Page 90 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 8: Series

If z = 1, then we can’t go any further with this, but I hope it’s clear that the series does not have Notes
a limit in case z = 1. Suppose now z  1. Then we have,

1 z n 1

s  1 z  1 z .
n


Now if |z| < 1, it should be clear that lim(z ) = 0, and so
n+1
 n  1
lim  z    lim s  .
j

n

 j 0  1 z

Or,
 1
j
 z  , for z  1.

j 0 1 z

There is a bit more to the story. First, note that if |z| > 1, then the Geometric series does not have
a limit (why?). Next, note that if |z|   < 1, then the Geometric series converges uniformly to
1 .
1 z

 n 
j
Notes    has a limit and appeal to the Weierstrass M-test.


 j 0 

Clearly, a power series will have a limit for some values of z and perhaps not for others. First,
note that any power series has a limit when z = z . Let’s see what else we can say. Consider a
0
 n 
0  Let
power series  c (z z ) .
j

 j 
 j 0 

c
 = lim sup   .
j
j
1
th
(Recall from 6 grade that lim sup(a ) = lim(sup{a : k  n}.) Now let R   . (We shall say R = 0
k
k
if l = , and R =  if  = 0. ) We are going to show that the series converges uniformly for all
|z – z |   < R and diverges for all |z – z | > R.
0
0
First, let’s show the series does not converge for |z – z | > R. To begin, let k be so that
0
1 1
z z 0  k  R  . 

c
There are an infinite number of c for which j c  k, otherwise lim sup    k. For each of
j
j
j
j
these c , we have,
j
j

c (z z )   j c z z 0  j  (k z z 0 j  1.


0
j
j
It is, thus, not possible for lim c (z z ) 0 n  0, and so the series does not converge.
n
n
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