Chapter Nine

Chapter Nine
SERIES
In this chapter we look at infinite sums, called
infinite series. We start in Section 9.1 with a
particular type of series, called a geometric
series. Section 9.2 considers general sequences
and series of constants and what it means for
such series to converge. The tests that allow us
to determine convergence are in
Sections 9.2 and 9.3. Section ??, introduces
power series, in which the terms are powers of
. These series converge for some -values and
not for others; the radius of convergence is
introduced to identify the interval on which the
series converges.
426
9.1
Chapter Nine SERIES
GEOMETRIC SERIES
or etc., are called terms in the series. To talk
The individual numbers, , This section introduces infinite series of constants, which are sums of the form
about the sum of the series, we must first explain how to add infinitely many numbers.
Let us look at the repeated administration of a drug. In this example, the terms in the series
represent each dose; the sum of the series represents the drug level in the body in the long run.
Repeated Drug Dosage
A person with an ear infection is told to take antibiotic tablets regularly for several days. Since the
drug is being excreted by the body between doses, how can we calculate the quantity of the drug
remaining in the body at any particular time?
To be specific, let’s suppose the drug is ampicillin (a common antibiotic) taken in 250 mg doses
four times a day (that is, every six hours). It is known that at the end of six hours, about
of the
drug is still in the body. What quantity of the drug is in the body right after the tenth tablet? The
fortieth?
Let
represent the quantity, in milligrams, of ampicillin in the blood right after the
tablet.
Then
!
!
#! "
$&%('
*),+- ./ 10 ),+- mg
) 2 +- 34 5
),+7 mg
26+3- 465
) ! .( 108 +- ) . +- ./ 90: +- 0&.( 10; +- ) 2 +- .( 90 3 4 +- .( 95 0 26+3- 46 5
),+7 mg
) ! .( 108 +- )=<>+- .( 90 +- .( 90: +- ?@./ 90& +- ),2 +- .( 90 +- 3A.(4 90 +- .( 15 0 2B+3A- 46 5
,) +7 7 mg
!
Remnants of first tablet
!
New tablet
Remnants of first and second tablets
New tablet
Remnants of first, second, and third tablets
!#C D) +- .(.( 1100FGH
! FE )D+-
New tablet
+- .(.( 1100JIK+
- 6 .( 10 .( 9+0:- .( 1 0; +- +++-
Looking at the pattern that is emerging, we guess that
Notice that there are 10 terms in this sum—one for every tablet—but that the highest power of 0.04
is the ninth, because no tablet has been in the body for more than 9 six-hour time periods. (Do you
see why?) Now suppose we actually want to find the numerical value of
. It seems that we have
to add 10 terms—and if we want the value of
, we would be faced with adding 40 terms:
! FE
E
!
! EL),+- .( 90 G +- ./ 10 I 6 +- .( 90: +- Fortunately, there’s a better way. Let’s start with ! FE .
! ME )D+- ./ 10 G +- ./ 10 I +- ./ 10JNH
6 +- .( 10 +- ./ 10& +- .( 90 ! FE from ! FE , a great many terms (all but two,
Notice the remarkable fact that if you subtract
in fact) drop out. First multiplying by
, we get
.( 10 ! ME)D+- .( 10 ME +- ./ 10 G +- .( 90 I O +- .( 10 +- .( 90 +- .( 10
9.1 GEOMETRIC SERIES
427
! ME*P (. 10 ! MEL),+- PQ+- ./ 10 ME Factoring ! ME on the left and solving for ! FE gives
! FE .J P 10 )D+- < P .(.( 990 0 FFE E ? ?
! MEL) +- < PP This is called the closed-form expression for ! FE . It is easy to evaluate on a calculator, giving
in closed-form by
! FE ),+7 + (to two decimal places). Similarly, ! E is given
?
(
.
1
0
E
! E ) +- < PP + , which is the same (to two decimal places) as
Evaluating this on a calculator shows ! E)R+7
! FE . Thus after ten tablets, the value ofME !#" appears
to have stabilized at just over 260 mg.
Looking at the closed-forms for !
and ! E , we can see that, in general, ! " must be given by
.S .( 10 " 0
! " ) +- P P What Happens as TVUXW ?
What does this closed-form for ! " predict about the long-run level of ampicillin in the body? As
$DY[Z , the quantity ./ 90 " Y . In the long run, assuming that 250 mg continue to be taken
every six hours, the level right after a tablet is taken is given by
.S .( 90 " 0 .F 10
! " ) +- P P Y +- P P ]\ +7 + Subtracting gives
The Geometric Series in General
^ )+- _ ) ^ ^1_ ^1_ ,
B ^1_ I 1^ _ G
^
^ )D+- _ ) In the previous example we encountered sums of the form
(with
and
). Such a sum is called a finite geometric series. A geometric series is one
in which each term is a constant multiple of the one before. The first term is , and the constant
multiplier, or common ratio of successive terms, is . (In our example,
and
.)
_
^ ^1_ ^1_ B ^1_ "9` ^1_ "9` A finite geometric series has the form
^ ^1_ ^1_ 6 ^_ "9` ^1_ " ` ^1_ " An infinite geometric series has the form
The “ ” at the end of the second series tells us that the series is going on forever—in other words,
that it is infinite.
Sum of a Finite Geometric Series
! FE
;
a
"
$
_ 9" ` a " ) ^ 1^ _ ^1_ 6 ^1_ "9` 1^ _ "9` The same procedure that enabled us to find the closed-form for
can be used to find the sum of
any finite geometric series. Suppose we write
for the sum of the first terms, which means up
to the term containing
:
428
Chapter Nine SERIES
a8" by _ :
_:a8" ) ^1_ ^1_ ^1_ 6 ^1_ " ` ^1_ " Now subtract _:a;" from a;" , which cancels out all terms except for two, giving
a.F " P _&0 a " ) ^ .JP ^1_ " " 0A
P _ a" ) ^ P _
Provided _c) b , we can solve to find a closed form for a " as follows:
Multiply
F. " 0
a " ) ^ 1^ _ ^1_ d
A ^1_ 9" ` ) ^ PP __ The sum of a finite geometric series is given by
Note that the value of
$
in the formula for
a8"
provided
_c) b .
is the number of terms in the sum
a;" .
$eYfZ !"
$
^ ^1_ 1^ _ 6 ^ _ "9` 1^ _ " a) ^
a ) ^ ^1_ a ) ^ ^1_ ^1_
To find the sum, a , of this infinite series, we consider the partial sum, a " , of the first $ terms. The
formula for the sum of a finite geometric series gives
.J " 0
a " ) ^ ^1_ ^_ B ^1_ " ` ) ^ PP __ , then _ " Y as $lYhZ ,
What happens to a " as $gYhZ ? It depends on the value of _ . If i _jik
so
monop
montp ^ .F P _ " 0 ^ .F P 0 ^ "qsr a;" ) "qsr P _ ) P _ ) P _
.J P _ 0 . When
Thus, provided i _jiLk
, as $uYvZ the partial sums a;" approach a limit of ^w
this happens, we define the sum of the infinite geometric series to be that limit and say the series
.J P _ 0 .
converges to ^w
Sum of an Infinite Geometric Series
!"
In the ampicillin example, we found the sum
and then let
We do the same here. The
sum
, which shows the effect of the first doses, is an example of a partial sum. The first three
partial sums of the series
are
For
i _ji k , the sum of the infinite geometric series is given by
a ) ^ ^ _ ^1_ B ^1_ "9` ^_ " ) P ^ _ i _jix _"
_ x Z ^gx $yYzZ
^) b _ k P If, on the other hand,
, then
and the partial sums have no limit as
(if
).
In this case, we say the series diverges. If
, the terms in the series become larger and larger in
magnitude, and the partial sums diverge to
(if
) or
(if
). When
, the
terms become larger in magnitude, the partial sums oscillate as
, and the series diverges.
P Z ^{k $|Y}Z
? The series is
What happens when _ )
9.1 GEOMETRIC SERIES
429
^ ^ ^ ^ ^) b and if
, the partial sums grow without bound, and the series does not converge. When
the series is
^ P ^ ^ P ^ ^ P and, if
^) b , the partial sums oscillate between ^
and 0, and the series does not converge.
_ )RP ,
430
Chapter Nine SERIES
Example 1
For each of the following infinite geometric series, find several partial sums and the sum (if it exists).
(a)
Solution
~ +
(b)
~ + Q L
(c)
7Pc+ + P + + + P ~ =

