Ebook Essential calculus Early transcendentals Part 2

8

SERIES
Infinite series are sums of infinitely many terms. (One of our aims in this chapter is to define
exactly what is meant by an infinite sum.) Their importance in calculus stems from Newton’s idea
of representing functions as sums of infinite series. For instance, in finding areas he often
integrated a function by first expressing it as a series and then integrating each term of the series.
We will pursue his idea in Section 8.7 in order to integrate such functions as eϪx . (Recall that we
have previously been unable to do this.) Many of the functions that arise in mathematical physics
and chemistry, such as Bessel functions, are defined as sums of series, so it is important to be
familiar with the basic concepts of convergence of infinite sequences and series.
Physicists also use series in another way, as we will see in Section 8.8. In studying fields as
diverse as optics, special relativity, and electromagnetism, they analyze phenomena by replacing a
function with the first few terms in the series that represents it.
2

8.1

SEQUENCES
A sequence can be thought of as a list of numbers written in a definite order:
a1 , a2 , a3 , a4 , . . . , an , . . .
The number a 1 is called the first term, a 2 is the second term, and in general a n is the
nth term. We will deal exclusively with infinite sequences and so each term a n will
have a successor a nϩ1 .
Notice that for every positive integer n there is a corresponding number a n and so
a sequence can be defined as a function whose domain is the set of positive integers.
But we usually write a n instead of the function notation f ͑n͒ for the value of the function at the number n.
NOTATION The sequence {a 1 , a 2 , a 3 , . . .} is also denoted by

͕a n ͖

or

ϱ

͕a n ͖ n෇1

EXAMPLE 1 Some sequences can be defined by giving a formula for the nth term.

In the following examples we give three descriptions of the sequence: one by using
the preceding notation, another by using the defining formula, and a third by writing
out the terms of the sequence. Notice that n doesn’t have to start at 1.
(a)

(b)
(c)
(d)
410

ͭ ͮ
ͭ
ͮ
n
nϩ1

ϱ

an ෇

n
nϩ1

an ෇

͑Ϫ1͒n͑n ϩ 1͒
3n

n෇1

͑Ϫ1͒n͑n ϩ 1͒
3n

{sn Ϫ 3 } ϱn෇3

a n ෇ sn Ϫ 3 , n ജ 3

ͭ ͮ

a n ෇ cos

n␲
cos
6

ϱ

n෇0

n␲
, nജ0
6

ͭ
ͭ

ͮ

1 2 3 4
n
, , , ,...,
,...
2 3 4 5
nϩ1

ͮ

2 3
4 5
͑Ϫ1͒n͑n ϩ 1͒
Ϫ , ,Ϫ ,
,...,
,...
3 9
27 81
3n

{0, 1, s2 , s3 , . . . , sn Ϫ 3 , . . .}

ͭ

1,

n␲
s3 1
, , 0, . . . , cos
,...
2 2
6

ͮ

■


SECTION 8.1

V EXAMPLE 2

SEQUENCES

■

411

Find a formula for the general term a n of the sequence

ͭ

ͮ

3

4
5
6
7
,Ϫ ,
,Ϫ
,
,...
5
25 125
625 3125

assuming that the pattern of the first few terms continues.
SOLUTION We are given that

a1 ෇

3
5

a2 ෇ Ϫ

4
25

a3 ෇

5
125

a4 ෇ Ϫ

6
625

a5 ෇

7
3125

Notice that the numerators of these fractions start with 3 and increase by 1 whenever
we go to the next term. The second term has numerator 4, the third term has numerator 5; in general, the nth term will have numerator n ϩ 2. The denominators are the
powers of 5, so a n has denominator 5 n. The signs of the terms are alternately positive and negative, so we need to multiply by a power of Ϫ1. In Example 1(b) the
factor ͑Ϫ1͒ n meant we started with a negative term. Here we want to start with a
positive term and so we use ͑Ϫ1͒ nϪ1 or ͑Ϫ1͒ nϩ1. Therefore,
a n ෇ ͑Ϫ1͒ nϪ1

nϩ2
5n

■

EXAMPLE 3 Here are some sequences that don’t have a simple defining equation.

(a) The sequence ͕ pn ͖, where pn is the population of the world as of January 1 in
the year n.
(b) If we let a n be the digit in the nth decimal place of the number e, then ͕a n ͖ is a
well-defined sequence whose first few terms are
͕7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, . . .͖
(c) The Fibonacci sequence ͕ fn ͖ is defined recursively by the conditions

f1 ෇ 1

f2 ෇ 1

fn ෇ fnϪ1 ϩ fnϪ2

nജ3

Each term is the sum of the two preceding terms. The first few terms are
a¡

a™ a£

1
2

0

͕1, 1, 2, 3, 5, 8, 13, 21, . . .͖

a¢

This sequence arose when the 13th-century Italian mathematician known as Fibonacci
solved a problem concerning the breeding of rabbits (see Exercise 41).
■

1

A sequence such as the one in Example 1(a), a n ෇ n͑͞n ϩ 1͒, can be pictured
either by plotting its terms on a number line as in Figure 1 or by plotting its graph as

in Figure 2. Note that, since a sequence is a function whose domain is the set of positive integers, its graph consists of isolated points with coordinates

FIGURE 1
an

1

͑1, a1 ͒

͑2, a2 ͒

͑3, a3 ͒

...

͑n, a n ͒

...

7

a¶= 8
0

1 2 3 4 5 6 7

FIGURE 2

n

From Figure 1 or 2 it appears that the terms of the sequence a n ෇ n͑͞n ϩ 1͒ are
approaching 1 as n becomes large. In fact, the difference
1Ϫ

n
1
෇
nϩ1
nϩ1


412

■

CHAPTER 8

SERIES

can be made as small as we like by taking n sufficiently large. We indicate this by
writing
n
lim
෇1
nlϱ n ϩ 1
In general, the notation
lim a n ෇ L

nlϱ

means that the terms of the sequence ͕a n ͖ approach L as n becomes large. Notice that
the following definition of the limit of a sequence is very similar to the definition of a
limit of a function at infinity given in Section 1.6.
1 DEFINITION

A sequence ͕a n ͖ has the limit L and we write
lim a n ෇ L

a n l L as n l ϱ

or

nlϱ

if we can make the terms a n as close to L as we like by taking n sufficiently
large. If lim n l ϱ a n exists, we say the sequence converges (or is convergent).
Otherwise, we say the sequence diverges (or is divergent).
Figure 3 illustrates Definition 1 by showing the graphs of two sequences that have
the limit L.
an

an

L

L

FIGURE 3

Graphs of two

sequences with
lim an= L

0

0

n

n

n  `

A more precise version of Definition 1 is as follows.
2 DEFINITION

A sequence ͕an ͖ has the limit L and we write
lim an ෇ L

■ Compare this definition with
Definition 1.6.7.

a n l L as n l ϱ

or

nlϱ

if for every ␧ Ͼ 0 there is a corresponding integer N such that
if

nϾN

Խa

then

n

Խ

ϪL Ͻ␧

Definition 2 is illustrated by Figure 4, in which the terms a 1 , a 2 , a 3 , . . . are plotted
on a number line. No matter how small an interval ͑L Ϫ ␧, L ϩ ␧͒ is chosen, there exists
an N such that all terms of the sequence from a Nϩ1 onward must lie in that interval.
a¡
FIGURE 4

0

a£

a™

aˆ

aN+1 aN+2
L-∑

L

L+∑

a˜

aß

a∞

a¢

a¶


SECTION 8.1

SEQUENCES

413

■

Another illustration of Definition 2 is given in Figure 5. The points on the graph of
͕an ͖ must lie between the horizontal lines y ෇ L ϩ ␧ and y ෇ L Ϫ ␧ if n Ͼ N . This
picture must be valid no matter how small ␧ is chosen, but usually a smaller ␧ requires
a larger N.
y

y=L+∑

L
y=L-∑
0

1 2 3 4

FIGURE 5

n

N

If you compare Definition 2 with Definition 1.6.7, you will see that the only difference between lim n l ϱ a n ෇ L and lim x l ϱ f ͑x͒ ෇ L is that n is required to be an
integer. Thus we have the following theorem, which is illustrated by Figure 6.
3 THEOREM If lim x l ϱ f ͑x͒ ෇ L and f ͑n͒ ෇ a n when n is an integer, then
lim n l ϱ a n ෇ L.

y

y=ƒ

L

0

FIGURE 6

x

1 2 3 4

In particular, since we know that limx l ϱ ͑1͞x r ͒ ෇ 0 when r Ͼ 0, we have
lim

4

nlϱ

1
෇0
nr

if r Ͼ 0

If a n becomes large as n becomes large, we use the notation lim n l ϱ a n ෇ ϱ. The
following precise definition is similar to Definition 1.6.8.

5 DEFINITION lim n l ϱ a n ෇ ϱ means that for every positive number M there
is an integer N such that

if

nϾN

then

an Ͼ M

If lim n l ϱ a n ෇ ϱ, then the sequence ͕a n ͖ is divergent but in a special way. We say
that ͕a n ͖ diverges to ϱ.

The Limit Laws given in Section 1.4 also hold for the limits of sequences and their
proofs are similar.


