Infinite Series
Objective: We will try to find the
sum of a series with infinitely many
terms.
Sums of Infinite Series
• Our first objective is to define what is meant by the
“sum” of infinitely many real numbers. We start with
a definition:
Sums of Infinite Series
• Our first objective is to define what is meant by the
“sum” of infinitely many real numbers. We start with
a definition:
Sums of Infinite Series
• Since it is impossible to add infinitely many numbers
directly, sums of infinite series are defined and
computed by an indirect limiting process. We will
consider the decimal 0.3333… This can be viewed as
the infinite series 0.3 + 0.03 + 0.003 + … or as
3
3
3
 2  3  ...
10 10 10
• Since this is the decimal expansion of 1/3, any
reasonable definition for the sum of this series should
yield 1/3.
Sums of Infinite Series
• To obtain such a definition, consider the following
sequence of (finite) sums.
s1 
s2 
s3 
s4 
3
 0 .3
10
3
3
 2  0.33
10 10
3
3
3
 2  3  0.333
10 10 10
3
3
3
3
 2  3  4  0.3333
10 10 10 10
Sums of Infinite Series
• The sequence of numbers s1 , s2 , s3 , s4 , … can be
viewed as a succession of approximations to the
“sum” of the infinite series, which we want to be 1/3.
As we progress through the sequence, more and
more terms of the infinite series are used, and the
approximations get better and better, suggesting
that the desired sum of 1/3 might be the limit. We
need to calculate the limit of the general term in the
sequence, namely
sn 
3
3
3
 2  ...  n
10 10
10
Closed Form
• In formulas such as
n
k
k 1
n(n  1)
1  2  ...  n 
2
• The left side is said to express the sum in open form
and the right side is said to express the sum in closed
form. The open form indicates the summands and
the closed form is an explicit formula for the sum. In
other words, in closed form, if you know the value of
n, you know the sum.
Closed Form
n
2
(
3

k
)
• Express 
in closed form.
k 1
Closed Form
n
2
(
3

k
)
• Express 
in closed form.
k 1
n

k 1
n
n
n
k 1
k 1
k 1
(3  k )   9   6k   k 2
2
Closed Form
n
2
(
3

k
)
• Express 
in closed form.
k 1
n

k 1
n
n
n
k 1
k 1
k 1
(3  k )   9   6k   k 2
2
n(n  1) n(n  1)( 2n  1)
9n  6

2
6
3
2
3
2
2
n

3
n

n
2
n

21
n
 73n
2
9n  3n  3n 

6
6
Sums of Infinite Series
3
3 
3
lim sn  lim   2  ...  n 
n
n  10 10
10 

• The problem with calculating
is complicated by the fact that both the last term and
the number of terms change with n. It is best to
rewrite such limits in a closed form in which the
number of terms does not vary, if possible. To do
this, we multiply both sides of the equation for the
general term by 1/10 to obtain:
1
3
3
3
sn  2  3  ...  n 1
10
10 10
10
Sums of Infinite Series
• Now subtract the new equation from the original.
3
3
3
sn   2  ...  n 
10 10
10
1
3
3
3
sn  2  3  ...  n 1 
10
10 10
10
Sums of Infinite Series
• Subtracting the new equation from the original gives
3
3
3
sn   2  ...  n 
10 10
10
1
3
3
3
sn  2  3  ...  n 1 
10
10 10
10
9
3
3
sn   n 1
10
10 10
9
3
1 
sn  1  n 
10
10  10 
Sums of Infinite Series
• The general term can now be written as
9
3
3
sn   n 1
10
10 10
9
3
1 
sn  1  n 
10
10  10 
1
1 
s n  1  n 
3  10 
Sums of Infinite Series
• Now, the limit becomes
1
1 
s n  1  n 
3  10 
1
1  1
lim sn  lim 1  n  
n
n 3
 10  3
Sums of Infinite Series
• Motivated by the previous example, we are now
ready to define the general concept of the “sum” of
an infinite series u1  u2  u3  ... uk  ... .
Sums of Infinite Series
• Motivated by the previous example, we are now
ready to define the general concept of the “sum” of
an infinite series u1  u2  u3  ... uk  ... .
• We begin with some terminology. Let sn denote the
sum of the initial terms of the series, up to and
including the term with index n. Thus,
s1  u1
s3  u1  u2  u3
s2  u1  u2
s4  u1  u2  u3  u4
n
sn  u1  u2  u3  ...  un   uk
k 1
Sums of Infinite Series
• The number sn is called the nth partial sum of the

