Page 737 - Advanced_Engineering_Mathematics o'neil

21.1 Power Series 717

converges. But then
∞

n
|c n (z − z 0 ) |
n=0
∞ n
converges, so n=0 c n (z − z 0 ) converges absolutely.
This theorem implies that there are exactly three possibilities for convergence of power
series.
1. It may be that the series does not converge for any points other than z 0 . In this case we
say that the power series has a radius of convergence zero, converging only at its center.
2. The power series may converge for all z. In this case, we say that it has infinite radius of
convergence.
3. The power series may converge for some points other than z 0 but also diverge at some
points (that is, Cases (1) and (2) do not hold). Let ζ be the closest point to z 0 at which the
series diverges, and let R =|ζ − z 0 |.
• If |z − z 0 | < R, then the power series must converge at z. Otherwise this open disk
would contain a point at which the series diverges, and this point would be closer to
z 0 than ζ.
• If |z − z 0 | > R, then the power series must diverge at z. For if it converged at such a z,
then it would converge at all points closer to z 0 than z, hence also at ζ, a contradiction.
Therefore, in this case, there is a number R such that the power series converges within the
disk |z − z 0 | < R, and diverges outside this disk. We call the open disk |z − z 0 | < R the open disk
of convergence of the power series, and R is the radius of convergence.
At specific points on the circle |z − z 0 |= R, the power series may converge or diverge. This
would have to be tested for each point and each power series.
These cases can be consolidated by setting R = 0 in Case (1) and R =∞ in Case (2). In this
case, the inequality

|z − z 0 | < ∞
is interpreted to mean the entire complex plane, since all points z are at a finite distance from z 0 .
Sometimes the radius of convergence of a power series can be computed using the ratio test.

EXAMPLE 21.1
The power series
∞ n
n 2 n
(−1) (z − 1 + 2i)
n + 1
n=0
has center 1 − 2i. Look at the magnitude of the ratio of successive terms:

(−1) n+1 (2 n+1 /(n + 1))(z − 1 + 2i) n+1 2(n + 1)
= (z − 1 + 2i)
(−1) (2 /(n + 1))(z − 1 + 2i) n + 2
n n n
→ 2|z − 1 + 2i| as n →∞.
By the ratio test, this real series converges if this limit is less than 1 and diverges if this limit is
greater than 1. Therefore, the power series converges if
1
|z − 1 + 2i| <
2
and diverges if |z − 1 + 2i| > 1/2. The radius of convergence is 1/2, and the open disk of
convergence is the disk |z − 1 + 2i| < 1/2.

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

October 14, 2010 15:35 THM/NEIL Page-717 27410_21_ch21_p715-728