1549901369-Elements_of_Real_Analysis__Denlinger_

8.6 P ower Series 507

the radius of convergence is 0, while if the series converges for all real numbers,

we say that the radius of convergence is +oo.

Example 8 .6.6 Find the interval of convergence and t he radius of convergence

00 ( 3)k

of the power series L x ~k

k=l

Solution: As we may recall from elementary calculus, the ratio test

(8.2.11) is useful here. We calculate the limit

I

(x _ 3)k+

1

k2k I k Ix_ 3 I Ix_ 3 I

L = }~~ (k + 1)2k+^1 (x - 3)k =kl~~ k + 1 -2- = -2-.

The ratio test tells us that the series converges a bsolutely if L < 1 and
diverges if L > 1. That is , the series converges absolutely if Ix - 31 < 2 and
diverges if Ix - 31 > 2. Thus the series converges absolutely in the interval
(1, 5) and diverges outside the interval [1, 5].

We test the endpoint x = 1:
00 (1-3)k 00 (-2)k 00 (-l)k
L k
2
k = L k
2
k = L -k-, the alternating harmonic series, which
k = l k=l k=l
converges.

We test the endpoint x = 5:
00 (5 - 3)k 00 (2)k 00 1
L k
2
k = L k
2
k = L k, the harmonic series, which diverges.
k=l k=l k=l

Therefore, the interval of convergence of the given power series is [1, 5).
The radius of convergence is 2. D

Notice that in testing the endpoints of the interval of convergence in Ex-
ample 8.6.6 we did not use the ratio test. In fact, we cannot use the ratio test
at the endpoints, since they are precisely the points for which the ratio test is
inconclusive.

Example 8 .6 .6 suggests the following formula for the radius of convergence
of a power series.

00
Theorem 8.6. 7 The radius of convergence of the power series L ak(x - c)k
k=O

is p = lim I~ I if this limit exists, finite or infinite.
k->oo ak+l

Proof. Exercise 2. •