series and the sequence sn n1 is called the
sequence of partial sums. As n increases, the partial
sum sn = u1 + u2 + …+ un + … includes more and more
terms of the series. Thus, if sn tends toward a limit as
n   , it is reasonable to view this limit as the
sum of all the terms in the series.
Sums of Infinite Series
• This suggests the following definition:
Example 1
• Determine whether the series 1 – 1 + 1 – 1 + 1- 1 +…
converges or diverges. If it converges, find the sum.
Example 1
• Determine whether the series 1 – 1 + 1 – 1 + 1- 1 +…
converges or diverges. If it converges, find the sum.
• It is tempting to conclude that the sum of the series is
zero by arguing that the positive and negative terms
will cancel. This is not correct. We will look at the
partial sums.
Example 1
• Determine whether the series 1 – 1 + 1 – 1 + 1- 1 +…
converges or diverges. If it converges, find the sum.
• It is tempting to conclude that the sum of the series is
zero by arguing that the positive and negative term
will cancel. This is not correct. We will look at the
partial sums.
s1  1
s2  1  1  0
s3  1 1  1  1
• Thus, the sequence of partial sums is 1, 0, 1, 0, 1, …
This is a divergent series and has no sum.
Geometric Series
• In many important series, each term is obtained by
multiplying the preceding term by some fixed
constant. Thus, if the initial term of the series is a
and each term is obtained by multiplying the
preceding term by r, then the series has the form

k
2
k
ar

a

ar

ar

...

ar
 ...(a  0)

k 0
Geometric Series
• In many important series, each term is obtained by
multiplying the preceding term by some fixed
constant. Thus, if the initial term of the series is a
and each term is obtained by multiplying the
preceding term by r, then the series has the form

k
2
k
ar

a

ar

ar

...

ar
 ...(a  0)

k 0
• Such series are called geometric series, and the
number r is called the ratio for the series.
Geometric Series
• Here are some examples of geometric series.
Geometric Series
• The following theorem is the fundamental result on
convergence or geometric series.
Example 2
• The series

5
5 5
5

5



...

 ...

k
2
k
4 4
4
k 0 4
is a geometric series with a = 5 and r = ¼. Since
|r|=1/4 < 1, the series converges and the sum is
a
5
20


1 r 1 1/ 4 3
Example 3
• Find the rational number represented by the repeating
decimal 0.784784784….
Example 3
• Find the rational number represented by the repeating
decimal 0.784784784….
• We can write this as 0.784 + 0.000784 + 0.000000784
so the given decimal is the sum of a geometric series
with a = 0.784 and r = 0.001. Thus, the sum is
a
0.784
.784 784



1  r 1  0.001 .999 999
Example 4
• In each part, determine whether the series
converges, and if so find its sum.

( a ) 3 5
k 1
2 k 1 k

(b) x k
k 0
Example 4
• In each part, determine whether the series
converges, and if so find its sum.


( a ) 3 5
(b) x k
2 k 1 k
k 1
k 0
• (a) This is a geometric series in a concealed form,
since we can rewrite it as

( a ) 3 5
k 1
2 k 1 k


9
9
  k 1   9 
k 1 5
k 1  5 
k
k 1
9
r   1; diverges
5
Example 4
• In each part, determine whether the series
converges, and if so find its sum.

( a ) 3 5
k 1
2 k 1 k

(b) x k
k 0
• (b) This is a geometric series with a = 1 and r = x, so
it converges if |x| < 1 and diverges otherwise. When
it converges its sum is

1
x 

1 x
k 0
k
Telescoping Sums

• Determine whether the series  1  1  1  1  ...
1 2 2  3 3  4
k 1 k (k  1)
converges or diverges.
• If it converges, find the sum.
Telescoping Sums

• Determine whether the series  1  1  1  1  ...
1 2 2  3 3  4
k 1 k (k  1)
converges or diverges.
• If it converges, find the sum.
• To write this in closed form, we will use the method
of partial fractions to obtain  1  1  1
 k (k  1)
k 1
k
k 1
Telescoping Sums

• Determine whether the series  1  1  1  1  ...
1 2 2  3 3  4
k 1 k (k  1)
converges or diverges.
• If it converges, find the sum.
1 
1
sn    

k 1
k 1  k
n
1 
 1 1 1 1 1  1
1             

 2   2 3   3 4   n n 1
Telescoping Sums

• Determine whether the series  1  1  1  1  ...
1 2 2  3 3  4
k 1 k (k  1)
converges or diverges.
• If it converges, find the sum.
1 
1
sn    

k 1
k 1  k
n
1 
 1 1 1 1 1  1
1             

 2   2 3   3 4   n n 1

1
1 

 lim sn  lim 1 
 1

n 
n 
 n 1
k 1 k (k  1)
Harmonic Series
• One of the most important of all diverging series is
the harmonic series  1
k
k 1
which arises in connection with the overtones
produced by a vibrating musical string. It is not
immediately evident that this series diverges.
However, the divergence will become apparent when
we examine the partial sums in detail.
Harmonic Series
• Because the terms in the series are all positive, the
partial sums
s1  1
1
s2  1 
2
1 1
s3  1  
2 3
1 1 1
s4  1   
2 3 4
form a strictly increasing sequence. Thus, by
Theorem 10.2.3 there is no constant M that is greater
than or equal to every partial sum, so the series
diverges.
Homework
• Pages 649-650
• 1-19 odd