+ +& ;+
and _ ) , so a ) .F 0 )+ .
which we can identify as a geometric series with ^ )
P w+
Let’s check this by finding the partial sums:
a)
a ) ~ + ) + )+P + a ) ~ + ) ),+ P a ) ~ + ) - )D+ P a C ) ~ + 7 ) 7 )+P 7 The formula for a " gives
P .6 0 "
"9`
a " ) P )+P +8
Thus, the partial sums are creeping up to the value a )+ , so a " Y + as $yYzZ .
and _ )+ ) grow without bound, so the
(b) The partial sums of this geometric series (with ^ )
series has no sum:
a *) ~
a ) ~ +sQ) a ) ~ + QL)D a ) ~ + QL) -
7s)
aC) +
The formula for a " gives
"
a;" ) PcPQ+ + )D+ " P (c) This is an infinite geometric series with ^ )7 and _ )RP . The partial sums,
a )7 a ) a \ 71 a \ a C \ -+ a; \ appear to be converging to - . This turns out to be correct because the sum is
a ) P . 7P w 10 ) - (a) This series may be written
9.1 GEOMETRIC SERIES
Regular Deposits into a Savings Account
6
-
431
People who save money often do so by putting some fixed amount aside regularly. To be specific,
is deposited every year in a savings account earning
a year, compounded annusuppose
ally. What is the balance,
, in dollars, in the savings account right after the
deposit?
As before, let’s start by looking at the first few years:

"
) 6 J. 08
) ) .J - 08
) .J - 08
$&%('
) 2 A3 .J4 - 5 0 2 63A4 5
) 2 6.F - 0 34 .J - 5 0 2 63A4 5
) 2 6.F - 0 3.J4 - 0 6.F - 5 0 2 A3 4 5
New deposit
Original deposit
First two deposits
New deposit
New deposit
First three deposits
" ) .F - 0 "9` .F - 0 "9` 6.J - 0&
and _ ) - . Thus we have
So " is a finite geometric series with ^ )
.S P .F - 0 " 0 )
"
P -
Observing the pattern, we see
.M.J - 0 " P B 0 -LP We can rewrite this so that both the numerator and denominator of the fraction are positive:
")
What Happens as
TVUXW
?
.F - 0 " Y
"
Common sense tells you that if you keep depositing
in an account and it keeps earning
interest, your balance grows without bound. This is what the formula for
shows also:
as
, so
has no limit. (Alternatively, observe that the infinite geometric series of which
is a partial sum has
, which is greater than 1, so the series does not converge.)
"
Z $yYzZ
"
_ ) -
Exercises and Problems for Section 9.1
Exercises
In Problems 1–10, decide which of the following are geometric series. For those which are, give the first term and the ratio
between successive terms. For those which are not, explain
why not.
1.
3.
5.
@
# Q
lA *

9c
2.
4.
6.
7.
9.
¡

:* *
¤ 6¤ ¤ Q¥ ¤
c

# s sc

# c £

A
8.
10.
@

| ¢£cA
¤ ¤¡§ ¤¡§
A ¦ ¨¦
11. Find the sum of the series in Problem 6.
12. Find the sum of the series in Problem 7.
13. Find the sum of the series in Problem 8.
14. Find the sum of the series in Problem 10.
& &L (© ª
Find the sum of the series in Problems 15–18.
15.
16.
432

¬B« ®

Chapter Nine SERIES
17.
18.
¬6« ®
ª
Problems
19. This problem shows another way of deriving the long-run
ampicillin level. Suppose that in the long run the ampicillin levels off to mg right after each tablet is taken.
Six hours later, right before the next dose, there will be
less ampicillin in the body. However, if stability has been
reached, the amount of ampicillin that has been excreted
must be exactly 250 mg because taking one more tablet
must raise the level back to mg. Use this to solve for
.
¯
¯
¯
° ±
20. Figure 9.1 shows the quantity of the drug atenolol in the
blood as a function of time, with the first dose at time
. Atenolol is taken in 50 mg doses once a day to
lower blood pressure.
²
(quantity, mg)
¯ª
¯©
¯
¯
¯
(a) If the half-life of atenolol in the blood is 6.3 hours,
what percentage of the atenolol present at the start
of a 24-hour period is still there at the end?
,
,
,
(b) Find expressions for the quantities ,
, and
shown in Figure 9.1. Write the expression for
in closed-form.
(c) Find expressions for the quantities , , ,
,
shown in Figure 9.1. Write the expression
and
for
in closed-form.
³
³
¯¯
¯ª¯©¯¯
³ © ³ ³ ´´´

¯
µ
¸
¶
·

³
µ ¶¸·
21. On page 426, you saw how to compute the quantity
mg of ampicillin in the body right after the
tablet of
250 mg, taken once every six hours.

(a) Do a similar calculation for
, the quantity of
ampicillin (in mg) in the body right before the
tablet is taken.
(b) Express
in closed form.
(c) What is
? Is this limit the same as
³ B ½ ³

B½
¹»º»¼
¹»º»¼ ¯ «
«
? Explain in practical terms why your an-
swer makes sense.
°¾±
³ ¿ ´ ´´
22. Draw a graph like that in Figure 9.1 for 250 mg of ampicillin taken every 6 hours, starting at time
. Put on
the graph the values of , ,
introduced in the
text on page 426 and the values of , ,
calculated in Problem 21.
¯ © ¯ ¯ 6 ³ ¿ ´© ´´ ³

(a) Find an expression for the height to which the ball
rises after it hits the floor for the
time.
(b) Find an expression for the total vertical distance the
ball has traveled when it hits the floor for the first,
second, third, and fourth times.
(c) Find an expression for the total vertical distance the
ball has traveled when it hits the floor for the
time. Express your answer in closed-form.
µ ¶¸·
Á
© Á
1 Â
(a) Show that a ball dropped from a height of feet
reaches the ground in
seconds.
(b) Show that the ball in Problem 23 stops bouncing af(time, days)
ter
Figure 9.1
´´´
A § =
± À ´
¦

À ´ ¦ § ± A ¦ § ± ´ 6

µ ¶¸·
24. You might think that the ball in Problem 23 keeps bouncing forever since it takes infinitely many bounces. This is
not true!
³© ³ ³ ³ °

23. A ball is dropped from a height of 10 feet and bounces.
Each bounce is of the height of the bounce before.
Thus after the ball hits the floor for the first time, the
ball rises to a height of
feet, and after it
hits the floor for the second time, it rises to a height of
feet.
A
Â

9Ã
6
Â {Ä @cÅ BÆ &ÇÉÈ

seconds
´
25. Suppose that c/ of every dollar spent in the US is spent
again in the US. If the Federal government pumps an
extra
billion into the economy, how much additional
spending occurs as a result?
Ê
6
26. This problem illustrates how banks create credit and can
thereby lend out more money than has been deposited.
Suppose that initially
is deposited in a bank. Experience has shown bankers that on average only
of the
money deposited is withdrawn by the owner at any time.
of their
Consequently, bankers feel free to lend out
deposits. Thus
of the original
is loaned out to
other customers (to start a business, for example). This
will become someone else’s income and, sooner or
later, will be redeposited in the bank. Then
of
,
or
, is loaned out again and eventu, the bank again loans out
ally redeposited. Of the
, and so on.
Ê
Ê¥
Ê¥

Ê¥ ¦ ´ ¥ § ± Ê ´ Ê ´
¥ Ë
A ¥ 6Ë
Ê
BË
¥ 6Ë Ê¥
(a) Find the total amount of money deposited in the
bank as a result of these transactions.
(b) The total amount of money deposited divided by the
original deposit is called the credit multiplier. Calculate the credit multiplier for this example and explain
what this number tells us.
Ë
9.2 CONVERGENCE OF SEQUENCES AND SERIES
A6
27. This problem deals with the question of estimating the
cumulative effect of a tax cut on a country’s economy.
Suppose the government proposes a tax cut totaling
million. We assume that all the people who have extra money to spend would spend 80 of it and save
20 . Thus, of the extra income generated by the tax cut,
million
million would be spent and
so become extra income to someone else. Assume that
Ë6 B §
Ê ¦´
9.2
Ê
Ë
± Ê

Ê ¦ ´ B §
433
these people also spend
of their additional income,
or
million, and so on. Calculate the total additional spending created by such a tax cut.
Ê
6
28. Suppose the government proposes a tax cut of
million as in Problem 27, but that economists now predict
that people will spend 90 of their extra income and save
only 10 . How much additional spending would be generated by the tax cut under these assumptions?
Ë
Ë
CONVERGENCE OF SEQUENCES AND SERIES
¬
We now consider general series in which each term
compactly using a
sign as follows
^"
is a number. The series can be written
¬ r ^1" ) ^ ^ ^ 6 ^9" "Ì For any particular values of ^ and _ , the geometric series is such a series, with general term ^ " )
^1_ "9` . As in Section 9.1, we investigate convergence of a series using the partial sum, a " , of the
first $ terms, which is defined as
" Í B ¬
Í
a;" ) Ì ^ ) ^ ^
^9"
a a a a " ,
Convergence of Sequences
a"
The partial sums form a sequence, or string of numbers.1 A sequence can be written
where
is called the general term. The limit of the sequence is defined in a way similar to the
limit of a function; see also Problem 20.
mtnop
The sequence a
a a a " has a limit Î , written "qsr a " ) Î , if we can make a "
as close to Î as we please for all sufficiently large $ .
Ï "ommoqntntpp r _ " ) i _jik Ï "qr wB$ ) To calculate the limit of a sequence, we use what we know about the limits of functions, including the properties in Theorem 2.1 and the following facts:
if
Example 1
`:"
a " ) ~{P ÐÐ :` "
Find the limits of the following sequences, if they exist.
(a)
a " ) ./ 0 "
(b)
(c)
a " ) ~ . P 60 "
1 In everyday English, the words “sequence” and “series” are used interchangeably. In mathematics, they have different
meanings and cannot be interchanged.
434
Solution
mtnop ./ 0 " /
.

