Materials for W9

2. Infinite Series
2.0 What is an infinite series?
Intuitively, a series is the sum of all the terms in a
sequence. Thus if we begin with an infinite sequence, we
will end up with an infinite series. If {an } is an infinite
sequence, then
∞

∑a
n =1
n = a1 + a 2 + a 3 + … + a k + …

is an infinite series. Since the terms in the series are

infinite, how do we find its sum?

2.1 Partial Sum

One way to add a list of numbers is to form subtotals until
the end of the list is reached. Therefore to give meaning
to the infinite sum
S = a1 + a 2 + a 3 + … + a k + …
we examine the subtotals or “partial sums” as follows:

S 1 = a1
S 2 = a1 + a 2
S 3 = a1 + a 2 + a 3

k
S k = a1 + a 2 + a 3 + … + a k = ∑ a n
n =1

k
Note: Sk = ∑ an is called the kth partial sum of the series.
n =1
Definition 2.1
(i) The sequence of partial sums
{Sk } = S1, S 2, S 3 , …, Sk , …
is called an infinite series or series for short.
(ii) If there exists a real number S such that
∞
lim Sk = S , in other words,
k →∞
∑a
n =1
n =S,

then we say that the series converges.

(iii) If lim Sk does not exist, or lim Sk = ±∞ , then we
k →∞ k →∞

say that the series diverges.

Note: Since series are based on sequences, all the theorems in

Chapter 1 hold.

Theorem 2.1
∞ ∞
Suppose the two series ∑a
n =1
n and ∑b
n =1
n converge, also

α and β are real numbers, then

∞
(i) ∑ (αa
n =1
n ± βbn ) converges and
∞ ∞ ∞
(ii) ∑ (αa
n =1
n ± βbn ) = α ∑ an ± β ∑ bn
n =1 n =1

Theorem 2.2
∞ ∞
If the series ∑a
n =1
n converges while the series ∑b
n =1
n diverges,
∞
then ∑ (a
n =1
n ± bn ) diverges.
2.2 Telescoping and Geometric Series

2.2.1 Telescoping Series

Definition 2.2
∞
A series ∑a
n =1
n is a telescoping series if there is a sequence {bn }

such that an = bn − bn +1 .

Example 1 Show that the following series is a telescoping

series.
∞ ∞ ∞
1 1 ⎛ 1⎞
(a) ∑ 2 (b) ∑ 2 (c) ∑ ln ⎜⎝⎜1 + n ⎠⎟⎟
n =1 n + n n =1 n + 3n + 2 n =1

Solution (a)
1
Given an = . Our aim is to find {bn } such that
n2 + n
1
an = bn − bn +1 . Try rewriting an = 2 is terms of its partial
n +n
fractions.
1 1 1 1
= = −
n2 + n n (n + 1) n n + 1
1 1
If we let bn = then bn +1 = . Thus
n n +1
1 1
− = bn − bn +1 = an
n n +1
Therefore {an } is a telescoping series.
Solution (b)
1
Given an = . Re-express as partial fractions.
n 2 + 3n + 2
1 1 1 1
an = 2 = = −
n + 3n + 2 (n + 1)(n + 2) n + 1 n + 2
1 1
Let bn = , hence bn +1 = . Therefore
n +1 n +2
1 1
an = − = bn − bn +1
n +1 n +2
1
Hence an = 2 is a telescoping series.
n + 3n + 2

Solution (c)
⎛ 1⎞
Given an = ln ⎜⎜1 + ⎟⎟ . Rewrite the expression using the
⎝ n⎠
appropriate properties of the logarithm functions.
⎛ 1⎞ ⎛ n + 1⎟⎞
an = ln ⎜⎜1 + ⎟⎟ = ln ⎜⎜ ⎟ = ln(n + 1) − ln(n )
⎝ n ⎠ ⎝ n ⎠
= [− ln(n )] − [− ln(n + 1)]
⎛ 1⎞
Let bn = − ln(n ) , thus bn +1 = − ln(n + 1) . Hence an = ln ⎜⎜1 + ⎟⎟
⎝ n⎠
is a telescoping series.

∞
2
Example 2 Determine whether the series ∑
n +1 n 2
+ 2n
is a

telescoping series.
Solution
2 2 1 1
an = = = −
n 2 + 2n n(n + 2) n n + 2
1 1
Let bn = then bn +1 = . We see that
n n +1
 1 1 1 1
bn − bn +1 = − ≠ − = an
n n +1 n n +2
∞
2
Therefore ∑
n +1 n 2
+ 2n
is not a telescoping series.

