n−1(1)n

n3=

∞∑

n=1

(−1)n−1

n3= −

∞∑

n=1

(−1)n

n3

By AST where

i) {1/n3} is a decreasing sequence, and

ii) limn→∞1n3 = 0

the series converges. Therefore, the interval of convergence is −1 ≤ x ≤ 1. �


∞∑

k=0

(−1)kx2k

(2k)!

27


limk→∞

∣∣∣∣

ak+1

ak

∣∣∣∣

= limk→∞

∣∣∣∣

(−1)k+1x2(k+1)/[2(k + 1)]!

x2k/(2k)!

∣∣∣∣

= limk→∞

∣∣∣∣

x2k+2

(2k + 2)!× (2k)!

x2k

∣∣∣∣

= limk→∞

(2k)!

(2k)!(2k + 1)(2k + 2)|x|2

= limk→∞

1

(2k + 1)(2k + 2)|x|2 = 0

Thus, the series converges for all x and the radius of convergence R = ∞ and intervalof convergence is (−∞,∞). �


∑∞n=1 an

where

an =(−2)nxn

4√n


limn→∞

∣∣∣∣

an+1

an

∣∣∣∣

= limn→∞

∣∣∣∣

(−2)n+1xn+1/ 4√n+ 1

(−2)nxn/ 4√n

∣∣∣∣

= limn→∞

∣∣∣∣− 2x 4

√n

n+ 1

∣∣∣∣

= limn→∞

∣∣∣∣− 2x 4

√

1

1 + 1n

∣∣∣∣

= | − 2x| = 2|x|

Thus, the series converges when 2|x| < 1 or |x| < 12. The radius of convergence R = 1

2.

At x = −12,

∞∑

n=1

(−2)n(−12)n

4√n

=

∞∑

n=1

(−2)n(−2)−n

n1

4

=

∞∑

n=1

1

n1

4

Since this is a p–series with p = 14≤ 1, it is divergent at x = −1

2. When x = 1

2,

∞∑

n=1

(−2)n(12)n

4√n

=

∞∑

n=1

(−2)n(2)−n

4√n

=

∞∑

n=1

(−1)n

4√n

By AST,

28

i)

{14√n

}

is a decreasing sequence, and

ii) limn→∞

14√n= 0

the series converges by AST test. Therefore, the interval of convergence is −12< x ≤ 1

2.

�

7.2 Power Series centered at x = a

If a is a constant, and if x is replaced with x− a, then the resulting series has the form

∞∑

n=0

cn(x− a)n = c0 + c1(x− a) + c2(x− a)2 + · · ·+ cn(x− a)n + · · ·

which is known as the power series centered at x = a or about x = a.E.g.,

i)

∞∑

n=0

(x− 1)n

n+ 1= 1 +

(x− 1)

2+

(x− 1)2

3+ · · ·+ (x− 1)n

n + 1+ · · ·

ii)

∞∑

k=0

(−1)k(x+ 3)k

k!= 1− (x+ 3) +

(x+ 3)2

2!− (x+ 3)3

3!+ · · ·

Theorem 5:Given a power series

∑∞n=0 cn(x − a)n, it is exactly one of the following statements is

true:

i) The series converges only when x = a.

ii) The series converges for all x.

iii) There is a positive number R such that the series converges if |x − a| < R anddiverges if |x− a| > R.

iv) The series may converges or diverges at either of these values x = a−R or x = a+R.

If the series converges for |x− a| < R, then the value a is at the center between a− Rand a+R:

Example 7.6:Find the radius of convergence and the interval of convergence for the series

∞∑

n=1

(x− 5)n

n2

29

a− R a+R

ba

x

RR


limn→∞

∣∣∣∣

(x− 5)n+1/(n+ 1)2

(x− 5)n/n2

∣∣∣∣

= limn→∞

∣∣∣∣

(x− 5)n+1

(n + 1)2× n2

(x− 5)n

∣∣∣∣

= limn→∞

(n

n + 1

)2

|x− 5|

= limn→∞

(1

1 + 1n

)2

|x− 5|

= |x− 5|

Thus, the series converges if |x − 5| < 1. The radius of convergence is R = 1. Theinterval is

|x− 5| < 1 ⇒ −1 < x− 5 < 1

5− 1 < x < 5 + 1

4 < x < 6

To check the end points:When x = 6,

∞∑

n=1

(x− 5)n

n2=

∞∑

n=1

(6− 5)n

n2=

∞∑

n=1

1

n2

This is a p–series with p = 2 > 1 and thus it is convergent.When x = 4,

∞∑

n=1

(x− 5)n

n2=

∞∑

n=1

(4− 5)n

n2=

∞∑

n=1

(−1)n

n2

By AST,

i)

{1

n2

}

is a decreasing sequence, and

ii) limn→∞

1

n2= 0.

Thus, this series is convergent by AST. Therefore, the interval of convergence is 4 ≤ x ≤6. �

30

Example 7.7:Find the radius of convergence and the interval of convergence for the series

∞∑

k=1

k(x+ 2)k

3k+1


limn→∞

∣∣∣∣

(k + 1)(x+ 2)k+1/3k+2

k(x+ 2)k/3k+1

∣∣∣∣

= limn→∞

∣∣∣∣

(k + 1)(x+ 2)k+1

3k+2× k(x+ 2)k

3k+1

∣∣∣∣

= limn→∞

∣∣∣∣

k + 1

k

∣∣∣∣×

∣∣∣∣

x+ 2

3

∣∣∣∣

=

∣∣∣∣

x+ 2

3

∣∣∣∣

By Ratio Test, for the series to converge

∣∣∣∣

x+ 2

3

∣∣∣∣< 1 , ∴ |x+ 2| < 3

Therefore, the radius of divergence is R = 3. Now,

|x+ 2| < 3 ⇒ −3 < x+ 2 < 3

−5 < x < 1

To check the endpoints:At x = −5,

∞∑

k=1

k(x+ 2)k

3k+1=

∞∑

k=1

k(−3)k

3k · 3 =

∞∑

k=1

k(−1)k · 3k3k · 3 =

1

3

∞∑

k=1

(−1)kk

This series is divergent.At x = 1,

∞∑

k=1

k(x+ 2)k

3k+1=

∞∑

k=1

k(3)k

3k · 3 =∞∑

k=1

k

3

This series is divergent. The interval of convergence is −5 < x < 1 or (−5, 1). �

∼ End ∼

31

Documents

Student Note - Series