# proof of Abel’s limit theorem

Without loss of generality we may assume $r=1$, because otherwise we can set $a^{\prime}_{n}:=a_{r}^{n}$, so that $\sum a^{\prime}_{n}x^{n}$ has radius $1$ and $\sum a^{\prime}$ is convergent if and only if $\sum a_{n}r^{n}$ is. We now have to show that the function $f(x)$ generated by $\sum a_{n}x^{n}$ (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

 $\lim_{x\to 1^{-}}f(x)=s.$

If $|x|<1$ we have:

 $\displaystyle s-f(x)$ $\displaystyle=\sum_{n=0}^{\infty}a_{n}-\sum_{n=0}^{\infty}a_{n}x^{n}$ $\displaystyle=\sum_{n=0}^{\infty}(1-x^{n})a_{n}$ $\displaystyle=(1-x)\sum_{n=1}^{\infty}(x^{n-1}+x^{n-2}+\dots+x+1)a_{n}$ $\displaystyle=(1-x)\sum_{n=0}^{\infty}(s-s_{n})x^{n}$

with $s_{n}:=\sum_{i=0}^{n}a_{i}$. Now, since $s-s_{n}\to 0$ as $n\to\infty$ we can choose an $N$ for every $\varepsilon>0$ such that $|s-s_{n}|<\frac{\varepsilon}{2}$ for all $m>N$. So for every $0 we have:

 $\displaystyle|s-f(x)|$ $\displaystyle<(1-x)\sum_{n=0}^{m}|r_{n}|x^{n}+\frac{\varepsilon}{2}(1-x)\sum_{% n=m+1}^{\infty}x^{n}$ $\displaystyle<(1-x)\sum_{n=0}^{m}|r_{n}|+\frac{\varepsilon}{2}.$

This is smaller than $\varepsilon$ for all $x<1$ sufficiently close to $1$, which proves

 $\lim_{x\to r^{-}}\sum a_{n}x^{n}=\sum a_{n}r^{n}=\sum\lim_{x\to r^{-}}a_{n}x^{% n}.$
Title proof of Abel’s limit theorem ProofOfAbelsLimitTheorem 2013-03-22 14:09:40 2013-03-22 14:09:40 mathwizard (128) mathwizard (128) 5 mathwizard (128) Proof msc 40A30 ProofOfAbelsConvergenceTheorem