2.2.2 The Sum of a Telescoping Series

Theorem 2.3
∞
Suppose that the series ∑a
n +1
n is a telescoping series with

an = bn − bn +1 for all n = 1,2,3, … . Then

∞

∑a
n +1
n = b1 − lim bn
n →∞

The telescoping series converges iff the sequence {bn } converges.

Example 3 Find the sum of the telescoping series

∞ ∞ ∞
1 1 ⎛ 1⎞
(a) ∑ 2 (b) ∑ 2 (c) ∑ ln ⎜⎜1 + ⎟⎟
n =1 n + n n =1 n + 3n + 2 n =1
⎝ n⎠

Solution (a)
1
From Example 1(a) we know that for this series, bn = . Thus
n
the sum of the series is
∞
1
∑
n =1 n 2
+ n
= b1 − lim bn
n →∞

1
= 1 − lim = 1 − 0 = 1
n →∞ n

Therefore the series converges.

Solution (b)
1
From Example 1(b) we know that bn = . Thus the sum of
n +1
the series is
∞
1 1 1 1
∑n
n =1
2
= − lim
+ 3n + 2 2 n →∞ n + 1 2
=

This is therefore a converging series.

Solution (c)
From Example 1(c) we have bn = − ln(n ) . Thus the sum of the
series is
∞
⎛ 1⎞
∑ ln ⎜⎜⎝1 + n ⎠⎟⎟ = − ln 2 + lim ln n
n =1
n →∞

= − ln 2 + ∞
Since the limit does not exist, the sum does not exist. Thus the
series diverges.

∞
2
Example 4 Find the sum of the series ∑
n +1 n 2
+ 2n
.

Solution
We know from Example 2 that this is not a telescoping series,
thus Theorem 2.3 cannot be used to find its sum. However we
proceed as follows.
 k
2 k
⎛1 1 ⎞⎟
∑
n =1 n 2
+ 2n
= ∑
n =1
⎜ −
⎜⎝ n n + 2 ⎠⎟⎟
1
= 1−
3
1 1
+ −
2 4
1 1
+ −
3 5
1 1
+ −
4 6

1 1
+ −
k −2 k
1 1
+ −
k −1 k +1
1 1
+ −
k k +2
1 1 1
= 1+ − −
2 k +1 k +2
Since
∞ k
2 2
∑ 2
n +1 n + 2n
= lim ∑ 2
k →∞
n +1 n + 2n

we have
∞
2 ⎛ 1 1 1 ⎞⎟ 3
∑
n +1 n 2
+ 2n
= lim ⎜
k →∞ ⎜
⎝
1 +
2
−
k + 1
−
k + 2
⎟⎟ =
⎠ 2
3
Therefore the series converges with it sum equalling .
2
2.2.3 The Geometric Series

∞
Definition 2.3 The series ∑r
n =0
n
is called a geometric series.

Theorem 2.4 If r < 1 , then the geometric series converges

and its sum is given by
∞
1
∑
n =0
rn =
1−r
.

If r ≥ 1 , then the series diverges.

Example 5 Show that the following series is a geometric

series; if its converges find its sum:
2 2 2
2 + + 2 + 3 +…
3 3 3
Solution
Find the series’ general term:
∞
2 2 2 2
2 + + 2 + 3 +… = ∑ n
3 3 3 n =0 3

From the general term we have

∞ ∞ ∞ n
2 1 ⎛ 1 ⎞⎟
∑
n =0 3 n
= 2∑ n = 2∑ ⎜⎜ ⎟
n =0 3
⎝ ⎠
n =0 3
1
Let r = we see that the series is a geometric series. Since
3
1
r = < 1 , the series is convergent, hence its sum is given by
3
⎛ ⎞⎟
⎜⎜
1 ⎟⎟⎟
∞ n
⎛ 1 ⎞⎟ ⎜
2∑ ⎜ ⎟ = 2 ⎜⎜
⎜ ⎟= 3
⎝ ⎠
n =0 3 ⎜⎜1 − 1 ⎟⎟⎟
⎜⎝ 3⎠
Example 6 Determine whether the following geometric series
converges or diverges. Find the sum of each convergent series.
∞ n ∞ n ∞
⎛ 4 ⎞⎟ ⎛ 3 ⎞⎟ 3
(a) ∑ ⎜⎜ ⎟ (b) ∑ ⎜⎜ ⎟ (c) ∑
⎝ ⎠
n =0 5
⎝ ⎠
n =0 2 n = 0 (−4)
n

Solution (a)
∞ n
⎛4⎞ 4 4
Given the series as ∑ ⎜⎜ ⎟⎟ , we have r = . Since r = < 1 ,
n =0 5
⎝ ⎠ 5 5
the series is convergent, hence its sum is given by
⎛ ⎞⎟
∞ n ⎜⎜ ⎟
⎛ 4 ⎞⎟ ⎜⎜ 1 ⎟⎟ = 5 .
∑ ⎜
⎜⎝ ⎠⎟
n =0 5
= ⎜⎜ 4 ⎟⎟⎟
⎜⎜⎝1 − ⎠⎟
5
Solution (b)
∞ n
⎛ 3 ⎞⎟ 3 3
Given the series as ∑ ⎜⎜ ⎟ , we have r = . Since r = ≥ 1 ,
n =0 2
⎝ ⎠ 2 2
the series is divergent, hence it has no sum.

