Section 9.8 Power Series Power Seriesstaff.katyisd.org/sites/thscalculusap/Larson... · 660 CHAPTER 9 Infinite Series EXAMPLE 1 Power Series a. The following power series is centered

SECTION 9.8 Power Series 659

Section 9.8 Power Series

• Understand the definition of a power series.• Find the radius and interval of convergence of a power series.• Determine the endpoint convergence of a power series.• Differentiate and integrate a power series.

Power Series

In Section 9.7, you were introduced to the concept of approximating functions byTaylor polynomials. For instance, the function can be approximated by itsMaclaurin polynomials as follows.

1st-degree polynomial

2nd-degree polynomial

3rd-degree polynomial

4th-degree polynomial

5th-degree polynomial

In that section, you saw that the higher the degree of the approximating polynomial,the better the approximation becomes.

In this and the next two sections, you will see that several important types offunctions, including

can be represented exactly by an infinite series called a power series. For example,the power series representation for is

For each real number it can be shown that the infinite series on the right convergesto the number Before doing this, however, some preliminary results dealing withpower series will be discussed—beginning with the following definition.

NOTE To simplify the notation for power series, we agree that even if x � c.�x � c�0 � 1,

ex.x,

ex � 1 � x �x2

2!�

x3

3!� . . . �

xn

n!� . . . .

ex

f�x� � ex

ex � 1 � x �x2

2!�

x3

3!�

x4

4!�

x5

5!

ex � 1 � x �x2

2!�

x3

3!�

x4

4!

ex � 1 � x �x2

2!�

x3

3!

ex � 1 � x �x2

2!

ex � 1 � x

f�x� � ex

Definition of Power Series

If is a variable, then an infinite series of the form

is called a power series. More generally, an infinite series of the form

is called a power series centered at c, where is a constant.c

��

n�0 an�x � c�n � a0 � a1�x � c� � a2�x � c�2 � . . . � an�x � c�n � . . .

��

n�0 anxn � a0 � a1x � a2x

2 � a3x3 � . . . � anxn � . . .

x

E X P L O R A T I O N

Graphical Reasoning Use a graphingutility to approximate the graph ofeach power series near (Usethe first several terms of each series.)Each series represents a well-knownfunction. What is the function?

a.

b.

c.

d.

e. ��

n�0 2n xn

n!

��

n�0 ��1�n x2n�1

2n � 1

��

n�0 ��1�n x2n�1

�2n � 1�!

��

n�0 ��1�n x2n

�2n�!

��

n�0 ��1�nxn

n!

x � 0.

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660 CHAPTER 9 Infinite Series

EXAMPLE 1 Power Series

a. The following power series is centered at 0.

b. The following power series is centered at

c. The following power series is centered at 1.

Radius and Interval of Convergence

A power series in can be viewed as a function of

where the domain of is the set of all for which the power series converges.Determination of the domain of a power series is the primary concern in this section.Of course, every power series converges at its center because

So, always lies in the domain of The following important theorem states that thedomain of a power series can take three basic forms: a single point, an intervalcentered at or the entire real line, as shown in Figure 9.17. A proof is given inAppendix A.

c,

f.c

� a0.

� a0�1� � 0 � 0 � . . . � 0 � . . .

f�c� � ��

n�0 an�c � c�n

c

xf

f�x� � ��

n�0 an�x � c�n

xx

��

n�1 1n

�x � 1�n � �x � 1� �12

�x � 1�2 �13

�x � 1�3 � . . .

��

n�0 ��1�n �x � 1�n � 1 � �x � 1� � �x � 1�2 � �x � 1�3 � . . .

�1.

��

n�0 xn

n!� 1 � x �

x2

2�

x3

3!� . . .

cx

A single point

The domain of a power series has only three basic forms: a single point, an intervalcentered at or the entire real line.Figure 9.17

c,

cx

R R

An interval

cx

The real line

THEOREM 9.20 Convergence of a Power Series

For a power series centered at precisely one of the following is true.