414

■

CHAPTER 8

SERIES

If ͕a n ͖ and ͕bn ͖ are convergent sequences and c is a constant, then

Limit Laws for Sequences

lim ͑a n ϩ bn ͒ ෇ lim a n ϩ lim bn

nlϱ

nlϱ

nlϱ

lim ͑a n Ϫ bn ͒ ෇ lim a n Ϫ lim bn

nlϱ

nlϱ

nlϱ

lim ca n ෇ c lim a n

nlϱ

lim c ෇ c

nlϱ

nlϱ

lim ͑a n bn ͒ ෇ lim a n ؒ lim bn

nlϱ

nlϱ

lim

lim a n
an
nlϱ
෇
bn
lim bn

nlϱ

nlϱ

if lim bn
nlϱ

nlϱ

[

lim a np ෇ lim a n

nlϱ

nlϱ

]

p

0

if p Ͼ 0 and a n Ͼ 0

The Squeeze Theorem can also be adapted for sequences as follows (see Figure 7).

If a n ഛ bn ഛ cn for n ജ n 0 and lim a n ෇ lim cn ෇ L, then lim bn ෇ L.

Squeeze Theorem for Sequences

nlϱ

cn

nlϱ

nlϱ

Another useful fact about limits of sequences is given by the following theorem,
whose proof is left as Exercise 45.

bn
an
0

Խ Խ

If lim a n ෇ 0, then lim a n ෇ 0.

6 THEOREM

nlϱ

nlϱ

n

FIGURE 7

The sequence ͕ bn ͖ is squeezed
between the sequences ͕ a n ͖
and ͕ cn ͖ .

EXAMPLE 4 Find lim

nlϱ

n
.
nϩ1

SOLUTION The method is similar to the one we used in Section 1.6: Divide numerator and denominator by the highest power of n that occurs in the denominator and
then use the Limit Laws.

lim

nlϱ

n
෇ lim
nlϱ
nϩ1

෇

This shows that the guess we made
earlier from Figures 1 and 2 was correct.
■

nlϱ

ln n

.
n

1ϩ

1
n

lim 1

nlϱ

෇

lim 1 ϩ lim

nlϱ

nlϱ

1
n

1
෇1
1ϩ0

Here we used Equation 4 with r ෇ 1.
EXAMPLE 5 Calculate lim

1

■


SECTION 8.1

SEQUENCES

■

415

SOLUTION Notice that both numerator and denominator approach infinity as n l ϱ.
We can’t apply l’Hospital’s Rule directly because it applies not to sequences but to
functions of a real variable. However, we can apply l’Hospital’s Rule to the related
function f ͑x͒ ෇ ͑ln x͒͞x and obtain

ln x
1͞x
෇ lim
෇0
xlϱ 1
x

lim

xlϱ

Therefore, by Theorem 3 we have

lim

nlϱ

an

ln n
෇0
n

■

EXAMPLE 6 Determine whether the sequence a n ෇ ͑Ϫ1͒ n is convergent or divergent.

1

SOLUTION If we write out the terms of the sequence, we obtain

0

1

2

3

4

͕Ϫ1, 1, Ϫ1, 1, Ϫ1, 1, Ϫ1, . . .͖

n

_1

The graph of this sequence is shown in Figure 8. Since the terms oscillate between 1
and Ϫ1 infinitely often, a n does not approach any number. Thus, lim n l ϱ ͑Ϫ1͒ n does
not exist; that is, the sequence ͕͑Ϫ1͒ n ͖ is divergent.
■

FIGURE 8

EXAMPLE 7 Evaluate lim

The graph of the sequence in
Example 7 is shown in Figure 9 and
supports the answer.
■

nlϱ

͑Ϫ1͒ n
if it exists.
n

SOLUTION

an
1

lim

nlϱ

Ϳ

͑Ϫ1͒ n
n

Ϳ

෇ lim

nlϱ

1
෇0
n

Therefore, by Theorem 6,
0

1

lim

n

nlϱ

͑Ϫ1͒ n

෇0
n

■

Discuss the convergence of the sequence a n ෇ n!͞n n, where
n! ෇ 1 ؒ 2 ؒ 3 ؒ и и и ؒ n.
V EXAMPLE 8

_1

FIGURE 9

SOLUTION Both numerator and denominator approach infinity as n l ϱ but here
we have no corresponding function for use with l’Hospital’s Rule (x! is not defined
when x is not an integer). Let’s write out a few terms to get a feeling for what
happens to a n as n gets large:

a1 ෇ 1

7

a2 ෇
an ෇

1ؒ2
2ؒ2

a3 ෇

1ؒ2ؒ3
3ؒ3ؒ3

1 ؒ 2 ؒ 3 ؒ иии ؒ n
n ؒ n ؒ n ؒ иии ؒ n

It appears from these expressions and the graph in Figure 10 that the terms are
decreasing and perhaps approach 0. To confirm this, observe from Equation 7 that
an ෇

1
n

ͩ

2 ؒ 3 ؒ иии ؒ n
n ؒ n ؒ иии ؒ n

ͪ


416

■

CHAPTER 8

SERIES

CREATING GRAPHS OF SEQUENCES

Some computer algebra systems have special commands that enable us to create
sequences and graph them directly. With
most graphing calculators, however,
sequences can be graphed by using parametric equations. For instance, the
sequence in Example 8 can be graphed
by entering the parametric equations
■

x෇t

y ෇ t!͞t t

Notice that the expression in parentheses is at most 1 because the numerator is less
than (or equal to) the denominator. So
0 Ͻ an ഛ

We know that 1͞n l 0 as n l ϱ. Therefore, a n l 0 as n l ϱ by the Squeeze
Theorem.
V EXAMPLE 9

and graphing in dot mode starting with
t ෇ 1, setting the t -step equal to 1. The
result is shown in Figure 10.
1

1
n
■

For what values of r is the sequence ͕r n ͖ convergent?

SOLUTION We know from Section 1.6 and the graphs of the exponential functions

in Section 3.1 that lim x l ϱ a x ෇ ϱ for a Ͼ 1 and lim x l ϱ a x ෇ 0 for 0 Ͻ a Ͻ 1.
Therefore, putting a ෇ r and using Theorem 3, we have
lim r n ෇

nlϱ

ͭ

ϱ
0

if r Ͼ 1
if 0 Ͻ r Ͻ 1

For the cases r ෇ 1 and r ෇ 0 we have
0

10

lim 1n ෇ lim 1 ෇ 1

nlϱ

lim 0 n ෇ lim 0 ෇ 0

and

nlϱ

nlϱ

nlϱ

FIGURE 10

Խ Խ

If Ϫ1 Ͻ r Ͻ 0, then 0 Ͻ r Ͻ 1, so

Խ Խ

Խ Խ

lim r n ෇ lim r

nlϱ

nlϱ

n

෇0

and therefore lim n l ϱ r n ෇ 0 by Theorem 6. If r ഛ Ϫ1, then ͕r n ͖ diverges as in
Example 6. Figure 11 shows the graphs for various values of r. (The case r ෇ Ϫ1 is
shown in Figure 8.)
an

an

r>1
1

1

_1
0

r=1

n

1
0

FIGURE 11

1

0
n

r<_1

The sequence an=r n

■

The results of Example 9 are summarized for future use as follows.

The sequence ͕r n ͖ is convergent if Ϫ1 Ͻ r ഛ 1 and divergent for all other
values of r.
8

lim r n ෇

nlϱ

ͭ

0
1

if Ϫ1 Ͻ r Ͻ 1
if r ෇ 1


SECTION 8.1

SEQUENCES

■

417

9 DEFINITION A sequence ͕a n ͖ is called increasing if a n Ͻ a nϩ1 for all
n ജ 1, that is, a1 Ͻ a2 Ͻ a3 Ͻ и и и . It is called decreasing if a n Ͼ a nϩ1 for all
n ജ 1. A sequence is monotonic if it is either increasing or decreasing.

EXAMPLE 10 The sequence

ͭ ͮ
3
nϩ5

is decreasing because

3
3
3
Ͼ
෇
nϩ5
͑n ϩ 1͒ ϩ 5
nϩ6

■ The right side is smaller because it
has a larger denominator.

and so a n Ͼ a nϩ1 for all n ജ 1.

■

EXAMPLE 11 Show that the sequence a n ෇

n
is decreasing.
n2 ϩ 1

SOLUTION We must show that a nϩ1 Ͻ a n , that is,

nϩ1
n
Ͻ 2
͑n ϩ 1͒ 2 ϩ 1
n ϩ1
■ Another way to do Example 11 is to
show that the function

f ͑x͒ ෇

x
x2 ϩ 1

xജ1

is decreasing because f Ј͑x͒ Ͻ 0 for
x Ͼ 1.

This inequality is equivalent to the one we get by cross-multiplication:
nϩ1
n
Ͻ 2
͑n ϩ 1͒ 2 ϩ 1
n ϩ1

&? ͑n ϩ 1͒͑n 2 ϩ 1͒ Ͻ n͓͑n ϩ 1͒2 ϩ 1͔
&?

n 3 ϩ n 2 ϩ n ϩ 1 Ͻ n 3 ϩ 2n 2 ϩ 2n

&?