0
"
) .
(a) Since
decreases to 0 as $ increases,
mtnop we have "qsr
&
`
"
Y " as |$ YÑZ , we have. "Bqs0 " r a " ) .
(b) Since Ð
.
6
0
P
) if $ is even and P )RP if $ is odd,
(c) Since
a *) a )D+ a ) a )+ and so on The sequence a " does not have a limit.
Chapter Nine SERIES
To show that a sequence has a limit, the following result is often useful.
monop

+ , is increasing and bounded above, then "qsr a "
If a sequence a " , for $ )
Theorem 9.1: Convergence of an Increasing, Bounded Sequence
a"
Ò
a "|Ó Ò
Ô Ô
a"
a"
exists.
To understand this theorem geometrically, see Figure 9.2. Saying that
is bounded above
means that the values of
are less than some number ; that is,
. Since
is increasing
and bounded above, the values of
must “pile up” at some number less than or equal to . This
number is the limit.2
Ô
a"
Ô
©
Figure 9.2: Values of
Ö

for
µ±
Ô
A ¿ ¿ ¿
Ò
Õ
Convergence of Series
To investigate the convergence of a series, we consider the sequence of partial sums
If
a;"
has a limit as
$yYzZ
a a a a " , then we define the sum of the series to be that limit.
om ntp
mtnop
¬ r ^ " converges and that its
)
"
"
a
exists,
say
a
a
,
then
we
say
the
series
"qr
"qsr
"Ì monop
r¬
)
sum is a . We write
"Ì ^9" a . If "qsr a;" does not exist, we say that the series diverges.
If
^ "g× Visualizing Series
^"
$
We can visualize the terms of a series as in Figure 9.3. In this figure, we assume
for all ,
so each rectangle has area . Then the series converges if the total area of the rectangles is finite
and the sum of the series is the total area of the rectangles. This is similar to an improper integral
, in which the area under the graph of can be finite, even on an infinite interval.
Ø EÚr Ù . _ 0 Û _
2 See
the online supplement for a proof.
Ù
Ô
ÔÔ
ÔÔ
©
Ô

9.2 CONVERGENCE OF SEQUENCES AND SERIES
É
...
Figure 9.3: Height and area of the
µ ¶¸·
µ
Ô
...
435

rectangle is
Here are some properties that are useful in determining whether or not a series converges.
¬ r ^ " and ¬ r " converge and if Ý is a constant, then
"Ì r
"Ì Ü
Ï ¬ . ^ " " 0 converges to ¬ r ^ " ¬ r " .
"r Ì Ü
"Ì "Ì Ü
r
Ï ¬ Ý ^ " converges to Ý ¬ ^ " .
"Ì "Ì 2. Changing a finite number of terms in a series does not change whether or not it converges,
although it may change the value of its sum if it does converge.
mtnop
r
¬
1
^
"
3. If
converges, then "qsr ^9" )
"r Ì ¬ ^ " diverges, then ¬ r Ý ^ " diverges if Ýy) b .
4. If
"Ì "Ì Theorem 9.2: Convergence Properties of Series
1. If
$
For proofs of these properties, see Problems 21–24. As for improper integrals, the convergence
of a series is determined by its behavior for large . (See the “behaves like” principle on page 374.)
From Property 2 we see that, if is a positive integer, then
and
either both
converge or both diverge. Thus, if all we care about is the convergence of a series, we can omit the
limits and write
.
Property 3 often is used to tell us that a series does not converge. If the terms do not go to 0,
the series has no chance of converging.
Warning: Knowing that
is not enough to ensure convergence. (See Example 3.)
Þ ^9"
Example 2
Solution
Ý
Þ "r Ì ^ "
Þ "r Ì£ß ^ "
om nop
"qsr ^9" ) ¬ F. cP Ð `:" 0 converge?
PàÐ `&"
Since the terms in the series, ^ " )
Does the series
Propery 3 of Theorem 9.2.
tend to 1, not 0, as
$DY[Z
, the series diverges by
436
Chapter Nine SERIES
Comparison of Series and Integrals
We investigate the convergence of some series by comparison with an improper integral. The harmonic series is the infinite series
~ ¡ +
$
Convergence of this sum would mean that the sequence of partial sums
a *) a ) ~ + a ) ~ + a;" ) ~ + ,
6 $ tends to a limit as $yYzZ . Let’s look at some values:
a *) a FE \ + a FEáE \ - o a FEEáE \ a MEEáEE \
The growth of these partial sums is slow, but they do in fact grow without bound, so the harmonic
series diverges. This is justified in the following example and in Problem 25.
Example 3
Solution
~ w + w * w diverges.
Ø r .F wB_ 0Û _ by a left-hand sum, where the terms w + w are
The idea is to approximate
heights of rectangles of base 1. In Figure 9.4, the sum of the areas of the rectangles is larger than
the area under the curve between _ )
and _ )
, and the same kind of relationship holds for the
first $ rectanges. Thus, we have
6 "ã Û moä . 60
~
a" ) +
$ xâ _ _ ) $
moä . ,60
Since $
gets arbitrarily large as $lYÑZ , so do the partial sums, a " . Thus, the partial sums
Show that the harmonic series
have no limit, so the series diverges.

å ± Æ
± ± Æ6
æ
± Æ © s© ç Ø © è¾© é ± ¹»ê
æ æ

Ø © ¦Æ § é
:
«
Rectangles showing
Area
Area
Area
Figure 9.4 Comparing the harmonic series to
Example 4
om ntp
mtnop Notice that the harmonic series diverges, even though "qsr ^ " ) "qsr
$ ) .
Ø r .J wB_ 0 Û _ , show that the following series converges:
By comparison with the improper integral ¬ r ) ~ "Ì $

9.2 CONVERGENCE OF SEQUENCES AND SERIES
± Æ

¥ 6ì ë â ª « é
Æ
± ¥ ± Æ A Æ66
B® Ø
Æ §
Æ
« © µ © « ¦ é
Shaded rectangles show
± Æ
±
æ æ
æ æ
:
Þ
r ¬
Since we want to show that
"Ì wB$ converges, we want to show that the partial sums of this series
Æ
Æ
Æ ¥
Area
Area
Area
Area
Figure 9.5 Comparing
Solution
437
to
tend to a limit. We do this by showing that the sequence of partial sums increases and is bounded
above, so Theorem 9.1 applies.
Each successive partial sum is obtained from the previous one by adding one more term in the
series. Since all the terms are positive, the sequence of partial sums is increasing.
r ¬
the right-hand sum repreTo show that the partial sums of
wB$ are bounded, we ) consider
"
Ì
sented by the area of the rectangles in Figure 9.5. We start at _
, since the area under the curve
is infinite for Ó _ Ó . The shaded rectangles in Figure 9.5 suggest that:
B r Û 7
$ Ó â _ _
The area under the graph is finite, since
r Û monop í Û monop P )
â _ _ ) í qsr â _ _ ) í qsr

Ü
To get a " , we add 1 to both side, giving
r a;" ) ~ 7 6 $ Ó ~ â _ Û _ )D+ Thus, the sequence of partial sums is bounded above by 2. Hence, by Theorem 9.1 the sequence of
partial sums converges, so the series converges.
Notice that we have shown that the series in the previous example converges, but we have not
found its sum. The integral gives us a bound on the partial sums, but it does not give us the limit of
the partial sums. Euler proved the remarkable fact that the sum is
.
The method of Examples 3 and 4 can be used to prove the following theorem. See Problem 28.
îw7
and Ù . _ 0 is a decreasing positive function, defined for all _ × , with
×
ï
.
0
^1" ) Ù $ r for. all0 Û $ .
Ï If â ð Ù _ _ converges, then Þ ^ " converges.
Ï If â ð r Ù . _ 0 Û _ diverges, then Þ ^ " diverges.
Theorem 9.3: The Integral Test
Suppose
438
Chapter Nine SERIES
The integral test can be used to show that
Þ "r Ì wB$:ñ
converges for
òx . See Problem 13.
Exercises and Problems for Section 9.2
Exercises
Ö

Problems 1–8 give expressions for , the general term of a
sequence. Find the limit of each sequence, if it exists.
1.
2.
3.
4.
5.
6.
7.
8.

