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James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

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Section 11.8 Power Series 747

In fact if we put x − 1 2 in the geometric series (2) we get the convergent series

n−0S ò 2D

1 n

− 1 1 1 2 1 1 4 1 1 8 1 1 16 1 ∙ ∙ ∙

but if we put x − 2 in (2) we get the divergent series

More generally, a series of the form

ò 2 n − 1 1 2 1 4 1 8 1 16 1 ∙ ∙ ∙

n−0

3

ò c n sx 2 ad n − c 0 1 c 1 sx 2 ad 1 c 2 sx 2 ad 2 1 ∙ ∙ ∙

n−0

Notice that

sn 1 1d! − sn 1 1dnsn 2 1d ? . . . ? 3 ? 2 ? 1

− sn 1 1dn!

is called a power series in sx 2 ad or a power series centered at a or a power series

about a. Notice that in writing out the term corresponding to n − 0 in Equations 1 and 3

we have adopted the convention that sx 2 ad 0 − 1 even when x − a. Notice also that

when x − a, all of the terms are 0 for n > 1 and so the power series (3) always converges

when x − a.

Example 1 For what values of x is the series ò n!x n convergent?

n−0

SOLUtion We use the Ratio Test. If we let a n , as usual, denote the nth term of the

series, then a n − n!x n . If x ± 0, we have

lim

nl`

Z

a n11

a n

Z − lim

nl`

Z

sn 1 1d!x n11

n!x n

Z − lim

nl`

sn 1 1d| x | − `

By the Ratio Test, the series diverges when x ± 0. Thus the given series converges only

when x − 0.

n

sx 2 3d

Example 2 For what values of x does the series ò

n

n−1 n

SOLUtion Let a n − sx 2 3d n yn. Then

Z

a n11

a n

Z − Z

sx 2 3d n11

n 1 1

?

n

sx 2 3d n

Z

converge?

− 1

1 1 1 n

| x 2 3 | l | x 2 3 | as n l `

By the Ratio Test, the given series is absolutely convergent, and therefore convergent,

when | x 2 3 | , 1 and divergent when | x 2 3 | . 1. Now

| x 2 3 |

, 1 &? 21 , x 2 3 , 1 &? 2 , x , 4

so the series converges when 2 , x , 4 and diverges when x , 2 or x . 4.

The Ratio Test gives no information when | x 2 3 | − 1 so we must consider x − 2

and x − 4 separately. If we put x − 4 in the series, it becomes o 1yn, the harmonic

series, which is divergent. If x − 2, the series is o s21d n yn, which converges by the

Alternating Series Test. Thus the given power series converges for 2 < x , 4. n

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