1. The series converges only at

2. There exists a real number such that the series converges absolutely forand diverges for

3. The series converges absolutely for all

The number is the radius of convergence of the power series. If the seriesconverges only at the radius of convergence is and if the seriesconverges for all the radius of convergence is The set of all values of

for which the power series converges is the interval of convergence of thepower series.x

R � �.x,R � 0,c,

R

x.�x � c� > R.�x � c� < R,

R > 0

c.

c,

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EXAMPLE 2 Finding the Radius of Convergence

Find the radius of convergence of

Solution For you obtain

For any fixed value of such that let Then

Therefore, by the Ratio Test, the series diverges for and converges only at itscenter, 0. So, the radius of convergence is



Solution For let Then

By the Ratio Test, the series converges if and diverges if Therefore, the radius of convergence of the series is



Solution Let Then

For any value of this limit is 0. So, by the Ratio Test, the series converges forall Therefore, the radius of convergence is R � �.x.

x,fixed

� limn→�

x2

�2n � 3��2n � 2�.

limn→�

�un�1

un � � limn→�

� ��1�n�1 x2n�3

�2n � 3�!

��1�n x2n�1 �2n � 1�! �

un � ��1�nx2n�1��2n � 1�!.

��

n�0 ��1�nx2n�1

�2n � 1�! .

R � 1.�x � 2� > 1.�x � 2� < 1

� �x � 2�. � lim

n→� �x � 2�

limn→�

�un�1


�3�x � 2�n�1

3�x � 2�n �un � 3�x � 2�n.x � 2,

��

n�0 3�x � 2�n.

R � 0.�x� > 0

� �.

� �x� limn→�

�n � 1�

limn→�

�un�1


��n � 1�!xn�1

n!xn �un � n!xn.�x� > 0,x

f�0� � ��

n�0 n!0n � 1 � 0 � 0 � . . . � 1.

x � 0,

��

n�0 n!xn.

STUDY TIP To determine the radius ofconvergence of a power series, use theRatio Test, as demonstrated in Examples2, 3, and 4.

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Endpoint Convergence

Note that for a power series whose radius of convergence is a finite number Theorem 9.20 says nothing about the convergence at the endpoints of the interval ofconvergence. Each endpoint must be tested separately for convergence or divergence.As a result, the interval of convergence of a power series can take any one of the sixforms shown in Figure 9.18.

EXAMPLE 5 Finding the Interval of Convergence

Find the interval of convergence of

Solution Letting produces

So, by the Ratio Test, the radius of convergence is Moreover, because theseries is centered at 0, it converges in the interval This interval,however, is not necessarily the interval of convergence. To determine this, you musttest for convergence at each endpoint. When you obtain the divergent harmonicseries

Diverges when

When you obtain the convergent alternating harmonic series

Converges when

So, the interval of convergence for the series is as shown in Figure 9.19.��1, 1�,

x � �1��

n�1 ��1�n

n� �1 �

12

�13

�14

� . . . .

x � �1,

x � 1��

n�1 1n

�11

�12

�13

� . . . .

x � 1,

��1, 1�.R � 1.

� �x�. � lim

n→� � nx

n � 1� limn→�

�un�1


� xn�1

�n � 1�

xn

n �un � xn�n

��

n�1 xn

n.

R,

Radius: R = 1

c = 0x

−1 1

Interval: [−1, 1)

Figure 9.19

cx

Radius: 0

cx

R

(c − R, c + R)

RRadius:

cx

(c − R, c + R]

R

cx

[c − R, c + R)

R

cx

[c − R, c + R]

R

Intervals of convergenceFigure 9.18

Radius: �

cx

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By the Ratio Test, the series converges if or So, theradius of convergence is Because the series is centered at it willconverge in the interval Furthermore, at the endpoints you have

Diverges when

and

Diverges when

both of which diverge. So, the interval of convergence is as shown in Figure9.20.




So, the radius of convergence is Because the series is centered at itconverges in the interval When you obtain the convergent series

Converges when

When you obtain the convergent alternating series

Converges when

Therefore, the interval of convergence for the given series is ��1, 1�.

x � �1��

n�1 ��1�n

n2 � �112 �

122 �

132 �

142 � . . . .

x � �1,

x � 1��

n�1 1n2 �

112 �

122 �

132 �

142 � . . . .

p-x � 1,��1, 1�.x � 0,R � 1.

� limn→�

� n2x�n � 1�2� � �x�.

limn→�

�un�1


�xn�1��n � 1�2

xn�n2 �un � xn�n2

��

n�1 xn

n2.