1 Ͻ n2 ϩ n

Since n ജ 1, we know that the inequality n 2 ϩ n Ͼ 1 is true. Therefore, a nϩ1 Ͻ a n
and so ͕a n ͖ is decreasing.
■

10 DEFINITION

A sequence ͕a n ͖ is bounded above if there is a number M

such that
an ഛ M

for all n ജ 1

It is bounded below if there is a number m such that
m ഛ an

for all n ജ 1

If it is bounded above and below, then ͕a n ͖ is a bounded sequence.
For instance, the sequence a n ෇ n is bounded below ͑a n Ͼ 0͒ but not above. The

sequence a n ෇ n͑͞n ϩ 1͒ is bounded because 0 Ͻ a n Ͻ 1 for all n.
We know that not every bounded sequence is convergent [for instance, the
sequence a n ෇ ͑Ϫ1͒n satisfies Ϫ1 ഛ a n ഛ 1 but is divergent from Example 6] and
not every monotonic sequence is convergent ͑a n ෇ n l ϱ͒. But if a sequence is


418

■

CHAPTER 8

SERIES

an

M
L

0 1 23

n

FIGURE 12

both bounded and monotonic, then it must be convergent. This fact is proved as
Theorem 11, but intuitively you can understand why it is true by looking at Figure 12. If ͕a n ͖ is increasing and a n ഛ M for all n, then the terms are forced to crowd
together and approach some number L.
The proof of Theorem 11 is based on the Completeness Axiom for the set ‫ ޒ‬of
real numbers, which says that if S is a nonempty set of real numbers that has an upper

bound M (x ഛ M for all x in S ), then S has a least upper bound b. (This means that
b is an upper bound for S, but if M is any other upper bound, then b ഛ M .) The
Completeness Axiom is an expression of the fact that there is no gap or hole in the real
number line.
11 MONOTONIC SEQUENCE THEOREM

Every bounded, monotonic sequence is

convergent.
PROOF Suppose ͕a n ͖ is an increasing sequence. Since ͕a n ͖ is bounded, the set

Խ

S ෇ ͕a n n ജ 1͖ has an upper bound. By the Completeness Axiom it has a least
upper bound L . Given ␧ Ͼ 0, L Ϫ ␧ is not an upper bound for S (since L is the least
upper bound). Therefore
aN Ͼ L Ϫ ␧

for some integer N

But the sequence is increasing so a n ജ a N for every n Ͼ N. Thus if n Ͼ N we have
an Ͼ L Ϫ ␧
0 ഛ L Ϫ an Ͻ ␧

so
since a n ഛ L . Thus

ԽL Ϫ a Խ Ͻ ␧
n

whenever n Ͼ N

so lim n l ϱ a n ෇ L .
A similar proof (using the greatest lower bound) works if ͕a n ͖ is decreasing.

■

The proof of Theorem 11 shows that a sequence that is increasing and bounded
above is convergent. (Likewise, a decreasing sequence that is bounded below is convergent.) This fact is used many times in dealing with infinite series in Sections 8.2
and 8.3.
Another use of Theorem 11 is indicated in Exercises 38– 40.

8.1

EXERCISES

1. (a) What is a sequence?

(b) What does it mean to say that lim n l ϱ a n ෇ 8?
(c) What does it mean to say that lim n l ϱ a n ෇ ϱ?
2. (a) What is a convergent sequence? Give two examples.

(b) What is a divergent sequence? Give two examples.

3. List the first six terms of the sequence defined by

an ෇

n
2n ϩ 1

Does the sequence appear to have a limit? If so, find it.


SECTION 8.1

5– 8 ■ Find a formula for the general term a n of the sequence,
assuming that the pattern of the first few terms continues.
6.

7. ͕2, 7, 12, 17, . . .͖
■

■

■

■

32. (a) If ͕a n ͖ is convergent, show that

lim a nϩ1 ෇ lim a n

{Ϫ 14 , 29 , Ϫ 163 , 254 , . . .}

nlϱ

■

■

■

■

■

■

■

■ Determine whether the sequence converges or diverges.
If it converges, find the limit.

9–28

■ Determine whether the sequence is increasing,
decreasing, or not monotonic. Is the sequence bounded?

33–36

3 ϩ 5n 2
n ϩ n2

10. a n ෇

nϩ1
3n Ϫ 1

33. a n ෇

11. a n ෇

2n
3 nϩ1

12. a n ෇

sn
1 ϩ sn

35. a n ෇ cos͑n␲͞2͒

13. a n ෇

͑n ϩ 2͒!
n!

14. a n ෇

n
1 ϩ sn

15. a n ෇

͑Ϫ1͒ n
n2 ϩ 1

16. a n ෇

͑Ϫ1͒ n
n 3 ϩ 2n 2 ϩ 1

9. a n ෇

nϪ1

17.

ͭ

e n ϩ e Ϫn
e 2n Ϫ 1

ͩ ͪ

24. a n ෇

sin 2n
1 ϩ sn

26. a n ෇

͑ln n͒ 2
n

27. a n ෇ ln͑2n 2 ϩ 1͒ Ϫ ln͑n 2 ϩ 1͒ 28. a n ෇

͑Ϫ3͒n
n!

23. a n ෇

1ϩ

2
n

25. ͕0, 1, 0, 0, 1, 0, 0, 0, 1, . . . ͖

■

■

■

■

■

■

■

■

■

■

an
3a n ϩ 1

■

■

■

■

■

■

■

a nϩ1 ෇ 3 Ϫ 1͞a n is increasing and a n Ͻ 3 for all n. Deduce
that ͕a n ͖ is convergent and find its limit.

40. Show that the sequence defined by

a1 ෇ 2

a nϩ1 ෇

1
3 Ϫ an

satisfies 0 Ͻ a n ഛ 2 and is decreasing. Deduce that the

sequence is convergent and find its limit.
■

30. Find the first 40 terms of the sequence defined by
1
2

■

1
n

39. Use induction to show that the sequence defined by a 1 ෇ 1,

then after n years the investment is worth a n ෇ 1000͑1.06͒n
dollars.
(a) Find the first five terms of the sequence ͕a n ͖.
(b) Is the sequence convergent or divergent? Explain.

ͭ

36. a n ෇ n ϩ

(a) By induction or otherwise, show that ͕a n ͖ is increasing
and bounded above by 3. Apply the Monotonic
Sequence Theorem to show that lim n l ϱ a n exists.
(b) Find lim n l ϱ a n .

29. If $1000 is invested at 6% interest, compounded annually,

a nϩ1 ෇

■

2n Ϫ 3
3n ϩ 4

38. A sequence ͕a n ͖ is given by a 1 ෇ s2 , a nϩ1 ෇ s2 ϩ a n .

22. ͕n cos n␲ ͖
n

■

34. a n ෇

{s2 , s2s2 , s2s2s2 , . . .}

20. ͕arctan 2n͖

cos 2n
2n

■

1
2n ϩ 3

37. Find the limit of the sequence
3

18. a n ෇ cos͑2͞n͒

19. ͕n 2e Ϫn ͖
21. a n ෇

■

n

ͮ

nlϱ

(b) A sequence ͕a n ͖ is defined by a 1 ෇ 1 and
a nϩ1 ෇ 1͑͞1 ϩ a n ͒ for n ജ 1. Assuming that ͕a n ͖ is
convergent, find its limit.

8. ͕5, 1, 5, 1, 5, 1, . . .͖
■

419

all its terms lie between the numbers 5 and 8. Explain why
the sequence has a limit. What can you say about the value
of the limit?

this sequence appear to have a limit? If so, find it. If not,
explain why.

{1, Ϫ 23 , 49 , Ϫ 278 , . . .}

■

31. Suppose you know that ͕a n ͖ is a decreasing sequence and

4. List the first nine terms of the sequence ͕cos͑n␲͞3͖͒. Does

5.

SEQUENCES

if a n is an even number
if a n is an odd number

and a1 ෇ 11. Do the same if a1 ෇ 25. Make a conjecture
about this type of sequence.

■

41. (a) Fibonacci posed the following problem: Suppose that

rabbits live forever and that every month each pair produces a new pair which becomes productive at age
2 months. If we start with one newborn pair, how many
pairs of rabbits will we have in the nth month? Show
that the answer is fn , where ͕ fn ͖ is the Fibonacci
sequence defined in Example 3(c).
(b) Let a n ෇ fnϩ1͞fn and show that a nϪ1 ෇ 1 ϩ 1͞a nϪ2.
Assuming that ͕a n ͖ is convergent, find its limit.
42. (a) Let a 1 ෇ a, a 2 ෇ f ͑a͒, a 3 ෇ f ͑a 2 ͒ ෇ f ͑ f ͑a͒͒, . . . ,

a nϩ1 ෇ f ͑a n ͒, where f is a continuous function. If
lim n l ϱ a n ෇ L, show that f ͑L͒ ෇ L.
(b) Illustrate part (a) by taking f ͑x͒ ෇ cos x, a ෇ 1, and
estimating the value of L to five decimal places.


420

■

CHAPTER 8

SERIES

43. We know that lim n l ϱ ͑0.8͒ n ෇ 0 [from (8) with r ෇ 0.8].

47. The size of an undisturbed fish population has been

modeled by the formula

Use logarithms to determine how large n has to be so that
͑0.8͒ n Ͻ 0.000001.
44. Use Definition 2 directly to prove that lim n l ϱ r n ෇ 0

pnϩ1 ෇

Խ Խ

when r Ͻ 1.