¦ ´ §
¾ 6§
¦ y´ó¡ô
õ ö ÷¦¸øµ §

À µµ *

§
µµ#Q ¦¦ * §
9.
10.
11.
12.
Problems

13. Use the integral test to show that
úüú û ç
¬B« ® ¹»êHµ
© µ
(a) Converges if
(b) Diverges if
14. Show that
¬B« ® Æ 9µ ù
©
16.
17.
18.
19.
Consider
tial sum.]
is divergent.
, where
definition for
.
similar to the
in Section 2.2. Instead of ,
you will need , a value of .
21. Show that if
stant, then
verge.
22. Let and ,
Ô
Ô

± ´
»
¹
»
º
¼
Ö ô © § «Ö µ¶¸· Ô

± ¬B« ®
©
ÞB½ Ô
¹»º ¼ (¦ Ö
¬6« ® «
©
converges, then
converge and if is a con
, and
con
be a positive integer. Show that if
for
and
either both converge, or
, then
both diverge. In other words, changing a finite number of
diverges.
[Hint:
, where
diverges and 24. Show that if

A
¥ 6 Q µ c
¥ À Q µ c
# # À c µ µ { A
FýJ FýJ FýJ Fýá Q µ FýJ Q
ù ù Q Bµ ½ ù cA ú ç
½ Ô § ¹»±dº»¼ þ Ö ±þ
ÿ¿
Ô
è ¹»Ôº ¼ ¦¸ « µ Ô
ÞÞ ¦ Ô § Þ Þ ¦ § Ô Þ

±
µ
Þ
Þ
20. Write a definition for
§
µ
23. Show that if
.
¬6« ® µ
© µ#
¬6« ® ó ô
ª
¬6« ® ó
©
¬6« ® µ8¦¸µ#
©Âµ
terms in a series does not change whether or not it converges.
By identifying each one as a left- or right-hand approximation
of an improper integral, decide whether the series in Problems 15–19 converge.
15.
Do the series in Problems 9–12 converge or diverge?
is the
par-
, then

; 8 # À ;
Æ
A 6
©(ªM¬ª ® ªMªMª ¿
© ¿
6 6
± ¦¸ § ± Æ
L
¿

¦¸ § ± Æ L ±

c µ ël¹»êHµ ´

A {A A ;L
¥
25. Consider the following grouping of terms in the harmonic series:
(a) Show that the sum of each group of fractions is more
.
than
(b) Explain why this shows that the harmonic series
does not converge.
26. Estimate the sum of the first
monic series,
terms of the har-
to the closest integer. [Hint: Use left- and right-hand
sums of the function
on the interval from
to
with .]

27. Although the harmonic series does not converge, the partial sums grow very, very slowly. Take a right-hand sum
of
with on the interval to
show that
¿µ
Ô

439
9.3 TESTS FOR CONVERGENCE
If a computer could add a million terms of the harmonic
series each second, estimate what the sum would be after
one year.
28. In this problem, you will justify the integral test. Suppose
and
is a decreasing positive function, defined
, with
for all .
for all
¦¸ § § ±
¸¦ µ

â « ¦¸ § é
(a) Suppose
9.3
Ô

Þ
angles under the curve, show that
[Hint: See Example 4 on page 436.]
(b) Suppose that
µ
â « ¦¸ § é
Ô
converges.

diverges. By drawing
Þ
rectangles above the curve, show that
verges. [Hint:See Example 3 on page 436.]
di-
converges. By drawing rect-
TESTS FOR CONVERGENCE
Comparison of Series
In Section 7.8, we compared two integrals to decide whether an improper integral converged. In
Theorem 9.3 we compared an integral and a series. Now we compare two series.
Ó ^ "ìÓ " for all $ Ï Þ " converges,
Ü then Þ ^ " converges.
Ü
Ï Þ ^ " diverges, then Þ " diverges.
Ü
Theorem 9.4: Comparison Test
Suppose
If
If
^9" Ó Ü "
Þ ^1"
^9"
Þ Ü"
©
Ô
Ô
©
Ü"
Þ Ü"
Þ ^9"
Since
, the plot of the lies under the plot of the . (See Figure 9.6.) The comparison
test says that if the total area for
is finite, then the total area for
is finite also. If the total
area for
is not finite, then neither is the total area for
.
Ô
Figure 9.6: Each
Example 1
Solution

Ô
Ô Ô Ô

}

...
µ
...
is represented by the area of a dark rectangle, and each
by a dark plus a light rectangle
r ¬
Use the comparison test to determine whether the series
"Ì w¡$ converges.
For $ × , we know that $ Ó $ , so
Ó Ó $ $
r
Thus, every term in the series Þ "Ì w¡$ is less than or equal to the corresponding term in
Þ "r Ì wB$ . Since we saw that Þ "r Ì w¡$ converges from page 436, we know that Þ "r Ì wB$
converges.
440
Chapter Nine SERIES
Example 2
Solution
Example 3
Solution
k{òk , then Þ "r Ì w¡$:ñ diverges.
Since ò k and $ × , we have $ üñ Ó $ , so wB$ ñ× wB$ . The harmonic series Þ Bw $
Þ wB$:ñ diverges also.
Use the comparison test to show that if
diverges, so
r 7 $ ,
¬
Decide whether the following series converge: (a)
(b)
"Ì + $ P .
(a) Since the convergence is determined by the behavior of the terms for large $ , we observe that
$ P Y $ ) as $yYzZ $
$ $ Since Þ
wB$ converges, we guess that Þ . $ P B0 w . $ l0 converges. To confirm this, we use
¬ r $ P
"Ì $
the comparison test. Since a fraction increases if its numerator is made larger or its denominator
is made smaller, we have
Ó $ P Ó $ ) for all $ × $
$ $
.
6
0
.

0
Thus, the series Þ $ P
w $ converges by comparison with Þ wB$ .
(b) First, we observe that
7 $ Y 7 $ ) as $YzZ +$ P +$ $
Since Þ
w¡$ diverges, so does Þ wB$ , and we guess that Þ . 7 $ 60 w . + $ P 60 diverges. To
confirm this, we use the comparison test. Since a fraction decreases if its numerator is made
smaller or its denominator is made larger, we have
Ó 7 $ Ó 7 $ +$ +$ P
so
Ó Ó 7 $ $ +$ P
.
B
0
.
6
0
w + $ P diverges by comparison with Þ Bw $ .
Thus, the series Þ 7 $
Series of Both Positive and Negative Terms
Þ ^"
Þ ^"
If
has both positive and negative terms, then its plot has rectangles lying both above and
below the horizontal axis. See Figure 9.7. The total area of the rectangles is no longer equal to
. However, it is still true that if the total area of the rectangles above and below the axis is
finite, then the series converges. The area of the
rectangle is
, so we have:
$&%('
Þ
Þ
i ^1"8i
Theorem 9.5: If !" converges, then so does #! .
Problem 44 shows how to prove this result.
ÔÔ
Ô ©
ÔÔ
Ô

9.3 TESTS FOR CONVERGENCE
...
µ
441
...
Figure 9.7: Representing a series with
positive and negative terms
Example 4
Explain how we know that the following series converges
Solution
Writing
¬ r . P 60 " ` ) P P "Ì $
^ " ) . P B0 9" ` Bw $ , we have
. 60 " ` i^ " i ) P $ ) $ w¡$ converges, so Þ . P 60 "9` w¡$ converges.
From Example 4 on page 436, we know that Þ
%$
$
$
$
$
$
$
$
$
$
Comparison with a Geometric Series: The Ratio Test
_
i _Ki:k We now develop a test for convergence by comparison with a geometric series. Recall that it is
easy to tell if a geometric series converges. If is the common ratio between successive terms of a
geometric series, then the series converges if
and diverges if
. The following test is
obtained by comparing the series
with a geometric series as
.
Þ i^ " i
Theorem 9.6: The Ratio Test
i _Ki × $YÑZ
Þ ^ " , suppose the sequence of ratios i ^ "ã i wi ^ " i has a limit:
montp i ^ "ã i "qr i ^1"8i ) Î
Ï If Î k , then Þ ^ " converges.
Ï If Î x , or if Î is infinite, then Þ ^ " diverges.
Ï If Î ) the test does not tell us anything about the convergence of Þ ^ " .

