1550251515-Classical_Complex_Analysis__Gonzalez_

(jair2018) #1

208 Chapter4


so that series converges absolutely if lzl < {IL;, and diverges if lzl > {IL;.

It follows that the radius of convergence of the original series is {IL, where

L = minrLr.


Corollary 4. 7 If

then


R = liminf n-+oo {'/'Yn'Yn+l · · · 'Yn+k-1 ( 4.11-9)

Proof This is a consequence of ( 4.11-8) since


I


~ I = I ~ 11 an+l I ·. · I an+k-1 I = 'Yn'Yn+l ... 'Yn+k-1
an+k an+l an+2 an+k
In particular, if the 'Yn+j are constants, denoted Cj, we have

r = {/coc1 · · · Ck-1 ( 4.11-10)

Examples



  1. For :Z:::::~ n!zn, we obtain
    I 1
    R = lim n. = lim --= 0
    n-+oo (n+l)! n-+oo n+I


2. F1 or '\"'00 uo -;:;r Zn , we h ave


R = lim ( n + 1) = oo
n-+oo

3. If :Z:::::;"' zn! = z + z^2 + z^6 + z^24 +··.,then formula ( 4.11-5) cannot be


applied. Writing the series as :Z:::::;"' amzm we have am = 1 if m = n! and
am = 0 if m i= n!. Formula (4.11-4) gives


R

l =limsup ~ = 1
m-+oo

so that .R = 1. It can be shown (Exercises 4.2, problem 22) that the series

diverges at every point of lzl = 1. ·



  1. For :Z:::nPzn (p constant) we have, by (4.11-5),
    . nP 1


R= lim = lim =1


n-+oo (n + l)P n-+oo (1 + l/n)P


Alternatively, by ( 4.11-4 ),


~ = lim sup npfn = lim npfn = lim epln n/n = e^0 = 1
R n-+oo n-;.oo n----;.oo
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