��3, 1�,

x � 1��

n�0 ��1�n�2�n

2n � ��

n�0 ��1�n

x � �3��

n�0 ��1�n ��2�n

2n � ��

n�0 2n

2n � ��

n�0 1

��3, 1�.x � �1,R � 2.

�x � 1� < 2.��x � 1��2� < 1

� �x � 12 �.

� limn→�

�2n�x � 1�2n�1 �

limn→�

�un�1


� ��1�n�1�x � 1�n�1

2n�1

��1�n�x � 1�n

2n �un � ��1�n�x � 1�n�2n

��

n�0 ��1�n�x � 1�n

2n .

Radius: R = 2

c = −1x

−3 −2 10

Interval: (−3, 1)

Figure 9.20

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Differentiation and Integration of Power Series

Power series representation of functions has played an important role in the develop-ment of calculus. In fact, much of Newton’s work with differentiation and integrationwas done in the context of power series—especially his work with complicatedalgebraic functions and transcendental functions. Euler, Lagrange, Leibniz, and theBernoullis all used power series extensively in calculus.

Once you have defined a function with a power series, it is natural to wonder howyou can determine the characteristics of the function. Is it continuous? Differentiable?Theorem 9.21, which is stated without proof, answers these questions.

Theorem 9.21 states that, in many ways, a function defined by a power seriesbehaves like a polynomial. It is continuous in its interval of convergence, and both itsderivative and its antiderivative can be determined by differentiating and integratingeach term of the given power series. For instance, the derivative of the power series

is

Notice that Do you recognize this function?f��x� � f�x�.

� f�x�.

� 1 � x �x2

2�

x3

3!�

x4

4!� . . .

f��x� � 1 � �2� x2

� �3� x2

3!� �4� x

3

4!� . . .

� 1 � x �x2

2�

x3

3!�

x4

4!� . . .

f�x� � ��

n�0 xn

n!

THEOREM 9.21 Properties of Functions Defined by Power Series

If the function given by

has a radius of convergence of then, on the interval isdifferentiable (and therefore continuous). Moreover, the derivative and anti-derivative of are as follows.

1.

2.

The radius of convergence of the series obtained by differentiating or integratinga power series is the same as that of the original power series. The interval ofconvergence, however, may differ as a result of the behavior at the endpoints.

� C � a0�x � c� � a1 �x � c�2

2� a2

�x � c�3

3� . . .

f �x� dx � C � ��

n�0 an

�x � c�n�1

n � 1

� a1 � 2a2�x � c� � 3a3�x � c�2 � . . .

f��x� � ��

n�1 nan�x � c�n�1

f

f�c � R, c � R�,R > 0,

� a0 � a1�x � c� � a2�x � c�2 � a3�x � c�3 � . . .

f�x� � ��

n�0 an�x � c�n

JAMES GREGORY (1638–1675)

One of the earliest mathematicians to workwith power series was a Scotsman, JamesGregory. He developed a power series methodfor interpolating table values—a method thatwas later used by Brook Taylor in the develop-ment of Taylor polynomials and Taylor series.

The

Gra

nger

Col

lect

ion

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EXAMPLE 8 Intervals of Convergence for and

Consider the function given by

Find the intervals of convergence for each of the following.

a. b. c.

Solution By Theorem 9.21, you have

and

By the Ratio Test, you can show that each series has a radius of convergence of Considering the interval you have the following.

a. For the series

Interval of convergence:

converges for and its interval of convergence is See Figure9.21(a).

b. For the series


converges for and diverges for So, its interval of convergence isSee Figure 9.21(b).

c. For the series


diverges for and its interval of convergence is See Figure 9.21(c).

From Example 8, it appears that of the three series, the one for the derivative,is the least likely to converge at the endpoints. In fact, it can be shown that if the

series for converges at the endpoints the series for will also converge there.

f�x�x � c ± R,f��x�f��x�,

��1, 1�.x � ±1,

��1, 1��

n�1 xn�1

f��x�,��1, 1�.

x � 1.x � �1

��1, 1��

n�1 xn

n

f�x�,

��1, 1�.x � ±1,

��1, 1��

n�1

xn�1

n�n � 1�

f�x� dx,

��1, 1�,R � 1.