©
For a series
3
)(
3 That
is, the sequence & '
& *)& '
& grows without bound.
442
Chapter Nine SERIES
Îàk Î om ntp i ^ "ã i
" qr i ^1"8i ) Î for all sufficiently large $ , say for all $ × Ý , we have
i ^ "ã i kà_ i^ " i
_
Proof The basic idea is to compare the series with a geometric series.
Suppose
. Let be a number between and , so that
Îàkà_yk . Since
Then,
i ^ ßAã i9kRi ^ ß i _ i ^ ßAã i9kRi ^ ßAã i _ykVi ^ ß i _ i ^ ßAã i9kRi ^ ßAã i _ykVi ^ ß i _ and so on. Thus, for a sufficiently large $ , we have
i ^ ß i i ^ ßAã i i ^ ßAã i i ^ ßã i kRi ^ ß i i ^ ß i _ i ^ ß i _ i ^ ß i _ g
converges by comparison with the convergent geometric
Thus, i ^ ß i i ^ ßAã i i ^ ßã i i ^ ßAã i
series i ^ ß i i ^ ß i _
i ^ ß i _ i ^ ß i _ . Hence, Þ i ^ " i converges, so Þ ^ " converges.
If Îàx , then we choose _ so that kà_kàÎ . Then, for sufficiently large $ , say $ ×
,
i ^9"ã i9xÉi ^9"8i _yxRi ^ üi mtnop
, is increasing. Thus, "qsr ^ " ) b , so Þ ^ " diverges (by
so the sequence i ^ i i ^ ã i i ^ ã i Theorem 9.2, property 3).
,+
.-
-
Example 5
-
-
Show that the following series converges:4
¬ r ) ~ +
"Ì $
w¡$ and ^ "ã ) w . $ B0 , we have
Since ^ " )
i ^ "ã i ) w . $ 60 ) . $ 60 ) $ . . $ P B0 B0.. $ PQ+60j0j
+ +
i^ " i
wB$
$
$ $ $P
. B0. $ PQ+ 0j
+ , giving
We cancel $ $ P
mtnop i ^ "ã i montp $
mtnop )
)
"qsr i ^9"8i "qr . $ B0 "qsr $ )
r ¬
Because the limit is 0, which is less than , the ratio test tells us that
"Ì wB$ converges.
0/
Solution
0/
/
/
/
/
0/
0/
/
0/
/
0/
4 We
define 132 (read two factorial) to be 1547698:1 . Similarly, ;32<8=;54>154?698=@ and A278:AB4?CDAFEG6IHJ4I4K4L15476 .
9.3 TESTS FOR CONVERGENCE
Example 6
Solution
443
What does the ratio test tell us about the convergence of the following two series?
¬r "Ì $
r . P B0 "9` ¬
and
"Ì $
monop
mtnop
.
B
0
"
P
)
)
i in both cases we have "qsrÉi ^9"ã w¡^9";i " qsrà$w . $ e 60 ) . Thus,
Because i
the ratio test does not tell us anything about the convergence of either series. In fact, the first series
is the harmonic series, which diverges. Example 7 will show that the second series converges.
Alternating Series
¬ r . P B0 "9` ) P P B . P 60 " ` ,
+
$
"Ì $
A series is called an alternating series if the terms alternate in sign. For example,
The convergence of an alternating series can often by determined using the following test:
Theorem 9.7: Alternating Series Test
¬ r . P 6 0 9" ` ^ Í ) ^ P ^ ^ P ^ 6,. P 60 "9` ^ " "Ì k ^ "ã ek ^ " for all $ and "moqntp r ^ " ) A series of the form
converges if
Although we do not prove this result, we can see why it is reasonable. The first partial sum,
a ) ^ , is positive. The second, a ) ^ P ) ^ , isP still positive,
^ kR^ , but a is smaller
^ , is since
^
^
than a . (See Figure 9.8.) The next sum, a
greater than a but smaller than
a . The partial sums oscillate back and forth, ÔÔ and since the distance between them tends to 0, they
eventually converge.
ÔÔ
å
©
å

Ö Ö
Ö Ö©
Ö© Ö Ö Ö
:
M
M
Figure 9.8 Partial sums,
Example 7
,
,
Show that the alternating harmonic series converges
,
¬ r . P 60 9" ` "Ì $
of an alternating series
444
^9" ) ¡w $
Chapter Nine SERIES
Solution
We have
^1"ã *) w . $ 60 . Thus,
)
^ "ã $ k $ ) ^ " for all $ ,
and
and
om ntp "qsr w¡$ )
Thus, the hypothesis of Theorem 9.7 is satisfied, so the alternating harmonic series converges.
monop
)
a
"qsr a;" . Then a is trapped between
a a a a;C so
a k a k kak ka kàa a " to approximate the true sum a is less than the distance from a;" to a8"ã ,
Thus, the error in using ;
which is ^9"ã . Stated symbolically, we have the following result:
a
Suppose is the sum of the alternating series, so
any two consecutive partial sums, say
and
or
and
Í
montp
"¬Í . 60 ` Í
P
Let a " )
^
be the $&%(' partial sum of an alternating series and let a ) " qr a " .
mtnop
Ì
"qsr ^9" ) . Then
Suppose that k ^1"ã kà^9" for all $ and
i a P a " i kà^ "ã Theorem 9.8: Error Bounds for Alternating Series
a"
Thus, the error in using
to approximate
series which is omitted in the approximation.
a
is less than the magnitude of the first term of the
r . 60 "9` ¬
Estimate the error in approximating the sum of the alternating harmonic series
P wB$
"
Ì
the sum of the first nine terms.
Example 8
by
a G ) P + P 6 ) -7
o . By Theorem 9.8, we know that the true value
The first term omitted is P w , with magnitude
o
of the sum differs from -7
by less than
.
Solution
The ninth partial sum is given by
Ô
Exercises and Problems for Section 9.3
Exercises
Ô
Ô
Decide if the statements in Problems 1–13 are true or false.
Give an explanation for your answer.

û
1. If
verges.

and
Þ

converges, then
Þ

con-
2. If

Ô
ûÔ û Þ Ô
û û Þ
Þ
Þ
3. If verges.
4. If
Ô
and
and
diverges, the
converges, then ON
Þ
Ô
converges, then
N
converges.
Þ

diverges.
con-
Ô
Ô
Þ
Þ
¬B« ® ¦ * § õö6÷¦¸øµ §

¬B« ® ¦ Q¦ § §

´
5. If PN
verge.
6. If
7.
8.
N
Ô

Ô Ô
6½ Þ Æ ± Þ
¹»º»¼ ©
«
converges, then QN
N
3(
converges, then
N
and ON N N N
N
con-
.
22.
is a series of nonnegative terms.
SR
10. The series
23.
LTU
converges.
Ô
Ô
*
Þ ¦ *§ §
Þ ¦
24.
converges.
11. If
converges, then
12. If ON
N
N
converges, then
converges.
N
converges.
13. To find the sum of the alternating harmonic seris to
within
of the true value, we should sum the first 100
terms.
Use the comparison test to confirm the statements in Problems 1–2.
15.
16.
21.
is an alternating series.
9. The series
14.
20.
¬B« ®
© µ
¬B« ®
¬B® µ
«©µ

¬6® µ
« © óµô
converges, so
diverges, so
¬6« ®
converges, so
converges.
diverges.
converges.
¬6« ®
©
¬6« ®
© µ ló
¬6« ® ô ¦¸µ# 6§§
© ¦¸µ#
¬6« ® µ
© µ
¬6« ® g
©µ
9.3 TESTS FOR CONVERGENCE
445
Use the ratio test to decide which of the series in Problems 25–
29 converge and which diverge.
25.
¬6« ® §
©¦µ
¬6« ® ¦¸µ § §
© ¦ µ
¬6« ®
ª µ
¬6« ®
© µó
¬6« ® µ ô
©

¬6« ® ¦ § ô ©
© *µs
¬6« ® ¦ § ô ©
© *Â µ
¬6« ® ¦ § ô © µ
© µ
IV
V
26.
27.
28.
29.
IV
Use the comparison test to determine whether the series in
Problems 17–24 converge.
Use the alternating series test to decide which of the series in
Problems 30–32 converge.
17.
30.
18.
19.
¬B« ®
© µ
¬B« ®
¬B® ¹ êHµ
« © µsµ
31.
32.
V
Problems
Use a computer or calculator to investigate the behavior of
the partial sums of the alternating series in Problems 33–35.
Which ones converge? Confirm convergence using the alternating series test. If a series converges, estimate its sum.
33.
34.
35.

@ì ~
* § A

@ ´ ´ ´ Q*Q § ¦ § ô Q
@ cQ¦ * § ¦¸ µ# c
QQ¦ µ Q
V
SV
SV
V

Determine which of the series in Problems 5–11 converge.
36.
37.
¬6« ®

446
38.
39.
40.

