let ${a_n}$ be a sequence of Real non negative numbers.
assume the following limit exists and is finite:
$$ L = \lim_{x\to1^-} \sum_{n=1}^\infty a_nx^n$$
prove that $\sum_{n=1}^\infty a_n$ converges to L
note: by $\lim_{x\to1^-}$ I mean that the limit is that x goes to 1 but $x < 1$.
The statement in the title is false. But it's true under the additional assumption that $a_n\ge0$. And it's actually quite easy.
Since $a_n\ge0$ there exists $S\in[0,\infty]$ such that $$\sum_{n=0}^\infty a_n=S.$$(Note we allowed the possibility $S=\infty$.) Now for a given $N$, $$\sum_{n=0}^Na_n=\lim_{x\to1}\sum_{n=0}^N a_nx^n\le L.$$So $$S=\lim_{N\to\infty}\sum_{n=0}^N a_n\le L.$$
For the other direction, say $\epsilon>0$. There exists $x\in(0,1)$ with $$\sum_{n=0}^\infty a_nx^n>L-\epsilon.$$But $$S\ge\sum_{n=0}^\infty a_nx^n.$$So $S>L-\epsilon$ for every $\epsilon>0$, hence $S\ge L$.