45. Prove Theorem 6.

[Hint: Use either Definition 2 or the Squeeze Theorem.]
46. (a) Show that if lim n l ϱ a 2n ෇ L and lim n l ϱ a2nϩ1 ෇ L ,

then ͕a n ͖ is convergent and lim n l ϱ a n ෇ L .
(b) If a 1 ෇ 1 and
1
anϩ1 ෇ 1 ϩ
1 ϩ an

find the first eight terms of the sequence ͕a n ͖. Then use
part (a) to show that lim n l ϱ a n ෇ s2 . This gives the
continued fraction expansion
1

s2 ෇ 1 ϩ
2ϩ

8.2

bpn
a ϩ pn

where pn is the fish population after n years and a and b are
positive constants that depend on the species and its environment. Suppose that the population in year 0 is p 0 Ͼ 0.
(a) Show that if ͕ pn ͖ is convergent, then the only possible
values for its limit are 0 and b Ϫ a.
(b) Show that pnϩ1 Ͻ ͑ b͞a͒ pn .

(c) Use part (b) to show that if a Ͼ b, then lim n l ϱ pn ෇ 0 ;
in other words, the population dies out.
(d) Now assume that a Ͻ b. Show that if p 0 Ͻ b Ϫ a, then
͕ pn ͖ is increasing and 0 Ͻ pn Ͻ b Ϫ a. Show also that
if p 0 Ͼ b Ϫ a, then ͕ pn ͖ is decreasing and pn Ͼ b Ϫ a.
Deduce that if a Ͻ b, then lim n l ϱ pn ෇ b Ϫ a.

1
2 ϩ иии

SERIES
If we try to add the terms of an infinite sequence ͕a n ͖ϱn෇1 we get an expression of the
form
a1 ϩ a2 ϩ a3 ϩ и и и ϩ an ϩ и и и

1

which is called an infinite series (or just a series) and is denoted, for short, by the
symbol
ϱ

͚a

n෇1

n

or

͚a

n

But does it make sense to talk about the sum of infinitely many terms?
It would be impossible to find a finite sum for the series
1 ϩ 2 ϩ 3 ϩ 4 ϩ 5 ϩ иии ϩ n ϩ иии

n

Sum of first n terms

1
2
3
4
5
6
7
10
15
20
25

0.50000000
0.75000000
0.87500000
0.93750000
0.96875000
0.98437500
0.99218750

0.99902344
0.99996948
0.99999905
0.99999997

because if we start adding the terms we get the cumulative sums 1, 3, 6, 10, 15,
21, . . . and, after the nth term, we get n͑n ϩ 1͒͞2, which becomes very large as n
increases.
However, if we start to add the terms of the series
1
1
1
1
1
1
1
ϩ ϩ ϩ
ϩ
ϩ
ϩ иии ϩ n ϩ иии
2
4
8
16
32
64
2
we get 2 , 4 , 8 , 16 , 32 , 64 , . . . , 1 Ϫ 1͞2 n, . . . . The table shows that as we add more and
more terms, these partial sums become closer and closer to 1. In fact, by adding sufficiently many terms of the series we can make the partial sums as close as we like
to 1. So it seems reasonable to say that the sum of this infinite series is 1 and to write

1

3

7

15 31 63

ϱ

͚

n෇1

1
1
1
1
1
1
෇ ϩ ϩ ϩ
ϩ иии ϩ n ϩ иии ෇ 1
2n
2
4
8
16
2

SECTION 8.2

SERIES

■

421

We use a similar idea to determine whether or not a general series (1) has a sum.
We consider the partial sums
s1 ෇ a 1
s2 ෇ a 1 ϩ a 2
s3 ෇ a 1 ϩ a 2 ϩ a 3
s4 ෇ a 1 ϩ a 2 ϩ a 3 ϩ a 4
and, in general,
n

sn ෇ a 1 ϩ a 2 ϩ a 3 ϩ и и и ϩ a n ෇

͚a

i

i෇1

These partial sums form a new sequence ͕sn ͖, which may or may not have a limit. If
lim n l ϱ sn ෇ s exists (as a finite number), then, as in the preceding example, we call it
the sum of the infinite series ͸ a n .
Given a series ͸ϱn෇1 a n ෇ a 1 ϩ a 2 ϩ a 3 ϩ и и и , let sn denote
its nth partial sum:

2 DEFINITION

n

sn ෇

͚a

i

෇ a1 ϩ a2 ϩ и и и ϩ an

i෇1

If the sequence ͕sn ͖ is convergent and lim n l ϱ sn ෇ s exists as a real number,
then the series ͸ a n is called convergent and we write
ϱ

a1 ϩ a2 ϩ и и и ϩ an ϩ и и и ෇ s

͚a

or

n

෇s

n෇1

The number s is called the sum of the series. If the sequence ͕sn ͖ is divergent,
then the series is called divergent.
■

Compare with the improper integral

y

ϱ

1

f ͑x͒ dx ෇ lim
tlϱ

y

t

1

f ͑x͒ dx

Thus the sum of a series is the limit of the sequence of partial sums. So when we
write ͸ϱn෇1 an ෇ s we mean that by adding sufficiently many terms of the series we can
get as close as we like to the number s. Notice that

To find this integral we integrate from 1
to t and then let t l ϱ. For a series, we
sum from 1 to n and then let n l ϱ.

ϱ

n

͚a

n

n෇1

෇ lim

͚a

n l ϱ i෇1

i

EXAMPLE 1 An important example of an infinite series is the geometric series

a ϩ ar ϩ ar 2 ϩ ar 3 ϩ и и и ϩ ar nϪ1 ϩ и и и ෇

ϱ

͚ ar

nϪ1

a

0

n෇1

Each term is obtained from the preceding one by multiplying it by the common
ratio r. (We have already considered the special case where a ෇ 12 and r ෇ 12 on
page 420.)
If r ෇ 1, then sn ෇ a ϩ a ϩ и и и ϩ a ෇ na l Ϯϱ. Since lim n l ϱ sn doesn’t
exist, the geometric series diverges in this case.
If r 1, we have
sn ෇ a ϩ ar ϩ ar 2 ϩ и и и ϩ ar nϪ1
and

rsn ෇

ar ϩ ar 2 ϩ и и и ϩ ar nϪ1 ϩ ar n


422

■

CHAPTER 8

SERIES

■ Figure 1 provides a geometric demonstration of the result in Example 1. If
the triangles are constructed as shown
and s is the sum of the series, then, by

similar triangles,

s
a
෇
a
a Ϫ ar

a
s෇
1Ϫr

so

Subtracting these equations, we get
sn Ϫ rsn ෇ a Ϫ ar n
sn ෇

3

a͑1 Ϫ r n ͒
1Ϫr

If Ϫ1 Ͻ r Ͻ 1, we know from (8.1.8) that r n l 0 as n l ϱ, so
ar#
ar@

lim sn ෇ lim

nlϱ

ar@

Խ Խ

ar
a-ar

nlϱ

a͑1 Ϫ r n ͒
a
a
a
෇
Ϫ
lim r n ෇
1Ϫr
1Ϫr
1 Ϫ r nlϱ
1Ϫr

ar
s

Thus when r Ͻ 1 the geometric series is convergent and its sum is a͑͞1 Ϫ r͒.
If r ഛ Ϫ1 or r Ͼ 1, the sequence ͕r n ͖ is divergent by (8.1.8) and so, by Equation 3,
lim n l ϱ sn does not exist. Therefore, the geometric series diverges in those cases. ■
We summarize the results of Example 1 as follows.

a

a

The geometric series

4

ϱ

͚ ar

a

nϪ1

෇ a ϩ ar ϩ ar 2 ϩ и и и

n෇1

Խ Խ

is convergent if r Ͻ 1 and its sum is

FIGURE 1

ϱ

͚ ar

In words: The sum of a convergent
geometric series is
■

first term
1 Ϫ common ratio

nϪ1

a
1Ϫr

෇

n෇1

ԽrԽ Ͻ 1

Խ Խ

If r ജ 1, the geometric series is divergent.
Find the sum of the geometric series

V EXAMPLE 2

5 Ϫ 103 ϩ 209 Ϫ 40
27 ϩ и и и
SOLUTION The first term is a ෇ 5 and the common ratio is r ෇ Ϫ 3 . Since
2

ԽrԽ ෇

2
3

Ͻ 1, the series is convergent by (4) and its sum is
5Ϫ

■ What do we really mean when we
say that the sum of the series in Example 2 is 3? Of course, we can’t literally
add an infinite number of terms, one by
one. But, according to Definition 2, the
total sum is the limit of the sequence
of partial sums. So, by taking the sum
of sufficiently many terms, we can get
as close as we like to the number 3.
The table shows the first ten partial
sums sn and the graph in Figure 2 shows
how the sequence of partial sums
approaches 3.

n

sn

1
2
3
4
5

6
7
8
9
10

5.000000
1.666667
3.888889
2.407407
3.395062
2.736626
3.175583
2.882945
3.078037
2.947975

10
20
40
5
5
ϩ
Ϫ
ϩ иии ෇
෇ 5 ෇3
2
3
9
27

1 Ϫ (Ϫ 3 )
3

■

sn

3

0

FIGURE 2

20 n


SECTION 8.2

SERIES

■

423

ϱ

EXAMPLE 3 Is the series

͚2

2n

3 1Ϫn convergent or divergent?

n෇1

SOLUTION Let’s rewrite the nth term of the series in the form ar nϪ1:
Another way to identify a and r is to
write out the first few terms:
■

4ϩ

16
3

ϱ

͚2

ϩ 649 ϩ и и и

2n

3 1Ϫn ෇

n෇1

ϱ

͚ ͑2 ͒ 3
2 n

Ϫ͑nϪ1͒

෇

n෇1

ϱ

͚

n෇1

ϱ
4n
nϪ1
෇
4 ( 43 )
͚
3 nϪ1
n෇1

We recognize this series as a geometric series with a ෇ 4 and r ෇ 43 . Since r Ͼ 1,
the series diverges by (4).
■
V EXAMPLE 4

Write the number 2.317 ෇ 2.3171717. . . as a ratio of integers.