¬B« ® ¦ § lô ©
© Â * µ ©
¬B« ® ¦ § ô
© µ
¬B« ® Å

l § ´

µ8¦¸µ
l § ± l
8µ ¦¸µ
µ
µ
¬B« ® Ô l ¬B« ® ´
µ d µ

Æ l § ± Æ Æ ¦¸µ § µ d ë,µ
¦¸µ ç µ
¬B« ®
Ô
Ô µ

* §
Þ
Þ ¦
Consider the infinite series
Since
For the first series
we have
44. Let
be a series such that ON N converges. In this
problem, you will show that
converges.
The second series is the divergent harmonic series. Since
both series diverge, their difference also diverges.
42. Show that if
conN
N converges, then
verges.
if
if
(b) Define a new series
it converges:
.
, as follows, and show that
YX
. Since
and so this series diverges by
comparison with the divergent harmonic series
converges.
(a) Define a new series
it converges:
we can write this series as
9.4
is even,
(b) Show that
converges.
(c) Use the results of parts (a) and (b) to show that
41. Is the following argument true or false? Give reasons for
your answer.

6 ±

l
;L µ ;L ¦¸µ § A µ ´

¨ cA
¾ Ô Ô Ô
cA
Þ
Þ Þ
ÞÔ Ô
Ô
±

´
ë

Ô
Þ
Ô
Ô
±
Ô
ë ´

Þ
Þ
Þ
43. (a) Show that if
, as follows, and show that
if
YX
if
(c) Use
POWER SERIES
and
to show that
converges.
P ku_,k on ä
Ð _
Þ ^1_ "
In Section 9.1 we saw that the geometric series
converges for
and diverges
otherwise. This section studies the convergence of more general series constructed from powers.
Chapter ?? shows how such power series are used to approximate functions such as )Z , [
, \7]^[ ,
and
.
moä
_
_
_ ) ^ is a sum of constants times powers of . _ P ^ 0 :
r
E . _ P ^ 0£ . _ P ^ 0 B " . _ P ^ 0 " ) ¬ E " . _ P ^ 0 " "Ì
A power series about
_
_
_
_
_
Þ " ._ P ^ 0"
a;" . _ 0 ) E . _ P ^ 0 . _ P
0^ 6 " . _ P ^ 0 "
a E . _ 0 a . _ 0 a . _ 0 a;" . _ 0 monop a fixed. 0 value of _ , if this sequence of partial sums converges to a limit Î , that is, if
For
"qsr a " _ ) Î , then we say that the power series converges to Î for this value of _ .
^
_
_
We think of as a constant. For any fixed , the power series
is a series of
numbers like those considered in Section 9.2. To investigate the convergence
of
a
power
series,
we
_
_
_
consider the _partial sums, which in this case are the polynomials
. As before, we consider the sequence5
ª
A power series may converge for some values of
5 Here we have chosen to call the first term in the sequence
`
_
©
and not for others.
CbaJH rather than `
CaJH to correspond to the power of CaEc'3H .
9.4 POWER SERIES
447
Numerical and Graphical View of Convergence
. _ P 6 0 P . _ P B 0 . _ P 6 0 P . _ P 60 , 6,. P B0 "9` . _ P B0 " +
$
To investigate the convergence of this series, we look at the sequence of partial sums graphed in Fig./ 0
ure 9.9. The graph suggests that the partial sums converge for _ in the interval + . In Examples 2
and 5, we show that the series converge for k,_ Ó + . This is called the interval of convergence
of this series.
, which is inside the interval, the series appears to converge quite rapidly:
At _ )
a C .F.F 1100 ) 7 ** a N .F.F 1100 ) 7- *
a; ) 7
aI ) 7 and _ )+ in the power series. For _ ) ,
Table 9.1 shows the results of using _ )
which is inside the interval of convergence but close to an endpoint, the series converges, though

rather slowly. For _ ) + , which is outside the interval of convergence, the series diverges: the
larger the value of $ , the more wildly the series oscillates. In fact, the contribution of the twenty . Figure 9.9
fifth term is about 28; the contribution of the hundredth term is about P+ shows the interval of convergence and the partial sums.
Ö © ¦¸ § Ö ¦¸ §
Table 9.1 Partial sums for series in
inside interval
å
Example 2 with _ )
of convergence
and _ )D+ outside

§
´ ¥ µ Ö ¦ ´ B§
µ
Ö
¦

´
´¥
¾ À À
Ö B¦¸ §
´
± Ö ©M© ¦¸ §

:
ëlüë
. Since the interval
Notice that the interval of convergence, k _ Ó + , is centered on _ )
As an example, consider the series
Interval
of
convergence
M
ed
2
0.495
2
0.455
5
0.69207
5
1.21589
8
0.61802
8
0.28817
11
0.65473
11
1.71710
14
0.63440
14
Figure 9.9 Partial sums for series in
Example 2 converge for
extends one unit on either side, we say the radius of convergence of this series is 1.
Intervals of Convergence
Each power series falls into one of the three following cases, characterized by its radius of convergence, f .
Ï
Ï
_) ^
) The series converges only for
; the radius of convergence is defined to be f
.
The series converges for all values of ; the radius of convergence is defined to be
f
.
There is a positive number f , called the radius of convergence, such that the series
converges for
gf and diverges for
hf . See Figure 9.10.
)
Ï Z
i _ P ^;i9k
_
i _ P ^;i9x
448
Chapter Nine SERIES
Series
diverges
Ô
å
Ôå
=i
Ô
i
Interval of convergence
M
M

