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

750 Chapter 11 Infinite Sequences and Series
In general, the Ratio Test (or sometimes the Root Test) should be used to determine
the radius of convergence R. The Ratio and Root Tests always fail when x is an endpoint
of the interval of convergence, so the endpoints must be checked with some other test.
Example 4 Find the radius of convergence and interval of convergence of the series
s23d
ò
n x n
n−0 sn 1 1
SOLUtion Let a n − s23d n x n ysn 1 1. Then
Z a n11
a n
Z − Z s23dn11 x n11
sn 1 2
? sn 1 1
s23d n x n Z
− Z 23xÎ n 1 1
n 1 2
Z
− 3Î 1 1 s1ynd
1 1 s2ynd | x | l 3 | x | as n l `
By the Ratio Test, the given series converges if 3 | x | , 1 and diverges if 3 | x | . 1.
Thus it converges if | x | , 1 3 and diverges if | x | . 1 3 . This means that the radius of
convergence is R − 1 3 .
We know the series converges in the interval s2 1 3 , 1 3 d, but we must now test for convergence
at the endpoints of this interval. If x − 2 1 3 , the series becomes
s23d
ò
n (2 1 3) n
n−0 sn 1 1
− ò
n−0
1
− 1 1 1 1 1 1 1 1 ∙ ∙ ∙
sn 1 1 s1 s2 s3 s4
which diverges. (Use the Integral Test or simply observe that it is a p-series with
p − 1 2 , 1.) If x − 1 3 , the series is
s23d
ò
n ( 1 3) n
n−0 sn 1 1
− ò
n−0
s21d n
sn 1 1
which converges by the Alternating Series Test. Therefore the given power series converges
when 2 1 3 , x < 1 3 , so the interval of convergence is s21 3 , 1 3 g.
n
Example 5 Find the radius of convergence and interval of convergence of the series
SOLUtion If a n − nsx 1 2d n y3 n11 , then
a
Z
n11
a n
nsx 1 2d
ò
n
n−0 3 n11
Z − Z
sn 1 1dsx 1 2d n11
3 n12 ?
−S1 1 1 nD | x 1 2 |
3
3 n11
nsx 1 2d n
l | x 1 2 |
3
Z
as n l `
Using the Ratio Test, we see that the series converges if | x 1 2 |
if | x 1 2 | y3 . 1. So it converges if | x 1 2 | , 3 and diverges if | x 1 2 |
the radius of convergence is R − 3.
y3 , 1 and it diverges
. 3. Thus
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