Solution (c)
∞ ∞
3 1 1
Given the series as ∑ n = 3∑ n , where r = .
n =0 (− 4 ) n =0 (− 4 ) − 4
1
Since r = < 1 , the series is convergent, and its sum is
4
⎛ ⎞⎟
⎜⎜
∞
3 ⎜ 1 ⎟⎟⎟ 12
∑
n =0 (− 4 )
n = 3⎜ ⎜
⎜⎜1 + ⎟⎟1 ⎟⎟ = .
5
⎜⎝ 4⎠
Example 7 Determine whether the series converges or
diverges.
∞ ⎡
1 ⎥⎤
n −1
⎛ 3 ⎞ 1
(a) ∑ ⎜ ⎟ +
⎢⎜ ⎟ −
⎢
n =1 ⎣
⎝ 2 ⎠ n + 1 n + 2 ⎥⎦
∞ ⎛ 1 ⎛ 4 ⎞
n −1 ⎞
(b) ∑ ⎜⎜ 2 + ⎜⎜ ⎟⎟ ⎟⎟⎟
⎜
n =1 ⎝ n + n
⎝ 5 ⎠ ⎠⎟

Solution (a)
∞ ⎡
1 ⎥⎤
n −1
⎛ 3 ⎞ 1
Consider ∑ ⎢⎜⎜ ⎟⎟ + − as
⎢
n =1 ⎣
⎝ 2 ⎠ n + 1 n + 2 ⎥
⎦
∞ ⎡⎛ 3 ⎞n −1 ⎛ 1 1 ⎞⎟⎤ ∞
⎛ ⎞
n −1 ∞
⎛ ⎞
∑ ⎢⎜⎜ ⎟⎟ + ⎜ − ⎟ ⎥ = ∑ ⎜⎜ ⎟⎟ + ∑ ⎜ 1 − 1 ⎟⎟
3
⎢⎝ ⎠
n =1 ⎣ 2 ⎝⎜ n + 1 n + 2 ⎟⎠⎥⎦ n =1 ⎝ 2 ⎠ ⎜
n =1 ⎝ n + 1 n + 2 ⎠⎟
∞ n −1
⎛ 3⎞
From our examples above ∑ ⎜⎜ ⎟⎟ is a divergent series and
⎝ ⎠
n =1 2
∞
⎛ 1 1 ⎞⎟
∑
n =1
⎜ −
⎜⎝ n + 1 n + 2 ⎠⎟⎟ is a convergent series. Since one of the term
in the series diverges, the whole series diverges. Therefore the
∞ ⎡
1 ⎥⎤
n −1
⎛ 3 ⎞ 1
series ∑ ⎢⎜⎜ ⎟⎟ + − diverges.
⎢
n =1 ⎣
⎝ 2 ⎠ n + 1 n + 2 ⎥
⎦

Solution (b)
∞ ⎛ n −1 ⎞ ∞ ∞ n −1
1 ⎛ 4 ⎞ 1 ⎛ 4 ⎞⎟
Rewrite ∑ ⎜⎜ 2 + ⎜⎜ ⎟⎟ ⎟⎟⎟ as ∑ + ∑ ⎜⎝⎜ 5 ⎠⎟ . From
⎜
n =1 ⎝ n + n
⎝ 5 ⎠ ⎠⎟ n =1 n 2
+ n n =1

our examples above both series are convergent, hence the series
∞ ⎛ n −1 ⎞
⎜⎜ 1 + ⎜⎛ 4 ⎞⎟ ⎟⎟ is also convergent with its sum being
∑ ⎜ 2
n =1 ⎝ n + n
⎜⎝ 5 ⎠⎟ ⎟⎟
⎠
∞ ⎛ n −1 ⎞
⎜⎜ 1 + ⎜⎛ 4 ⎞⎟ ⎟⎟ = 1 + 5 = 6 .
∑ ⎜ 2
n =1 ⎝ n + n
⎜⎝ 5 ⎠⎟ ⎟⎟
⎠