i
Series
diverges
Ô
±
i
Figure 9.10: Radius of convergence, , determines an interval, centered at
which the series converges
, in
There are some series whose radius of convergence we already know. For example, the geometric series
~ _ _ ¡ _ " and diverges for i _ji × , so its radius of convergence is 1. Similarly, the
converges for i _Ki k
series
~ _ _ ¡ _ " and diverges for i _;w i × , so its radius of convergence is 3.
converges for i _;w i9k
The next theorem gives a method of computing the radius of convergence for many series. To
r . 0"
¬
" _ P ^ converges, we use the ratio test.
find the values of _ for which the power series
E
"
Ì
" . _ P ^ 0 " and assuming "g) b and _Q) b ^ , we have
Writing ^9" )
montp i ^ "ã i monop i "ã . _ P ^ 0 "ã i monop i "ã ioi _ P ^;i
montp i "ã i "qr i ^9"8i ) "qsr i " . _ P ^ 0 " i ) "qsr i "£i ) i _ P ^;i "qsr i "£i
Case 1. Suppose i ^ "ã i wi ^ " i is unbounded. Then the ratio test shows that the power series con.
verges only for _ ) montp ^ . The radius of convergence is )
Case 2. Suppose "qr i ^ "ã i wi ^ " i )
. Then the ratio test shows that the power series converges
mtnop
for all _ . The radiusmtnop of convergence is ) Z .
Case 3. Suppose "qsr i ^9"ã i wi ^9"8i ) Òài _ P ^&i , where "qsr i "ã i wi "8i ) Ò . In Case 1, Ò
does not exist; in Case 2, Ò )
) wÒ . Then we have montp . Thus, we can assume Ò exists and Ò]) b , and we can define
i ) Òei _ P ^;i ) i _ P ^;i i
9
^
"
ã
"qsr i ^ " i
so the ratio test tells us that the power series:
Ï Converges for i _ P ^;i k ; that is, for i _ P ^;i9k
Ï Diverges for i _ P ^;i x ; that is, for i _ P ^;i9x .
_
_
_
_
_
_
_
_
_
f
f
_
_
f
f
gf
f
jf
f
The results are summarized in the following theorem.
r . 0"
¬
" _ P ^ , use the ratio
To calculate the radius of convergence, , for the power series
E
"
Ì
. 0" .
test withmtnop ^ " )
.
Ï If "mtqsnop r i ^ "ã " i w_ i ^ P " i^ is unbounded,
then )
Ï If "mtqsnop r i ^ "ã i wi ^ " i ) , then ) Z .
Ï If "qsr i ^9"ã i wi ^9"8i ) Òei _ P ^;i , where Ò is finite and nonzero, when ) w¡Ò .
Theorem 9.9: Method for Computing Radius of Convergence
_
f
_
f
f
f
montp
"
"qsrRi ^9"ã i wi ^9"8i
449
9.4 POWER SERIES
Note that the ratio
test does not tell us anything if
fails to exist, which can
_
occur if some of the
s are zero.
A proof that a power series has a radius of convergence and of Theorem 9.9 is given in the
online theory supplement. To understand these facts informally, we can think of a power series as
being like a geometric series whose coefficients vary from term to _ term. The radius of convergence
depends on the behaviour of the coefficients: if there are constants and such that for larger and
larger ,
_
_
Ò
$
"
"
i
i
Ò
\
"" 9_ " and Þ Ò " _ " ) Þ . Òy_ 0 " converge or diverge together. The
then it is plausible that Þ
.
0
Òl_ converges
w¡Ò . We have
geometric series Þ
for i Òl_ji9k , that is, for i _jik
.
0
.
0
"
ã
"
ã
"
ã
i ^ "ã i ) i "ã ioi . _ P ^ 0 " i \ Ò " i . _ P ^ 0 " i ) Òài _ P ^;i i ^ " i i " ioi _ P ^ i
Ò i_ P ^ i
Show that the following power series converges for all _ :
~ _ _ _ B _ " +
$
wB$ , none of the " s are zero and we can use the ratio test:
Because " )
mtnop i ^ "ã i mtnop i "ã i mtnop w . $ 60
mtnop $
mtnop "qsr i ^9"£i ) i _ji "qsr i "£i ) i _ji "qsr wB$ ) i _ji "qsr . $ 60 ) i _ji "qsr $ )
This gives ) Z , so the series converges for all _ . We see in Chapter ?? that it converges to Ð .
_
_
_
_
_
_
_
Example 1
_
/
Solution
_
/
0/
_
0/
_
/
_
0/
0/
/
)Z
f
Example 2
Solution
. _ P 6 0 P . _ P B0 . _ P 60 P . _ P 60 ,
6,. P B0 "9` . _ P B0 " +
$
What does this tell us about the interval of convergence of this series?
. B0 if $ is odd and P . _ P B0 " wB$ if $ is even, so " )
The general term of the series is _ P
. P B0 "9` w¡$ , and we can use the ratio test." WewB$ have
mtnop i ^ "ã i montp i "ã i montp i . P 60 " w . $ 60 i mtnop $
"qsr i ^9"£i ) i _ P i "qr i "i ) i _ P i "qsr i . P 60 "9` wB$¾i ) i _ P i "qsr $ ) i _ P i . The power series converges for i _ P i:k and diverges for
The radius of convergence is )
that the radius of convergence does
i _ P iHx ,moä so the series converges for k) _D k + . Notice
not tell us what happens at the endpoints, _
and _ )+ . We see in Chapter ?? that the series
converges to _ for _ in its interval of convergence.
montp
" . _ P ^ 0 " . What happens if some
The ratio test requires "q#Ìr i ^9"ã i wi ^1";i to exist for ^1" )
of the coefficients " are zero? Then we use the fact that an infinite series can be written in several
ways and pick one in which the terms are nonzero. For example, we think of the power series
C N
"9` _ P _ _ - P _ 6,. P B0 "9` . + _ $ P 60 , so ^ " ) . P B0 "9` _ "9` w . + $ P B0 . With this
as the series with ^ ) _ and ^ )ÚP _ w
choice of ^ " , all ^ " ) b , so we can use the ratio test.
B½
©
©
«
Determine the radius of convergence of the series
_
_
_
f
_
_
/
/
/
/
/
6 We
do not take '
8:a.kL'
8=l3kL'
8mEna
*o;32pkL'
8=l3kKq>qKq because then rpspt
/
)(
6
&'
& *)& '
& does not exist.
450
Chapter Nine SERIES
Example 3
C N
_ P _ _ - P _ 6,. P B0 "9` /
/
Solution
"9` . + _ $ P 6 0 Find the radius and interval of convergence of the series
/
/
_. "9` 60 +$ P
so that, replacing $ by $
, we have
"ã ` . 6 0 " _ "ã .
6
0
_
"
ã
`
*
)
P
^1"ã
. + . $ 60 P 60 ) P . + $ B0
^9" ) . P B0 "9` We take
Kv
u
>v
u
/
/
/
Thus,
i ^9"ã i ) . . P 6 06"90 "` "ã ) . . P B06"90 "` _ " ã"9` . + . $ P B060 ) . . P 60 6_ 0 ) . _ 60 P _ +$
+$ +$ +$ +$
i^ " i P
" ` Because
montp i ^ "ã i mtnop _ )
)
.
B
0
"
s
q
r
"
s
q
r
+
+
i^ " i $ $
montp
)
)
"qsr i ^ "ã i wi ^ " i k for all _ Thus, the ratio test guarantees that the power
we have Î
series converges for every _ . The radius of convergence isnoä infinite and the interval of convergence
is all _ . We see in Chapter ?? that the series converges to
_.
u Z3w
$
$
$
$
$
$
U3x^y
>vLz
$
U|{^y $$
>vLz
uZ w
$
/
$
$
$
$
$
$
$
$
$
$
$
/ $
$
$
$
$
$
$
[
Example 4
Solution
~ + _ + _ ^ " ,) + " _ " for $ even and ^9" )
If we take 9
+ _ ¡ + " _ " for $ odd, "moqsntp r i ^9"ã i wi ^1"8i does not exist. Therefore,
for this series we take
)D+ " _ " "
^
, we have
so that, replacing $ by $
^ "ã )D+ "ã _ "ã ),+ "ã _ "ã Thus,
i ^1"ã i ) + "ã " _ "" ã ) + _ ) _ i^ " i
i+ _ i
Because
montp i ^9"ã i ) _ "
q
r
"
i
^
i
montp
) _ . Thus, the ratio test guarantees
we have
that the power series converges
if _ k Î ) that is,i ^1if"ãi _j i9i wk i ^9 "8.i The
radius of convergence is . The series converges for P ke_k , all the terms in the series
and diverges for _lk or _yx P . At _ )
are 1, so the series diverges
(by Theorem 9.2 Property 3). Thus, the interval of convergence is P kà_k .
Find the radius and interval of convergence
>v
u
$
u
Kv
$
$
$
}
$
$
$
$
451
9.4 POWER SERIES
What Happens at the Endpoints of the Interval of Convergence?
_) ^
The ratio test does not tell us whether a power series converges at the endpoints of its interval of
} f . There is no simple theorem that answers this question. Since substituting
convergence,
} f converts the power series to a series of numbers, the tests in Sections 9.2 and 9.3 are
often useful. See Examples 4 and 5.
_ ) ^
. _ P 6 0 P . _ P B0 . _ P 60 P . _ P 60 , 6,. P B0 "9` . _ P B0 " +
$
; it converges for k,_Qk + and
In Example 2 on page 449, we showed that this series has )
diverges for _k
or _x + . We need to determine whether it converges at the endpoints of the
and _ ),+ . At _ ),+ , we have the series
interval of convergence, _ )
P P B . P 60 " ` +
$
.
,
6
0
This is an alternating series with ^9" )
, we have the series w $ , so by the alternating series test (Theorem ??), it
converges. At _ )
P P + P P P P $ P This is the negative of the harmonic series, so it diverges. Therefore, the interval of convergence is
k_ Ó + . The right endpoint is included and the left endpoint is not.
Example 5
Determine the interval of convergence of the series
Solution
f
Ô
Exercises and Problems for Section 9.4
Exercises

¢ ©(ª ©

6§ ü ¾c|¦¸ ¦¸ 6§ § {A¦¸

¨
Which of the series in Problems 1–4 are power series?
1.
3.
2.

4. e~
6.
7.
8.
SV
SV
JV
SV
V
SV
11.
12.
13.
V
V
~
SV
JV
Ô

¦¸ § ¦¸ § ¸¦ § c

¦¸ § ¦¸ 6§ ¦¸ 6§ ¸¦ § yA
V
|V

L l > § g § Q §
ú9 ú&¦ gú gú&¦ ú ¦ ú l QA
@ ¦¸ § ¦¸ § ¦¸ § ¢

Q
§ ¦¸ y § ¦¸ l § ¦¸ l §
¦¸
g
|V
10.
Ô
~

SV
Find the radius of convergence of the series in Problems 12–
15.
Find an expression for the general term of the series in Problems 5–10. Give the starting value of the index ( or for
example).
5.
9.
Ô

V
14.
¬6« ® ¦ §
ª
¬6« ® µ
ª ¿ ç
¬6« ® ¦¸µ §
ª yµ
¬6« ® ¦¸ §
ª µ

>
452
Chapter Nine SERIES
Use the ratio test to find the radius of convergence of the
power series in Problems 15–21.
15.
17.

j6 Q ¥ 1
¥

6 A
B L c
18.
19.
16.
SV
JV
V
20.
SV
21.

* § ©
QQ¦ ô µ c ´
Problems
22. (a) Determine the radius of convergence of the series
23. Show that the series
*
± Æ
¬B« ® ¦ §
© µ

ë 6Æ
± Æ6
converges for N N

Investigate whether the series converges for
.
.
ú
g §
g § §

ú
&
¦
ú

ª
±À
JV
i
25. Show that if
i
:
:
converges
for N N
with given by the radius of convergence
.
test, then so does

26. Suppose that the power
and diverges when

V
JV
JV
V
V
|V
~
converges when
. Which of the following
A
±±
±±
(a)
(b)
(c)
(d)
The power series converges when
The power series converges when
The power series diverges when
The power series diverges when
.
.
.
.