SOLUTION

2.3171717. . . ෇ 2.3 ϩ

Module 8.2 explores a series
that depends on an angle ␪ in
a triangle and enables you to
see how rapidly the series
converges when ␪ varies.

17
17
17
ϩ иии
3 ϩ
5 ϩ
10
10
10 7

After the first term we have a geometric series with a ෇ 17͞10 3 and r ෇ 1͞10 2.
Therefore
17
17
3
10
1000
2.317 ෇ 2.3 ϩ
෇ 2.3 ϩ

1
99
1Ϫ 2
10
100
23
17
1147
෇
ϩ
෇
10
990
495

■

ϱ

EXAMPLE 5 Find the sum of the series

͚ x , where Խ x Խ Ͻ 1.
n

n෇0

SOLUTION Notice that this series starts with n ෇ 0 and so the first term is x 0 ෇ 1.

(With series, we adopt the convention that x 0 ෇ 1 even when x ෇ 0.) Thus
ϱ

͚x

n

෇ 1 ϩ x ϩ x2 ϩ x3 ϩ x4 ϩ и и и

n෇0

Խ Խ Խ Խ

This is a geometric series with a ෇ 1 and r ෇ x. Since r ෇ x Ͻ 1, it converges
and (4) gives
ϱ

͚x

5

n෇0
ϱ

EXAMPLE 6 Show that the series

͚

n෇1

n

෇

1
1Ϫx

1
is convergent, and find its sum.
n͑n ϩ 1͒

SOLUTION This is not a geometric series, so we go back to the definition of a convergent series and compute the partial sums.
n

sn ෇

͚

i෇1

1
1
1
1
1
෇
ϩ
ϩ
ϩ иии ϩ
i͑i ϩ 1͒
1ؒ2
2ؒ3

3ؒ4
n͑n ϩ 1͒

We can simplify this expression if we use the partial fraction decomposition
1
1
1
෇ Ϫ
i͑i ϩ 1͒
i
iϩ1

■


424

■

CHAPTER 8

SERIES

(see Section 6.3). Thus we have
n

sn ෇

Notice that the terms cancel in pairs.
This is an example of a telescoping

sum: Because of all the cancellations,
the sum collapses (like a pirate’s collapsing telescope) into just two terms.
■

͚

i෇1

n
1
෇͚
i͑i ϩ 1͒
i෇1

1
1
Ϫ
i
iϩ1

1
2

1
1
Ϫ
2
3

ϩ

1
nϩ1
lim sn ෇ lim

and so

nlϱ

nlϱ

1
1
Ϫ
3
4

ϩ

ͩ

1Ϫ

1
nϩ1

ͪ

ϩ иии ϩ

ͩ

1
1
Ϫ
n
nϩ1

ͪ

෇1Ϫ0෇1

Therefore, the given series is convergent and
ϱ

͚

n෇1

1

ͪ

ͩ ͪ ͩ ͪ ͩ ͪ

෇ 1Ϫ
෇1Ϫ

■ Figure 3 illustrates Example 6 by
showing the graphs of the sequence

of terms a n ෇ 1͞[n͑n ϩ 1͒] and the
sequence ͕sn ͖ of partial sums. Notice
that a n l 0 and sn l 1. See Exercises 36 and 37 for two geometric interpretations of Example 6.

ͩ

1
෇1
n͑n ϩ 1͒

■

Show that the harmonic series

V EXAMPLE 7

͕sn ͖

ϱ

͚

n෇1

1
1
1
1
෇ 1 ϩ ϩ ϩ ϩ иии
n

2
3
4

is divergent.
SOLUTION For this particular series it’s convenient to consider the partial sums s2 ,
s4 , s8 , s16 , s32 , . . . and show that they become large.

͕a n ͖
0

n

s2 ෇ 1 ϩ 12
FIGURE 3

s4 ෇ 1 ϩ 12 ϩ ( 13 ϩ 14 ) Ͼ 1 ϩ 12 ϩ ( 14 ϩ 14 ) ෇ 1 ϩ 22
s8 ෇ 1 ϩ 12 ϩ ( 13 ϩ 14 ) ϩ ( 15 ϩ 16 ϩ 17 ϩ 18 )
Ͼ 1 ϩ 12 ϩ ( 14 ϩ 14 ) ϩ ( 18 ϩ 18 ϩ 18 ϩ 18 )
෇ 1 ϩ 12 ϩ 12 ϩ 12 ෇ 1 ϩ 32
s16 ෇ 1 ϩ 12 ϩ ( 13 ϩ 14 ) ϩ ( 15 ϩ и и и ϩ 18 ) ϩ ( 19 ϩ и и и ϩ 161 )
Ͼ 1 ϩ 12 ϩ ( 14 ϩ 14 ) ϩ ( 18 ϩ и и и ϩ 18 ) ϩ ( 161 ϩ и и и ϩ 161 )
෇ 1 ϩ 12 ϩ 12 ϩ 12 ϩ 12 ෇ 1 ϩ 42
Similarly, s32 Ͼ 1 ϩ 52 , s64 Ͼ 1 ϩ 62 , and in general
s2 n Ͼ 1 ϩ

The method used in Example 7 for
showing that the harmonic series diverges is
due to the French scholar Nicole Oresme
(1323–1382).

■

n
2

This shows that s2 n l ϱ as n l ϱ and so ͕sn ͖ is divergent. Therefore, the harmonic
series diverges.
■
ϱ

6 THEOREM

If the series

͚a

n෇1

n

is convergent, then lim an ෇ 0.
nlϱ


SECTION 8.2

■

SERIES

425

PROOF Let sn ෇ a 1 ϩ a 2 ϩ и и и ϩ a n . Then a n ෇ sn Ϫ snϪ1. Since ͸ a n is convergent, the sequence ͕sn ͖ is convergent. Let lim n l ϱ sn ෇ s. Since n Ϫ 1 l ϱ as
n l ϱ, we also have lim n l ϱ snϪ1 ෇ s. Therefore

lim a n ෇ lim ͑sn Ϫ snϪ1 ͒ ෇ lim sn Ϫ lim snϪ1 ෇ s Ϫ s ෇ 0

nlϱ

nlϱ

nlϱ

■

nlϱ

NOTE 1 With any series ͸ a n we associate two sequences: the sequence ͕sn ͖ of its
partial sums and the sequence ͕a n ͖ of its terms. If ͸ a n is convergent, then the limit of
the sequence ͕sn ͖ is s (the sum of the series) and, as Theorem 6 asserts, the limit of the
sequence ͕a n ͖ is 0.

|

NOTE 2 The converse of Theorem 6 is not true in general. If lim n l ϱ a n ෇ 0, we
cannot conclude that ͸ a n is convergent. Observe that for the harmonic series ͸ 1͞n
we have a n ෇ 1͞n l 0 as n l ϱ, but we showed in Example 7 that ͸ 1͞n is divergent.

If lim a n does not exist or if lim a n

7 THE TEST FOR DIVERGENCE

nlϱ

ϱ

then the series

͚a

n

0,

nlϱ

is divergent.

n෇1

The Test for Divergence follows from Theorem 6 because, if the series is not divergent, then it is convergent, and so lim n l ϱ a n ෇ 0.
ϱ

n2
diverges.
5n 2 ϩ 4

͚

EXAMPLE 8 Show that the series

n෇1

SOLUTION

lim a n ෇ lim

nlϱ

nlϱ

n2
1
1
෇ lim
෇
n l ϱ 5 ϩ 4͞n 2
5n ϩ 4
5

0

2

■

So the series diverges by the Test for Divergence.

NOTE 3 If we find that lim n l ϱ a n
0, we know that ͸ a n is divergent. If we find

that lim n l ϱ a n ෇ 0, we know nothing about the convergence or divergence of ͸ a n .
Remember the warning in Note 2: If lim n l ϱ a n ෇ 0, the series ͸ a n might converge or
it might diverge.
8 THEOREM If ͸ a n and ͸ bn are convergent series, then so are the series
͸ ca n (where c is a constant), ͸ ͑a n ϩ bn ͒, and ͸ ͑a n Ϫ bn ͒, and
ϱ

(i)

͚

ca n ෇ c

n෇1

͚ ͑a

n෇1

͚

ϱ

an

(ii)

n෇1

ϱ

(iii)

ϱ

n

Ϫ bn ͒ ෇

͚

n෇1
ϱ

͚a

n෇1

n

Ϫ

͑a n ϩ bn ͒ ෇

ϱ

͚

n෇1

an ϩ

ϱ

͚b

n

n෇1

ϱ

͚b

n

n෇1

These properties of convergent series follow from the corresponding Limit Laws
for Sequences in Section 8.1. For instance, here is how part (ii) of Theorem 8 is
proved:


