Page 91 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Next, we show that the series does converge uniformly for |z – z |   < R. Let k be so that
0
1 1
   k  .
R 

Now, for j large enough, we have    k. Thus, for |z – z |  , we have
c
j
j
0
j

c (z z )   j c z z 0 j   (k z z 0 j  (k ) . j



j
0
j
 n 
j
The geometric series  (k )  converges because k < 1 and the uniform convergence of



 j 0 

 n j 
0  follows from the M-test.
 c (z z ) 

j

 j 0 

Example:
 n 1 
j
c
j
Consider the series  z j  .  Let’s compute R = 1/ lim sup   = lim sup ( j !). Let K be any

j
 j 0 !j 
K 2K
positive integer and choose an integer m large enough to insure that 2m > . Now consider
K
(2 )!
! n , where n = 2K + m:
K n
n! (2K m)! (2K m)(2K m 1)...(2K 1)(2K





K n  K 2K m  KmK 2K

(2K)!
 2m  1
K 2K
n
Thus n!  K. Reflect on what we have just shown: given any number K, there is a number n
n

j!
,
j
j
such that n! is bigger than it. In other words, R = lim sup    and so the series  1 z  
n
 j 0 j! 
  
converges for all z.
 n 
0  there is a number
j
Let’s summarize what we have. For any power series  c (z z ) ,

 j 
 j 0 

1
R  such that the series converges uniformly for |z – z |   < R and does not
0
c
lim sup  
j
j
converge for |z – z | > R.
0
Notes We may have R = 0 or R = .
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