Þ ¦¸ § Þ
Þ Þ

¬6« ® ¦¸ µ §
©
± ±
Decide if the statements in Problems 27–31 are true or false.
Give an explanation for your answer.
and
24. For constant , find the radius of convergence of the binomial power series:7
SV
V
|V
QA
¦ 6 § ¦ § c
c
are true, false or not possible to determine? Give reasons
for your answers.
What does this tell us about the convergence of this
series?
(b) Investigate convergence at the end points of the interval of convergence of this series.

À
¥
L ¦ § ¦ §
L À :¥

À Q
27.

and
have the same radii of con-
vergence.
and
have the same radii of conver28. If

and , must be equal.
gence, then the coefficients,
29.
is a power series.
30. A geometric series is a power series.
i
i

,
converges at one
31. If a power series with
endpoint of its interval of convergence, then it converges
at the other endpoint also.
REVIEW PROBLEMS FOR CHAPTER NINE
Exercises

Use the comparison test to confirm the statements in Problems 1–2.
¬B« ®
©
¬B« ®
©µ
1.
2.
¬6« ® µ
© µ
¬B« ®
© µ*÷Mº»ê µ
converges, so
diverges, so
converges.
diverges.
Use the comparison test to decide which of the series in Problems 3–4 converge and which diverge.
7 For
an explanation of the name, see Section 10.2.
3.
4.
¬6« ® µ
©µ
¬6« ® µ |µs
© µ

Determine which of the series in Problems 5–11 converge.
5.
W
6.

µ
7.
8.
¬B« ®
© ÷Mº»êHµ
B¬ « ® §
©¦µ
¬B« ® ¦ µ § §
© ¦¸µ*
¬B« ® ¦ § ô ©
© Â µ#
¬« ® ¹»ê
©
REVIEW PROBLEMS FOR CHAPTER NINE

Find the radius of convergence of the series in Problems 12–
15.
12.
IV
10.
13.
V
14.
11.
¬6« ® µ
ª
¬6« ® ¦ µ § §
ª ¦¸µ
¬6« ® ¦ µ §
ª
¬6« ®
ªµ
IV
IV
9.
453
15.
Problems
16. A repeating decimal can always be expressed as a fraction. This problem shows how writing a repeating decimal as a geometric series enables you to find the fraction.
Consider the decimal
BB
´ 66 ´A´´´ ±

66
´ gA ´ ´´´ ´ ´
6B ± BÆ
´ ´´´ ¥6¥
(a) Use the fact that
to write 0.232323. . . as a geometric
series.
(b) Use the formula for the sum of a geometric series to
.
show that
17. Cephalexin is an antibiotic with a half-life in the body of
0.9 hours, taken in tablets of 250 mg every six hours.

V
V
Ê
6
sell
worth of Christmas trees once a year forever,
with the first sale in the immediate future. What is the
present value of this Christmas tree business? Assume
that the interest rate is 4 per year, compounded continuously.
Ë
20. Before World War I, the British government issued what
are called consols, which pay the owner or his heirs a
fixed amount of money every year forever. (Cartoonists
of the time described aristocrats living off such payments
as “pickled in consols.”) What should a person expect to
pay for a consol which pays 10 a year forever? Assume
the first payment is one year from the date of purchase
and that interest remains 4% per year, compounded annually. ( denotes pounds, the British unit of currency.)
(a) What percentage of the cephalexin in the body at the
start of a six-hour period is still there at the end (assuming no tablets are taken during that time)?
(b) Write an expression for
,
,
,
, where
mg, is the amount of cephalexin in the body right
after the
tablet is taken.
,
in closed-form and evaluate them.
(c) Express
(d) Write an expression for
and put it in closedform.
(e) If the patient keeps taking the tablets, use your answer to part (d) to find the quantity of cephalexin in
the body in the long run, right after taking a pill.
Problems 21–23 are about bonds, which are issued by a government to raise money. An individual who buys a $1000 bond
gives the government $1000 and in return receives a fixed sum
of money, called the coupon, every six months or every year
for the life of the bond. At the time of the last coupon, the
individual also gets the $1000, or principal back.
18. Around January 1, 1993, Barbra Streisand signed a conmillion a year for 10
tract with Sony Corporation for
years. Suppose the first payment was made on the day of
signing and that all other payments are made on the first
day of the year. Suppose also that all payments are made
into a bank account earning
a year, compounded annually.
22. What is the present value of a $1000 bond which pays
$50 a year for 10 years, starting one year from now? Assume the interest rate is 4% per year, compounded annually.
¯ µ ¶ · ¯
¯©¯¯¯

¯
¯
Ê
Ë
(a) How much money was in the account
(i) On the night of December 31, 1999?
(ii) On the day the last payment is made?
(iii) What was the present value of the contract on
the day it was signed?
19. One way of valuing a company is to calculate the present
value of all its future earnings. Suppose a farm expects to
21. What is the present value of a $1000 bond which pays
$50 a year for 10 years, starting one year from now? Assume interest rate is 6% per year, compounded annually.
23. (a) What is the present value of a $1000 bond which
pays $50 a year for 10 years, starting one year from
now? Assume the interest rate is 5% per year, compounded annually.
(b) Since $50 is 5% of $1000, this bond is often called a
5% bond. What does your answer to part (a) tell you
about the relationship between the principal and the
present value of this bond when the interest rate is
5%?
454
Chapter Nine SERIES
Ë
(c) If the interest rate is more than 5 per year, compounded annually, which is larger: the principal or
the value of the bond? Why do you think the bond is
then described as trading at discount?
(d) If the interest rate is less than 5 per year, compounded annually, why is the bond described as
trading at a premium?
Ë
±
24. Suppose that the power

B¬ « ® ¦¸ 6 §
ª ±

converges
when
and diverges when
. Which of the following are true, false or not possible to determine? Give
reasons for your answers.
(a)
(b)
(c)
(d)
(e)
± ±d À
± ± Ô ´ Ô
±
Ô
û û
.
The power series converges when
The power series diverges when
The power series converges when
The power series diverges when
The power series converges when
Ô
Ô
Ô
Þ

, show that
25. (a) For a series
N
N.
(b) Use part (a) to show that if QN
converges.
Þ
.
.
.
.
N
N
N
converges, then
PROJECTS
Exercises
1. The Fly and the Train
In an old puzzle, there are two trains, each moving
at 10 km/hr toward one another. Initially the trains
are 30 kilometers apart. At the same moment a fly,
whose velocity is 20 km/hr, starts at one train and
flies till it meets the other, then turns around and
flies back till it meets the first train, and so on.
(a) How far has the fly traveled the first time
it turns around? The second time? The third
time? The fourth time?
(b) How far has the fly traveled by the
time
it turns around? Write your answer in closedform.
(c) Use you answer to part (b) to decide how far
the fly has traveled by the time the trains meet
in the middle and squash it.
(d) How long does it take the trains to meet in
the middle? Use this to answer part (c) without
summing a series.
$&%('
2. Probability of Winning in Sports
In certain sports, winning a game requires a
lead of two points. That is, if the score is tied you
have to score two points in a row to win.
(a) For some sports (e.g. tennis), a point is scored
every play. Suppose your probability of scoring the next point is always . Then, your opponent’s probability of scoring the next point is
always
.
Pò
ò
(i) What is the probability that you win the
next two points?
(ii) What is the probability that you and your
opponent split the next two points, that is,
that neither of you wins both points?
(iii) What is the probability that you split the
next two points but you win the two after
that?
(iv) What is the probability that you either win
the next two points or split the next two
and then win the next two after that?
(v) Give a formula for your probability of
winning a tied game.
(vi) Compute your probability of winning a
; when
;
tied game when
when
; when
. Comment
on your answers.
(vii) In other sports (e.g. volleyball), you can
score a point only if it is your turn, with
turns alternating until a point is scored.
Suppose your probability of scoring a
point when it is your turn is , and your
opponent’s probability of scoring a point
when it is her turn is .
A. Find a formula for the probability
that you are the first to score the next
point, assuming it is currently your
turn.
B. Suppose that if you score a point, the
next turn is yours. Using your answers to part (a) and your formula for
, compute the probability of winning a tied game (if you need two
points in a row to win).
Assume
and
and
it is your turn.
) - ò ) 7
ò
ò )
ò )
ò
a
a
Ï
ò ) -
) -
Ï Assume ò ) 7
and
) -
PROJECTS
and
it is your turn.
455

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