proof of Abel’s limit theorem

Without loss of generality we may assume r=1, because otherwise we can set an:=arn, so that anxn has radius 1 and a is convergentMathworldPlanetmathPlanetmath if and only if anrn is. We now have to show that the function f(x) generated by anxn (with r=1)is continuousMathworldPlanetmath from below at x=1 if it is defined there. Let s:=an. We have to show that


If |x|<1 we have:

s-f(x) =n=0an-n=0anxn

with sn:=i=0nai. Now, since s-sn0 as n we can choose an N for every ε>0 such that |s-sn|<ε2 for all m>N. So for every 0<x<1 we have:

|s-f(x)| <(1-x)n=0m|rn|xn+ε2(1-x)n=m+1xn

This is smaller than ε for all x<1 sufficiently close to 1, which proves

Title proof of Abel’s limit theorem
Canonical name ProofOfAbelsLimitTheorem
Date of creation 2013-03-22 14:09:40
Last modified on 2013-03-22 14:09:40
Owner mathwizard (128)
Last modified by mathwizard (128)
Numerical id 5
Author mathwizard (128)
Entry type Proof
Classification msc 40A30
Related topic ProofOfAbelsConvergenceTheorem