Theorem : Suppose $\sum^\infty_{n=1}a_n$ converges.Then
$lim_{r\to1,r<1}\sum^\infty_{n=1}r^na_n$=$\sum^\infty_{n=1}a_n$.
My Attempt:
I thought to apply summation by parts.
So I get $\sum^N_{n=1}r^na_n=\sum^{N-1}_{n=1}A_n(r^n-r^{n+1})+A_Nr^N$
Where $A_N=\sum^N_{n=1}a_n$
As $n\to \infty, A_n\to A$
.
Here I wanted to use fact that $r\to 1$
I could not able to collect ideas to give proof.
Any Help will be appreciated
If a power series converges at a point $x_0=R$ then it converges uniformly in $[0,R]$. So in your case $\sum^\infty_{n=0}a_nx^n$ converges uniformly in $[0,1]$. And it is a known theorem that if a sequence of continuous functions converge uniformly then the limit function is continuous as well. In your example you have the sequence $f_k(x)=\sum^k_{n=0}a_nx^n$, and $f(x)=\sum^\infty_{n=0}a_nx^n$ is the limit function, hence it is continuous at the point $x=1$ when you take $x \to 1$, $x<1$.