426

■

CHAPTER 8

SERIES

Let
ϱ

n

sn ෇

͚a

͚a

s෇

i

i෇1

n

͚b

tn ෇

n

t෇

i

n෇1

i෇1

ϱ

͚b

n

n෇1

The nth partial sum for the series ͸ ͑a n ϩ bn ͒ is
n

͚ ͑a

un ෇

ϩ bi ͒

i

i෇1

and, using Equation 5.2.10, we have
n

lim u n ෇ lim

nlϱ

i

n l ϱ i෇1

ϩ bi ͒ ෇ lim

nlϱ

n

෇ lim

n

ai ϩ

i෇1

͚b

ͪ

i

i෇1

n

͚a

i

n l ϱ i෇1

͚ͩ
n

͚ ͑a

͚b

ϩ lim

n l ϱ i෇1

i

෇ lim sn ϩ lim tn ෇ s ϩ t
nlϱ

nlϱ

Therefore, ͸ ͑a n ϩ bn ͒ is convergent and its sum is
ϱ

͚ ͑a

n

ϱ

͚a

ϩ bn ͒ ෇ s ϩ t ෇

n෇1

ϩ

n෇1
ϱ

͚

EXAMPLE 9 Find the sum of the series

n෇1

SOLUTION The series

n

ͩ

ϱ

͚b

n

■

n෇1

ͪ

3
1
ϩ n .
n͑n ϩ 1͒
2

͸ 1͞2 n is a geometric series with a ෇ 12 and r ෇ 12 , so
ϱ

͚

n෇1

1
1
2
෇
෇1
2n
1 Ϫ 12

In Example 6 we found that

ϱ

͚

n෇1

1
෇1
n͑n ϩ 1͒

So, by Theorem 8, the given series is convergent and
ϱ

͚

n෇1

ͩ

3
1
ϩ n
n͑n ϩ 1͒
2

ͪ

෇3

ϱ

ϱ
1
1
ϩ ͚ n ෇3ؒ1ϩ1෇4
n͑n ϩ 1͒
n෇1 2

͚

n෇1

■

NOTE 4 A finite number of terms doesn’t affect the convergence or divergence of
a series. For instance, suppose that we were able to show that the series
ϱ

͚

n෇4

n
n3 ϩ 1

is convergent. Since
ϱ

͚

n෇1

ϱ
n
1
2
3
n
෇ ϩ ϩ
ϩ ͚ 3
n ϩ1
2
9
28
n෇4 n ϩ 1
3

it follows that the entire series ͸ϱn෇1 n͑͞n 3 ϩ 1͒ is convergent. Similarly, if it is known
that the series ͸ϱn෇Nϩ1 a n converges, then the full series
ϱ

͚a

n෇1

is also convergent.

N

n

෇

͚a

n෇1

n

ϩ

ϱ

͚

n෇Nϩ1

an


SECTION 8.2

8.2

(b) What is a convergent series? What is a divergent series?

27.

͚

n෇1

3– 8 ■ Determine whether the geometric series is convergent or
divergent. If it is convergent, find its sum.

ϱ

29.

͚

n෇0

ϩ 209 Ϫ 40
27 ϩ и и и

■

4. 1 ϩ 0.4 ϩ 0.16 ϩ 0.064 ϩ и и и
ϱ

5.

2 nϪ1
3

6.

n෇1

7.

n෇0

͚

␲
3 nϩ1

■

■

■

͚

n෇1

8.
■

cos n x
2n
■

■

■

n෇1

1

͚ (s2 )

n

■

■

10.

͚

k2
2
k Ϫ1

12.

ϱ

1ϩ2
3n

͚
ϱ

11.

k෇2

13.

ϱ

1
2n

n෇1

͚

n෇1

ϱ

ϱ

n

14.

■

n

16.

■

■

■

■

■

ϱ

1ϩ3
2n

n

nϪ1

͚ ͑cos 1͒

Ϫ ͑0.3͒ n ͔

k

■

n෇2
ϱ

21.

͚

n෇1
■

20.

3
n͑n ϩ 3͒

22.

■

23–26

ϱ

2
n Ϫ1
2

■

■

͚

n෇1

■

■

■

2
n 2 ϩ 4n ϩ 3

ϱ

n

͚ ln n ϩ 1

n෇1
■

■

■

■

■

■

■

■

34. A certain ball has the property that each time it falls from a

■

height h onto a hard, level surface, it rebounds to a height
rh, where 0 Ͻ r Ͻ 1. Suppose that the ball is dropped from
an initial height of H meters.
(a) Assuming that the ball continues to bounce indefinitely,
find the total distance that it travels.
(b) Calculate the total time that the ball travels. (Use the
fact that the ball falls 12 tt 2 meters in t seconds.)
(c) Suppose that each time the ball strikes the surface
with velocity v it rebounds with velocity Ϫk v, where

Express the number as a ratio of integers.

24. 0.73 ෇ 0.73737373 . . .
25. 3.417 ෇ 3.417417417 . . .
26. 6.254 ෇ 6.2545454 . . .
■

■

■

■

■

■

■

■

■

■

receive the money also spend some of it. The people receiving some of the twice-spent money will spend some of that,
and so on. Economists call this chain reaction the multiplier
effect. In a hypothetical isolated community, the local government begins the process by spending D dollars. Suppose
that each recipient of spent money spends 100c% and saves
100s% of the money that he or she receives. The values c
and s are called the marginal propensity to consume and the
marginal propensity to save and, of course, c ϩ s ෇ 1.
(a) Let Sn be the total spending that has been generated after
n transactions. Find an equation for Sn .
(b) Show that lim n l ϱ Sn ෇ kD, where k ෇ 1͞s. The number
k is called the multiplier. What is the multiplier if the
marginal propensity to consume is 80%?
Note: The federal government uses this principle to justify
deficit spending. Banks use this principle to justify lending a large percentage of the money that they receive in
deposits.

■

23. 0.2 ෇ 0.2222 . . .

■

nϪ1
nϩ1

find a n and ͸ϱn෇1 a n .

■ Determine whether the series is convergent or divergent by expressing sn as a telescoping sum (as in Example 6).
If it is convergent, find its sum.

͚

■

1
n

33. When money is spent on goods and services, those that

͚ ͓͑0.8͒

■

1ϩ

k͑k ϩ 2͒

͑k ϩ 3͒2

19–22

19.

■

32. If the nth partial sum of a series ͸ϱn෇1 a n is sn ෇ 3 Ϫ n 2 Ϫn,

k෇1
■

■

31. If the nth partial sum of a series ͸ϱn෇1 a n is

find a n and ͸ϱn෇1 a n .

ϱ

18.

n෇1

■

■

ͩ ͪ

sn ෇

n෇1

͚ arctan n

■

nϩ1
2n Ϫ 3

ϱ

͚ s2

■

͚

n෇1

ϱ

■

͚

k෇1

n෇1

17.

͚

n෇1

ϱ

15.

■

is another series with this property.

■

■ Determine whether the series is convergent or divergent. If it is convergent, find its sum.

9.

■

ϱ

9–18

ϱ

2 n͑x ϩ 1͒ n

n෇0

͚ ln

n෇0
■

͚

28.

whose terms approach 0. Show that

͑Ϫ6͒
5 nϪ1

nϪ1

ϱ

n

■

■

ϱ

xn
3n

30. We have seen that the harmonic series is a divergent series
ϱ

͚ 5( )
ϱ

427

27–29 ■ Find the values of x for which the series converges.
Find the sum of the series for those values of x.
ϱ

2. Explain what it means to say that ͸ϱn෇1 a n ෇ 5.

10
3

■

EXERCISES

1. (a) What is the difference between a sequence and a series?

3. 5 Ϫ

SERIES

428

■

CHAPTER 8

SERIES

0 Ͻ k Ͻ 1. How long will it take for the ball to come
to rest?
ϱ

35. What is the value of c if

͚

39. What is wrong with the following calculation?

0 ෇ 0 ϩ 0 ϩ 0 ϩ иии

͑1 ϩ c͒Ϫn ෇ 2?

෇ ͑1 Ϫ 1͒ ϩ ͑1 Ϫ 1͒ ϩ ͑1 Ϫ 1͒ ϩ и и и

n෇2

෇ 1 Ϫ 1 ϩ 1 Ϫ 1 ϩ 1 Ϫ 1 ϩ иии

n

; 36. Graph the curves y ෇ x , 0 ഛ x ഛ 1, for n ෇ 0, 1, 2, 3,

4, . . . on a common screen. By finding the areas between
successive curves, give a geometric demonstration of the
fact, shown in Example 6, that
ϱ

͚

n෇1

෇ 1 ϩ ͑Ϫ1 ϩ 1͒ ϩ ͑Ϫ1 ϩ 1͒ ϩ ͑Ϫ1 ϩ 1͒ ϩ и и и
෇ 1 ϩ 0 ϩ 0 ϩ 0 ϩ иии ෇ 1
(Guido Ubaldus thought that this proved the existence of
God because “something has been created out of nothing.”)

1
෇1
n͑n ϩ 1͒

40. Suppose that ͸ϱn෇1 a n ͑a n
37. The figure shows two circles C and D of radius 1 that touch

at P. T is a common tangent line; C1 is the circle that
touches C, D, and T ; C2 is the circle that touches C, D,
and C1; C3 is the circle that touches C, D, and C2. This
procedure can be continued indefinitely and produces an
infinite sequence of circles ͕Cn ͖. Find an expression for
the diameter of Cn and thus provide another geometric
demonstration of Example 6.

0͒ is known to be a convergent
series. Prove that ͸ϱn෇1 1͞a n is a divergent series.

41. Prove part (i) of Theorem 8.
42. If ͸ a n is divergent and c

0, show that ͸ ca n is divergent.

