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Math 311-001 201110 A proof of "Cesaro = Abel"
 

Summary: Math 311-001 201110
A proof of "Cesaro = Abel"
Suppose that an converges Cesaro to L. That means that n L,
where
n =
s0 + s1 + + sn-1
n
, sk =
k
j=0
aj.
The first question is whether the series anxn
converges for x < 1. The
answer to this is yes, since
an = (n + 1)(n+1 - n) + n - n-1 ;
using that the sequence {n} is bounded (by c, say), we get (if 0 x < 1)
|an| xn
2c
n
(n + 1)xn

  

Source: Argerami, Martin - Department of Mathematics and Statistics, University of Regina

 

Collections: Mathematics