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[parent] proof of Abel's limit theorem (Proof)

Without loss of generality we may assume $ r=1$, because otherwise we can set $ a^\prime_n:=a_r^n$, so that $ \sum a^\prime_nx^n$ has radius $ 1$ and $ \sum a^\prime$ is convergent if and only if $ \sum a_nr^n$ is. We now have to show that the function $ f(x)$ generated by $ \sum a_nx^n$ (with $ r=1$)is continuous from below at $ x=1$ if it is defined there. Let $ s:=\sum a_n$. We have to show that

$\displaystyle \lim_{x\to1^-}f(x)=s.$
If $ \vert x\vert<1$ we have:
$\displaystyle s-f(x)$ $\displaystyle =\sum_{n=0}^\infty a_n-\sum_{n=0}^\infty a_nx^n$    
  $\displaystyle =\sum_{n=0}^\infty(1-x^n)a_n$    
  $\displaystyle =(1-x)\sum_{n=1}^\infty(x^{n-1}+x^{n-2}+\dots+x+1)a_n$    
  $\displaystyle =(1-x)\sum_{n=0}^\infty(s-s_n)x^n$    

with $ s_n:=\sum_{i=0}^na_i$. Now, since $ s-s_n\to0$ as $ n\to\infty$ we can choose an $ N$ for every $ \varepsilon>0$ such that $ \vert s-s_n\vert<\frac{\varepsilon}{2}$ for all $ m>N$. So for every $ 0<x<1$ we have:
$\displaystyle \vert s-f(x)\vert$ $\displaystyle <(1-x)\sum_{n=0}^m\vert r_n\vert x^n+\frac{\varepsilon}{2}(1-x)\sum_{n=m+1}^\infty x^n$    
  $\displaystyle <(1-x)\sum_{n=0}^m\vert r_n\vert+\frac{\varepsilon}{2}.$    

This is smaller than $ \varepsilon$ for all $ x<1$ sufficiently close to $ 1$, which proves
$\displaystyle \lim_{x\to r^-}\sum a_nx^n=\sum a_nr^n=\sum\lim_{x\to r^-}a_nx^n.$



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Cross-references: continuous, generated by, function, convergent, radius, without loss of generality

This is version 2 of proof of Abel's limit theorem, born on 2004-02-16, modified 2004-02-16.
Object id is 5583, canonical name is ProofofAbelsLimitTheorem.
Accessed 2286 times total.

Classification:
AMS MSC: 40A30 (Sequences, series, summability :: Convergence and divergence of infinite limiting processes :: Convergence and divergence of series and sequences of functions)
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