43. If ͸ a n is convergent and ͸ bn is divergent, show that

the series ͸ ͑a n ϩ bn ͒ is divergent. [Hint: Argue by
contradiction.]

44. If ͸ a n and ͸ bn are both divergent, is ͸ ͑a n ϩ bn ͒ neces-

sarily divergent?
45. Suppose that a series ͸ a n has positive terms and its partial

P

sums sn satisfy the inequality sn ഛ 1000 for all n. Explain
why ͸ a n must be convergent.

C£
C™

1

1

46. The Fibonacci sequence was defined in Section 8.1 by the

C

D

C¡

equations
T

Խ

Խ

38. A right triangle ABC is given with ЄA ෇ ␪ and AC ෇ b.

CD is drawn perpendicular to AB, DE is drawn perpendicular to BC, EF Ќ AB, and this process is continued indefinitely as shown in the figure. Find the total length of all the
perpendiculars

f1 ෇ 1,

(a)

1
1
1
෇
Ϫ
fnϪ1 fnϩ1

fnϪ1 fn
fn fnϩ1

(b)

͚

1
෇1
fnϪ1 fnϩ1

ϱ

fn
෇2
fnϪ1 fnϩ1

ϱ

͚

n෇2

D

¨

F
H

B

n෇2

(c)

A

b

G

E

C

fn ෇ fnϪ1 ϩ fnϪ2

nജ3

Show that each of the following statements is true.

Խ CD Խ ϩ Խ DE Խ ϩ Խ EF Խ ϩ Խ FG Խ ϩ и и и
in terms of b and ␪.

f2 ෇ 1,

47. The Cantor set, named after the German mathematician

Georg Cantor (1845–1918), is constructed as follows. We

start with the closed interval [0, 1] and remove the open
interval ( 13 , 23 ). That leaves the two intervals [0, 13 ] and [ 23, 1]
and we remove the open middle third of each. Four intervals
remain and again we remove the open middle third of each
of them. We continue this procedure indefinitely, at each
step removing the open middle third of every interval that
remains from the preceding step. The Cantor set consists of


SECTION 8.3

the numbers that remain in [0, 1] after all those intervals
have been removed.
(a) Show that the total length of all the intervals that are
removed is 1. Despite that, the Cantor set contains infinitely many numbers. Give examples of some numbers
in the Cantor set.
(b) The Sierpinski carpet is a two-dimensional counterpart
of the Cantor set. It is constructed by removing the center one-ninth of a square of side 1, then removing the
centers of the eight smaller remaining squares, and
so on. (The figure shows the first three steps of the
construction.) Show that the sum of the areas of the
removed squares is 1. This implies that the Sierpinski
carpet has area 0.

THE INTEGRAL AND COMPARISON TESTS

■

429

49. Consider the series
ϱ

͚

n෇1

n
͑n ϩ 1͒!

(a) Find the partial sums s1, s2, s3, and s4. Do you recognize
the denominators? Use the pattern to guess a formula
for sn .
(b) Use mathematical induction to prove your guess.
(c) Show that the given infinite series is convergent, and
find its sum.
50. In the figure there are infinitely many circles approaching

the vertices of an equilateral triangle, each circle touching
other circles and sides of the triangle. If the triangle has
sides of length 1, find the total area occupied by the circles.

48. (a) A sequence ͕a n ͖ is defined recursively by the equation

a n ෇ 12 ͑a nϪ1 ϩ a nϪ2 ͒ for n ജ 3, where a 1 and a 2 can be
any real numbers. Experiment with various values of a 1
and a 2 and use your calculator to guess the limit of the
sequence.
(b) Find lim n l ϱ a n in terms of a 1 and a 2 by expressing
a nϩ1 Ϫ a n in terms of a 2 Ϫ a 1 and summing a series.

8.3

THE INTEGRAL AND COMPARISON TESTS
In general, it is difficult to find the exact sum of a series. We were able to accomplish
this for geometric series and the series ͸ 1͓͞n͑n ϩ 1͔͒ because in each of those cases
we could find a simple formula for the nth partial sum sn . But usually it isn’t easy
to compute lim n l ϱ sn . Therefore, in this section and the next we develop tests that
enable us to determine whether a series is convergent or divergent without explicitly
finding its sum.
In this section we deal only with series with positive terms, so the partial sums are
increasing. In view of the Monotonic Sequence Theorem, to decide whether a series
is convergent or divergent, we need to determine whether the partial sums are bounded
or not.

n

n

sn ෇

͚

i෇1

5
10
50
100
500

1000
5000

1
i2

1.4636
1.5498
1.6251
1.6350
1.6429
1.6439
1.6447

TESTING WITH AN INTEGRAL

Let’s investigate the series whose terms are the reciprocals of the squares of the positive integers:
ϱ
1
1
1
1
1
1
͚
2 ෇
2 ϩ
2 ϩ
2 ϩ
2 ϩ

2 ϩ иии
n
1
2
3
4
5
n෇1
There’s no simple formula for the sum sn of the first n terms, but the computergenerated table of values given in the margin suggests that the partial sums are approaching a number near 1.64 as n l ϱ and so it looks as if the series is convergent.


430

■

CHAPTER 8

SERIES

We can confirm this impression with a geometric argument. Figure 1 shows the
curve y ෇ 1͞x 2 and rectangles that lie below the curve. The base of each rectangle is
an interval of length 1; the height is equal to the value of the function y ෇ 1͞x 2 at the
right endpoint of the interval. So the sum of the areas of the rectangles is
ϱ
1
1
1
1
1
1

2 ϩ
2 ϩ
2 ϩ
2 ϩ
2 ϩ иии ෇ ͚
2
1
2
3
4
5
n
n෇1

y

y=

1
≈

area= 1
1@
0

1

2

area= 1

2@

FIGURE 1

3

area= 1
3@

4

area= 1
4@

x

5

area= 1
5@

If we exclude the first rectangle, the total area of the remaining rectangles is
smaller than the area under the curve y ෇ 1͞x 2 for x ജ 1, which is the value of the
integral x1ϱ ͑1͞x 2 ͒ dx. In Section 6.6 we discovered that this improper integral is convergent and has value 1. So the picture shows that all the partial sums are less than
1
ϱ 1
ϩ y 2 dx ෇ 2
1 x
12
Thus the partial sums are bounded and the series converges. The sum of the series (the

limit of the partial sums) is also less than 2:
ϱ

1
1
1
1
1
ϩ иии Ͻ 2
2 ෇
2 ϩ
2 ϩ
2 ϩ
n
1
2
3
42

͚

n෇1

[The exact sum of this series was found by the Swiss mathematician Leonhard Euler
(1707–1783) to be ␲ 2͞6, but the proof of this fact is beyond the scope of this book.]
Now let’s look at the series
ϱ

͚

n෇1

n

n

sn ෇

͚

i෇1

5
10
50
100
500
1000
5000

1
1
1
1
1
1
෇
ϩ
ϩ
ϩ

ϩ
ϩ иии
sn
s1
s2
s3
s4
s5

The table of values of sn suggests that the partial sums aren’t approaching a finite
number, so we suspect that the given series may be divergent. Again we use a picture
for confirmation. Figure 2 shows the curve y ෇ 1͞sx , but this time we use rectangles
whose tops lie above the curve.

1
si

3.2317
5.0210
12.7524
18.5896
43.2834
61.8010
139.9681

FIGURE 2

y

y= 1

x
œ„

0

1

2

area= 1
1
œ„

3

area= 1
2
œ„

4

area= 1
3
œ„

5

area= 1
4
œ„

x


SECTION 8.3

THE INTEGRAL AND COMPARISON TESTS

■

431

The base of each rectangle is an interval of length 1. The height is equal to the value
of the function y ෇ 1͞sx at the left endpoint of the interval. So the sum of the areas
of all the rectangles is
ϱ
1
1
1
1
1
1
ϩ
ϩ
ϩ
ϩ
ϩ иии ෇ ͚
s1
s2
s3

s4
s5
n෇1 sn

This total area is greater than the area under the curve y ෇ 1͞sx for x ജ 1, which is
equal to the integral x1ϱ (1͞sx ) dx . But we know from Section 6.6 that this improper
integral is divergent. In other words, the area under the curve is infinite. So the sum of
the series must be infinite, that is, the series is divergent.
The same sort of geometric reasoning that we used for these two series can be used
to prove the following test. (The proof is given at the end of this section.)
THE INTEGRAL TEST Suppose f is a continuous, positive, decreasing function
on ͓1, ϱ͒ and let a n ෇ f ͑n͒. Then the series ͸ϱn෇1 a n is convergent if and only if
the improper integral x1ϱ f ͑x͒ dx is convergent. In other words:
ϱ

(a) If y f ͑x͒ dx is convergent, then
1

ϱ

(b) If y f ͑x͒ dx is divergent, then
1

ϱ

͚a

n

is convergent.

n෇1
ϱ

͚a

n

is divergent.

n෇1

NOTE When we use the Integral Test it is not necessary to start the series or the integral at n ෇ 1. For instance, in testing the series
ϱ

͚

n෇4

1
͑n Ϫ 3͒2

y

we use

ϱ

4

1
dx
͑x Ϫ 3͒2

Also, it is not necessary that f be always decreasing. What is important is that f be
ultimately decreasing, that is, decreasing for x larger than some number N. Then
͸ϱn෇N a n is convergent, so ͸ϱn෇1 a n is convergent by Note 4 of Section 8.2.
ϱ

V EXAMPLE 1

Determine whether the series

͚

n෇1

ln n
converges or diverges.
n

SOLUTION The function f ͑x͒ ෇ ͑ln x͒͞x is positive and continuous for x Ͼ 1

because the logarithm function is positive and continuous there. But it is not obvious
whether or not f is decreasing, so we compute its derivative:
f Ј͑x͒ ෇
■ In order to use the Integral Test we
need to be able to evaluate x1ϱ f ͑x͒ dx
and therefore we have to be able to find
an antiderivative of f . Frequently this is

difficult or impossible, so we need other
tests for convergence too.

x͑1͞x͒ Ϫ ln x
1 Ϫ ln x
෇
x2
x2

Thus f Ј͑x͒ Ͻ 0 when ln x Ͼ 1, that is, x Ͼ e. It follows that f is decreasing when
x Ͼ e and so we can apply the Integral Test:

y

ϱ

1

ln x
t ln x
͑ln x͒2
dx ෇ lim y
dx ෇ lim
tlϱ 1
tlϱ
x
x
2

ͬ

t

1

͑ln t͒
෇ϱ
2
2

෇ lim

tlϱ

Since this improper integral is divergent, the series ͸ ͑ln n͒͞n is also divergent by
■
the Integral Test.