� C �x2

1 � 2�

x3

2 � 3�

x4

3 � 4� . . . .

f �x� dx � C � ��

n�1

xn�1

n�n � 1�

� 1 � x � x2 � x3 � . . .

f��x� � ��

n�1 xn�1

f��x�f�x� f�x� dx

f�x� � ��

n�1 xn

n� x �

x2

2�

x3

3� . . . .

f�x� dxf��x�,f�x�,

Radius: R = 1

c = 0

x

−1 1

Interval: [−1, 1]Radius: R = 1

c = 0

x

−1 1

Interval: [−1, 1)Radius: R = 1

c = 0

x

−1 1

Interval: (−1, 1)

(a) (b) (c)Figure 9.21

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E x e r c i s e s f o r S e c t i o n 9 . 8 See www.CalcChat.com for worked-out solutions to odd-numbered exercises.

In Exercises 1–4, state where the power series is centered.

1. 2.

3. 4.

In Exercises 5–10, find the radius of convergence of the powerseries.

5. 6.

7. 8.

9. 10.

In Exercises 11–34, find the interval of convergence of thepower series. (Be sure to include a check for convergence at theendpoints of the interval.)

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

31.

32.

33.

34.

In Exercises 35 and 36, find the radius of convergence of thepower series, where and is a positive integer.

35. 36.

In Exercises 37– 40, find the interval of convergence of thepower series. (Be sure to include a check for convergence at theendpoints of the interval.)

37. 38.

39.

40.

In Exercises 41–44, write an equivalent series with the index ofsummation beginning at

41. 42.

43. 44.

In Exercises 45–48, find the intervals of convergence of (a) (b) (c) and (d) Include a check for convergence at the endpoints of the interval.

45.

46.

47.

48.

Writing In Exercises 49–52, match the graph of the first 10terms of the sequence of partial sums of the series

with the indicated value of the function. [The graphs are labeled(a), (b), (c), and (d).] Explain how you made your choice.

(a) (b) Sn

n2

2468

1012

4 6 8

Sn

n2

2

4 6 8

1

3

g�x� � ��

n�0 x

3�n

f �x� � ��

n�1 ��1�n�1�x � 2�n

n

f �x� � ��

n�0 ��1�n�1�x � 1�n�1

n � 1

f �x� � ��

n�1 ��1�n�1 �x � 5�n

n5n

f �x� � ��

n�0 �x

2�n

f �x� dx.f � �x�,f��x�,f �x�,

��

n�0 ��1�n x2n�1

2n � 1��

n�0

x2n�1

�2n � 1�!

��

n�0 ��1�n�1�n � 1�xn�

�

n�0 xn

n!

n � 1.

��

n�1

n!�x � c�n

1 � 3 � 5 . . . �2n � 1�

k ≥ 1��

n�1 k�k � 1��k � 2� . . . �k � n � 1�xn

n!,

��

n�1 ��1�n�1�x � c�n

ncnk > 0��

n�0 �x

k�n

,

��

n�0 �n!�k xn

�kn�!��

n�1 �x � c�n�1

c n�1

kc > 0

��

n�1

n!�x � 1�n

1 � 3 � 5 . . . �2n � 1�

��

n�1 ��1�n�1 3 � 7 � 11 . . . �4n � 1��x � 3�n

4n

��

n�1 � 2 � 4 � 6 . . . 2n

3 � 5 � 7 . . . �2n � 1�� x2n�1

��

n�1 2 � 3 � 4 . . . �n � 1�xn

n!

��

n�1 n!xn

�2n�!��

n�0

x2n�1

�2n � 1�!

��

n�0 ��1�n x2n

n!��

n�1

nn � 1

��2x�n�1

��

n�0 ��1�n x2n�1

2n � 1��

n�1 �x � 3�n�1

3n�1

��

n�1 (�1�n�1�x � 2�n

n2n��

n�0 ��1�n�1�x � 1�n�1

n � 1

��

n�0

�x � 2�n�1

�n � 1�4n�1��

n�1 ��1�n�1�x � 5�n

n5n

��

n�0 ��1�n n!�x � 4�n

3n��

n�1 ��1�n�1xn

4n

��

n�0

��1�n xn

�n � 1��n � 2��

n�0 �2n�! �x

2�n

��

n�0 �3x�n

�2n�!��

n�0 xn

n!

