proof of Abel’s limit theorem

Without loss of generality we may assume $r=1$, because otherwise we can set $a_n^{\prime }:=a_r^n$, so that $\sum a_n^{\prime }{x}^n$ 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

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

If we have:

$s-f(x)$$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n-{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{x}^n$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}(1-x^n)a_n$

$=(1-x){\displaystyle \sum _{n=1}^{\mathrm{\infty }}}(x^{n-1}+x^{n-2}+\mathrm{\dots }+x+1)a_n$

$=(1-x){\displaystyle \sum _{n=0}^{\mathrm{\infty }}}(s-s_n)x^n$

with $s_n:={\sum }_{i=0}^n{a}_i$. Now, since $s-s_n\to 0$ as $n\to \mathrm{\infty }$ we can choose an $N$ for every $\epsilon >0$ such that for all $m>N$. So for every we have:

$|s-f(x)|$

This is smaller than $\epsilon$ for all sufficiently close to $1$, which proves

$$\underset{x\to r^{-}}{lim}\sum a_n{x}^n=\sum a_n{r}^n=\sum \underset{x\to r^{-}}{lim}{a}_n{x}^n.$$