432

■

CHAPTER 8

SERIES
ϱ

V EXAMPLE 2

For what values of p is the series

͚

n෇1

1
convergent?
np

SOLUTION If p Ͻ 0, then lim n l ϱ ͑1͞n ͒ ෇ ϱ. If p ෇ 0, then lim n l ϱ ͑1͞n p ͒ ෇ 1.
p

In either case lim n l ϱ ͑1͞n p ͒ 0, so the given series diverges by the Test for
Divergence [see (8.2.7)].
If p Ͼ 0, then the function f ͑x͒ ෇ 1͞x p is clearly continuous, positive, and
decreasing on ͓1, ϱ͒. We found in Chapter 6 [see (6.6.2)] that

■ Exercises 29–34 show how to
estimate the sum of a series that is
convergent by the Integral Test.

y

ϱ

1

1
dx

xp

converges if p Ͼ 1 and diverges if p ഛ 1

It follows from the Integral Test that the series ͸ 1͞n p converges if p Ͼ 1 and
diverges if 0 Ͻ p ഛ 1. (For p ෇ 1, this series is the harmonic series discussed in
Example 7 in Section 8.2.)
■
The series in Example 2 is called the p-series. It is important in the rest of this
chapter, so we summarize the results of Example 2 for future reference as follows.
ϱ

1

The p-series

͚

n෇1

1
is convergent if p Ͼ 1 and divergent if p ഛ 1.
np

For instance, the series
ϱ

͚

n෇1

1
1
1
1
1
ϩ иии
3 ෇
3 ϩ
3 ϩ
3 ϩ
n
1
2
3
43

is convergent because it is a p-series with p ෇ 3 Ͼ 1. But the series
ϱ

͚

n෇1

ϱ
1
1
1
1
1

෇
෇ 1 ϩ 3 ϩ 3 ϩ 3 ϩ иии
͚
1͞3
3
n
s2
s3
s4
n෇1 sn

is divergent because it is a p-series with p ෇ 13 Ͻ 1.
TESTING BY COMPARING

The series
ϱ

2

͚

n෇1

1
2n ϩ 1

reminds us of the series ͸ϱn෇1 1͞2 n, which is a geometric series with a ෇ 12 and r ෇ 12
and is therefore convergent. Because the series (2) is so similar to a convergent series,
we have the feeling that it too must be convergent. Indeed, it is. The inequality
1

1
Ͻ n
2n ϩ 1
2
shows that our given series (2) has smaller terms than those of the geometric series
and therefore all its partial sums are also smaller than 1 (the sum of the geometric
series). This means that its partial sums form a bounded increasing sequence, which is
convergent. It also follows that the sum of the series is less than the sum of the geometric series:
ϱ
1
Ͻ1
͚
n
2
ϩ1
n෇1


SECTION 8.3

THE INTEGRAL AND COMPARISON TESTS

■

433

Similar reasoning can be used to prove the following test, which applies only to
series whose terms are positive. The first part says that if we have a series whose terms
are smaller than those of a known convergent series, then our series is also convergent.
The second part says that if we start with a series whose terms are larger than those

of a known divergent series, then it too is divergent.
THE COMPARISON TEST Suppose that ͸ a n and ͸ bn are series with positive
terms.
(a) If ͸ bn is convergent and a n ഛ bn for all n, then ͸ a n is also convergent.
(b) If ͸ bn is divergent and a n ജ bn for all n, then ͸ a n is also divergent.

■ It is important to keep in mind the
distinction between a sequence and a
series. A sequence is a list of numbers,
whereas a series is a sum. With every
series ͸ a n there are associated two
sequences: the sequence ͕a n ͖ of terms
and the sequence ͕sn ͖ of partial sums.

Standard Series for Use with the Comparison Test

PROOF

(i) Let
n

sn ෇

͚a

n

tn ෇

i

i෇1

͚b

t෇

i

i෇1

ϱ

͚b

n

n෇1

Since both series have positive terms, the sequences ͕sn ͖ and ͕tn ͖ are increasing
͑snϩ1 ෇ sn ϩ a nϩ1 ജ sn ͒. Also tn l t, so tn ഛ t for all n. Since a i ഛ bi , we have
sn ഛ tn . Thus sn ഛ t for all n. This means that ͕sn ͖ is increasing and bounded above
and therefore converges by the Monotonic Sequence Theorem. Thus ͸ a n converges.
(ii) If ͸ bn is divergent, then tn l ϱ (since ͕tn ͖ is increasing). But a i ജ bi so
sn ജ tn . Thus sn l ϱ. Therefore, ͸ a n diverges.
■
In using the Comparison Test we must, of course, have some known series ͸ bn for
the purpose of comparison. Most of the time we use one of these series:
■
■

A p -series [͸ 1͞n p converges if p Ͼ 1 and diverges if p ഛ 1; see (1)]
A geometric series [͸ ar nϪ1 converges if r Ͻ 1 and diverges if r ജ 1;
see (8.2.4)]

Խ Խ
ϱ

V EXAMPLE 3

Determine whether the series

͚

n෇1

diverges.

Խ Խ

5
converges or
2n ϩ 4n ϩ 3
2

SOLUTION For large n the dominant term in the denominator is 2n 2, so we compare

the given series with the series ͸ 5͑͞2n 2 ͒. Observe that
5
5

Ͻ
2n 2 ϩ 4n ϩ 3
2n 2

because the left side has a bigger denominator. (In the notation of the Comparison
Test, a n is the left side and bn is the right side.) We know that
ϱ

͚

n෇1

5
5
2 ෇
2n
2

ϱ

͚

n෇1

1
n2

is convergent ( p-series with p ෇ 2 Ͼ 1). Therefore
ϱ

͚

n෇1

5
2n 2 ϩ 4n ϩ 3

is convergent by part (a) of the Comparison Test.

■


434

■

CHAPTER 8

SERIES

Although the condition a n ഛ bn or a n ജ bn in the Comparison Test is given for all
n, we need verify only that it holds for n ജ N, where N is some fixed integer, because
the convergence of a series is not affected by a finite number of terms. This is illustrated in the next example.
ϱ

V EXAMPLE 4

Test the series

͚

n෇1

ln n
for convergence or divergence.
n

SOLUTION We used the Integral Test to test this series in Example 1, but we can

also test it by comparing it with the harmonic series. Observe that ln n Ͼ 1 for
n ജ 3 and so
ln n
1
Ͼ
nജ3
n
n
We know that ͸ 1͞n is divergent ( p-series with p ෇ 1). Thus the given series is
divergent by the Comparison Test.

■

NOTE The terms of the series being tested must be smaller than those of a convergent series or larger than those of a divergent series. If the terms are larger than the
terms of a convergent series or smaller than those of a divergent series, then the Comparison Test doesn’t apply. Consider, for instance, the series
ϱ

͚

n෇1

1
2 Ϫ1
n

The inequality
1
1
Ͼ n
2 Ϫ1
2
n

is useless as far as the Comparison Test is concerned because ͸ bn ෇ ͸ ( 12 ) is convergent and a n Ͼ bn . Nonetheless, we have the feeling that ͸ 1͑͞2 n Ϫ 1͒ ought to be
n
convergent because it is very similar to the convergent geometric series ͸ ( 12 ) . In such
cases the following test can be used.
n

THE LIMIT COMPARISON TEST Suppose that ͸ a n and ͸ bn are series with

positive terms. If
■ Exercises 38 and 39 deal with the
cases c ෇ 0 and c ෇ ϱ.

lim

nlϱ

an
෇c

bn

where c is a finite number and c Ͼ 0, then either both series converge or
both diverge.
PROOF Let m and M be positive numbers such that m Ͻ c Ͻ M. Because a n ͞bn is
close to c for large n, there is an integer N such that

mϽ
and so

an
ϽM
bn

mbn Ͻ a n Ͻ Mbn

when n Ͼ N
when n Ͼ N

If ͸ bn converges, so does ͸ Mbn . Thus ͸ a n converges by part (i) of the Comparison Test. If ͸ bn diverges, so does ͸ mbn and part (ii) of the Comparison Test shows
■
that ͸ a n diverges.