��

n�0 ��1�n�1�n � 1�xn�

�

n�1 ��1�n xn

n

��

n�0 �x

5�n

��

n�0 �x

2�n

��

n�0 �2n�!x2n

n!��

n�0 �2x�2n

�2n�!

��

n�0 ��1�n xn

2n��

n�1 �2x�n

n2

��

n�0 �2x�n�

�

n�0 ��1�n

xn

n � 1

��

n�0

��1�n�x � ��2n

�2n�!��

n�1

�x � 2�n

n3

��

n�1

��1�n1 � 3 . . . �2n � 1�2nn!

xn��

n�0nxn

332460_0908.qxd 11/4/04 3:30 PM Page 666


(c) (d)

49. 50.

51. 52.

Writing In Exercises 53–56, match the graph of the first 10terms of the sequence of partial sums of the series

with the indicated value of the function. [The graphs are labeled(a), (b), (c), and (d).] Explain how you made your choice.

(a) (b)

(c) (d)

53. 54.

55. 56.

63. Let and

(a) Find the intervals of convergence of and

(b) Show that

(c) Show that

(d) Identify the functions and

64. Let

(a) Find the interval of convergence of

(b) Show that

(c) Show that

(d) Identify the function

In Exercises 65–70, show that the function represented by thepower series is a solution of the differential equation.

65.

66.

67.

68.

69.

70.

71. Bessel Function The Bessel function of order 0 is

(a) Show that the series converges for all

(b) Show that the series is a solution of the differential equation

(c) Use a graphing utility to graph the polynomial composed ofthe first four terms of

(d) Approximate accurate to two decimal places.10 J0 dx

J0.

x2 J0 � x J0� � x2 J0 � 0.

x.

J0�x� � ��

k�0 ��1�k x2k

22k �k!�2 .

y � x 2 y � 0y � 1 � ��

n�1

��1�n x 4n

22n n! � 3 � 7 � 11 . . . �4n � 1�,

y � xy� � y � 0y � ��

n�0

x2n

2n n!,

y � y � 0y � ��

n�0

x2n

�2n�!,

y � y � 0y � ��

n�0

x2n�1

�2n � 1�!,

y � y � 0y � ��

n�0 ��1�n x2n

�2n�!

y � y � 0y � ��

n�0 ��1�n x2n�1

�2n � 1�! ,

f.

f �0� � 1.

f��x� � f �x�.f.

f �x� � ��

n�0 xn

n!.

g.f

g��x� � �f �x�.f��x� � g�x�.

g.f

g�x� � ��

n�0 ��1�n x2n

�2n�! .f �x� � ��

n�0 ��1�n x2n�1

�2n � 1�!

g�38�g� 9

16�

g��18�g�1

8�

1 2 3 4 5 6 7 8 9

3

6

9

12

15

18

Sn

n−1 1 2 3 4 5 6 7 8 9

0.25

0.50

0.75

1.00

Sn

n

Sn

n−1 1 2 3 4 5 6 7 8 9

0.5

1.0

1.5

2.0

1 2 3 4 5 6 7 8 9

1

2

3

4

Sn

n

g�x� � ��

n�0 �2x�n

g��2�g�3.1�g�2�g�1�

Sn

n2 4 6 8

1

1

1

3

2

4

4

Sn

n2

2

4 6 8

1

Writing About Concepts (continued)60. Describe how to differentiate and integrate a power series

with a radius of convergence Will the series resultingfrom the operations of differentiation and integration havea different radius of convergence? Explain.

61. Give examples that show that the convergence of a powerseries at an endpoint of its interval of convergence may beeither conditional or absolute. Explain your reasoning.

62. Write a power series that has the indicated interval of convergence. Explain your reasoning.

(a) (b) (c) (d) ��2, 6��1, 0��1, 1��2, 2�

R.

Writing About Concepts57. Define a power series centered at

58. Describe the radius of convergence of a power series.Describe the interval of convergence of a power series.

59. Describe the three basic forms of the domain of a powerseries.

c.

