Many quantities that arise in applications cannot be computed exactly.
We cannot write down an exact decimal expression for the number π
or for values of the sine function such as sin 1. However, sometimes
these quantities can be represented as infinite sums. For example,
using Taylor series (Section 11.7), we can show that
Infinite sums of this type are called infinite series.
What precisely does Eq. (1) mean? It is impossible to add up infinitely
many numbers, but what we can do is compute the partial sums SN,
defined as the finite sum of the terms up to and including Nth term.
Here are the first five partial sums of the infinite series for sin 1:
Compare these values with
the value obtained from a
calculator:
We see that S5 differs from sin 1 by less than 10−9. This suggests that
the partial sums converge to sin 1, and in fact, we will prove that
sin1  lim S N
N 
(Example 2 in Section 11.7). So although we cannot add up infinitely
many numbers, it makes sense to define the sum of an infinite series as
a limit of partial sums.
In general, an infinite series is an expression of the form
where {an} is any sequence. For example,
The Nth partial sum SN is the finite sum of the terms up to and
including aN:
DEFINITION Convergence of an Infinite Series An infinite series
converges to the sum S if its partial sums converge to S:
In this case, we write
• If the limit does not exist, we say that the infinite series diverges.
• If the limit is infinite, we say that the infinite series diverges to
infinity.
Telescoping Series Investigate numerically:
Then compute the sum S using the identity:
Numerically :
The values of the partial sums suggest convergence to S = 1. To prove
this, we observe that because of the identity, each partial sum collapses
down to just two terms:
Telescoping Series Investigate numerically:
Then compute the sum S using the identity:
1 1
S1  
1 2
1
1
1 1   1 1 
 SN  1 
S2          1 
N 1
1
2
2
3
3

 

1 


S

lim
S

lim
1

N

 1
1
1 1   1 1   1 1 
N 
N 
 N 1 
S3              1 
1 2   2 3   3 4 
4
The values of the partial sums suggest convergence to S = 1. To prove
this, we observe that because of the identity, each partial sum collapses
down to just two terms:
In most cases (apart from telescoping series and the geometric series
introduced below), there is no simple formula like Eq. (2) for the partial
sum SN. Therefore, we shall develop techniques that do not rely on
formulas for SN.
It is important to keep in mind the difference between a sequence {an}
and an infinite series
Sequences versus Series Discuss the difference between {an} and

1
an , where an 
.

n  n  1
n 1
1
1
1
,
,
, ....
The sequence is the list of numbers
1 2  2  3  3  4 
1
This sequence converges to zero: lim an  lim
0
n 
n  n  n  1
The infinite series is the sum of the numbers an, defined formally as the
limit of the partial sums. This sum is not zero. In fact, the sum is equal
to 1 by Example 1:


1
1
1
1
an  



   1

1 2  2  3  3  4 
n 1
n 1 n  n  1
Make sure you understand the difference between sequences and
series.
• With a sequence, we consider the limit of the individual terms an.
• With a series, we are interested in the sum of the terms
a1 + a2 + a3 +…
which is defined as the limit of the partial sums.
THEOREM 1 Linearity of Infinite Series
a
n
and
b
n
converge    an  bn  and
 ca
converge (c is any constant), and
 a  b  a  b 
 a  b  a  b 
 ca  c a ,  c is any constant 
n
n
n
n
n
n
n
n
n
n
n
can also
A main goal in this chapter is to develop techniques for determining
whether a series converges or diverges. It is easy to give examples of
series that diverge:

•
S  1
n 1

•
diverges to infinity (the partial sums increase without bound):
S1 = 1, S2 = 1 + 1 = 2, S3 = 1 + 1 + 1 = 3, S4 = 1 + 1 + 1 + 1 = 4,…
S    1
n 1
diverges (the partial sums jump between 1 and 0):
n 1
S1 = 1, S2 = 1 − 1 = 0, S3 = 1 − 1 + 1 = 1, S4 = 1 − 1 + 1 − 1 = 0,…
Next, we study the geometric series, which converge or diverge
depending on the common ratio r.
A geometric series with common ratio r  0 is a series
defined by a geometric sequence crn, where c  0. If the
series begins at n = 0, then

S   cr n  c  cr  cr 2  cr 3  cr 4  cr 5  
n 0
For r  1/ 2 and c = 1, we can visualize the geometric series
starting at n = 1 (Figure 1):
Adding up the terms corresponds to moving
stepwise from 0 to 1, where each step is a move
to the right by half of the remaining distance.
Thus S = 1.