332460_0908.qxd 11/4/04 3:30 PM Page 667


72. Bessel Function The Bessel function of order 1 is

(a) Show that the series converges for all

(b) Show that the series is a solution of the differential equation

(c) Use a graphing utility to graph the polynomial composed ofthe first four terms of

(d) Show that

In Exercises 73–76, the series represents a well-known function.Use a computer algebra system to graph the partial sum andidentify the function from the graph.

73. 74.

75.

76.

77. Investigation In Exercise 11 you found that the interval of

convergence of the geometric series is

(a) Find the sum of the series when Use a graphing utility to graph the first six terms of the sequence of partialsums and the horizontal line representing the sum of theseries.

(b) Repeat part (a) for

(c) Write a short paragraph comparing the rate of convergenceof the partial sums with the sum of the series in parts (a)and (b). How do the plots of the partial sums differ as theyconverge toward the sum of the series?

(d) Given any positive real number there exists a positiveinteger such that the partial sum

Use a graphing utility to complete the table.

78. Investigation The interval of convergence of the series

is

(a) Find the sum of the series when Use a graphing utili-ty to graph the first six terms of the sequence of partial sumsand the horizontal line representing the sum of the series.

(b) Repeat part (a) for

(c) Write a short paragraph comparing the rate of convergenceof the partial sums with the sum of the series in parts (a)and (b). How do the plots of the partial sums differ as theyconverge toward the sum of the series?

(d) Given any positive real number M, there exists a positiveinteger N such that the partial sum

Use a graphing utility to complete the table.

True or False? In Exercises 79–82, determine whether thestatement is true or false. If it is false, explain why or give anexample that shows it is false.

79. If the power series converges for then it also

converges for

80. If the power series converges for then it also

converges for

81. If the interval of convergence for is then the

interval of convergence for is

82. If converges for then

83. Prove that the power series

has a radius of convergence of if p and q are positiveintegers.

84. Let where the coef-ficients are and for

(a) Find the interval of convergence of the series.

(b) Find an explicit formula for

85. Let where for

(a) Find the interval of convergence of the series.

(b) Find an explicit formula for

86. Prove that if the power series has a radius of conver-

gence of R, then has a radius of convergence of

87. For let and Prove that if the interval of

convergence of the series is

then the series converges conditionally at x0 � R.

�x0 � R, x0 � R�,��

n�0 cn�x � x0�n

cn > 0.R > 0n > 0,

�R.��

n�0 cn x2n

��

n�0 cn x n

f �x�.

n ≥ 0.cn�3 � cnf �x� � ��

n�0 cn xn,

g�x�.

n ≥ 0.c2n�1 � 2c2n � 1g�x� � 1 � 2x � x2 � 2x3 � x4 � . . . ,

R � �

��

n�0

�n � p�!n!�n � q�! xn

��

n�0

an

n � 1.1

0 f �x� dx �

�x� < 2,f �x� � ��

n�0 an x

n

�0, 2�.��

n�0 an �x � 1�n

��1, 1�,��

n�0 an x

n

x � �1.

x � 2,��

n�0 an x

n

x � �2.

x � 2,��

n�0 an x

n

�N

n�0 �3 �

23�

n

> M.

x � �16.

x �16.

��13, 13�.�

�

n�0 �3x�n

�N

n�0 �3

2�n

> M.

NM,

x � �34.

x �34.

��2, 2�.��

n�0 �x

2�n

�1 ≤ x ≤ 1f �x� � ��

n�0 ��1�n

x2n�1

2n � 1,

�1 < x < 1f �x� � ��

n�0 ��1�n xn,

f �x� � ��

n�0 ��1�n

x2n�1

�2n � 1�!f �x� � ��

n�0 ��1�n

x2n

�2n�!

S10

J0��x� � �J1�x�.J1.

x2 J1 � x J1� � �x2 � 1� J1 � 0.

x.

J1�x� � x ��

k�0

��1�k x2k

22k�1 k!�k � 1�!.

10 100 1000 10,000

N

M

10 100 1000 10,000

N

M

332460_0908.qxd 11/4/04 3:30 PM Page 668

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Section 9.8 Power Series Power Seriesstaff.katyisd.org/sites/thscalculusap/Larson... · 660 CHAPTER 9 Infinite Series EXAMPLE 1 Power Series a. The following power series is centered