1 1 1 1 1
S   n        1
2 4 8 16
n 1 2
There is a simple device for computing the partial sums of a geometric
series:
2
3
N
S N  c  cr  cr  cr    cr
rS N  cr  cr 2  cr 3    cr N  cr N 1
S N  rS N  c  cr N 1
S N 1  r   c 1  r N 1 
If r  1, we may divide by (1 − r) to obtain
S N  c  cr  cr 2  cr 3    cr N 
c 1  r N 1 
1 r
This formula enables us to sum the geometric series.
THEOREM 2 Sum of a Geometric Series Let c  0.
If |r| < 1, then

c
cr  c  cr  cr  cr   

1 r
n 0
n
2
3

M
cr
n
M
M 1
M 2
M 3
cr

cr

cr

cr

cr
  

1 r
nM
If |r| ≥ 1, then the geometric series diverges.
S N  c  cr  cr 2  cr 3    cr N 
c 1  r N 1 
1 r

n

1
1
c
1
5


n
n


Evaluate  5 . 1  c  1 & r    5 
5
1 r 4 / 5 4
n 0
n 0  5 
n 0


c
cr  c  cr  cr  cr   

1 r
n 0
n
2
3
Evaluate

M
cr
n
M
M 1
M 2
M 3
cr

cr

cr

cr

cr
  

1 r
nM
3
 3
7 
n
M

cr
27
4
 3

7  



4  1 r
7/4
16
n 3 
2  3n
Evaluate S   n .
5
n 0

Write S as a sum of two geometric series. This is valid by Theorem 1
because both geometric series converge:
 1 3 
1
 3
S    2        2      


5
5
5
5








n 0 
n

0
n

0

2
1
5 5


  5
4/5 2/5 2 2

n
n

n

n
CONCEPTUAL INSIGHT Sometimes, the following incorrect
argument is given for summing a geometric series:
1 1 1
S
   
2 4 8
1 1 1
2 S  1       1  S
2 4 8
Thus, 2S = 1 + S, or S = 1. The answer is correct, so why is the
argument wrong? It is wrong because we do not know in advance
that the series converges. Observe what happens when this
argument is applied to a divergent series:
S  1  2  4  8  16  
2S 
2  4  8  16    S  1
This would yield 2S = S − 1, or S = −1, which is absurd because S
diverges. We avoid such erroneous conclusions by carefully
defining the sum of an infinite series as the limit of partial sums.
The infinite series
diverges because the Nth partial sum SN = N diverges to infinity. It is
less clear whether the following series converges or diverges:
We now introduce a useful test that allows us to conclude that this
series diverges.
THEOREM 3 Divergence Test If the nth term an does not converge
to zero, then the series


a
n 1
n
diverges.
a
n 1
n
converges  lim an  0
n 
The Divergence Test (also called the nth-Term Test) is often stated as follows:

n
Prove the divergence of S  
.
n 1 4n  1
1
lim an   S diverges.
n 
4
Determine the convergence or divergence of
n
n 1
 1, &  1 diverges  S diverges
n 1
The Divergence Test tells only part of the story. If an does not tend to
zero, then
certainly diverges. But what if an does tend to zero? In
this case, the series may converge or it may diverge. In other words,
is a necessary condition of convergence, but it is not
sufficient. As we show in the next example, it is possible for a
series to diverge even though its terms tend to zero.
Sequence Tends to Zero, yet the Series Diverges Prove the
divergence of
1/ N  0, but because each term in the sum SN is greater than or
equal to 1/ N , we have
This shows that S N  N . But N increases without bound (Figure 2).
Therefore SN also increases without bound. This proves that the series
diverges.
chosen in a similar fashion relative to AB and E is chosen
relative to BC , then
Geometric series were used as
early as the third century BCE
by Archimedes in a brilliant
argument for determining the
area S of a “parabolic segment”
(shaded region in Figure 3).
Given two points A and C on a
parabola, there is a point B
between A and C where the
tangent line is parallel to AC
(apparently, Archimedes knew
the Mean Value Theorem more
than 2000 years before the
invention of calculus). Let T be
the area of triangle ΔABC.
Archimedes proved that if D is
This construction of triangles can be continued. The next
step would be to construct the four triangles on the
of total area
segments
. Then
construct eight triangles of total area
, etc. In this way,
we obtain infinitely many triangles that completely fill up
the parabolic segment. By the formula for the sum of a
geometric series,
For this and many other achievements, Archimedes is
ranked together with Newton and Gauss as one of the
greatest scientists of all time.
The modern study of infinite series began in the
seventeenth century with Newton, Leibniz, and their
contemporaries. The divergence of
(called the harmonic series) was known to the medieval
scholar Nicole d’Oresme (1323–1382), but his proof was
lost for centuries, and the result was rediscovered on more
than one occasion. It was also known that the sum of the
reciprocal squares
converges, and in the 1640s,
the Italian Pietro Mengoli put forward the challenge of finding its sum. Despite the efforts of the
best mathematicians of the day, including Leibniz and the Bernoulli brothers Jakob and Johann,
the problem resisted solution for nearly a century. In 1735, the great master Leonhard Euler (at
the time, 28 years old) astonished his contemporaries by proving that
This formula, surprising in itself, plays a role in a variety of mathematical fields. A theorem
from number theory states that two whole numbers, chosen randomly, have no common factor
with probability 6/π2 ≈ 0.6 (the reciprocal of Euler’s result). On the other hand, Euler’s result
and its generalizations appear in the field of statistical mechanics.
Archimedes (287 BCE–212 BCE), who discovered the law of the lever, said “Give me a place to
stand on, and I can move the earth” (quoted by Pappus of Alexandria c. AD 340).
Archimedes showed that the area S
of the parabolic segment is
where T is the area of ΔABC.