Suppose $\{ a_n \}$ is a sequence of non-negative real numbers and $\sum a_nx^n$ converges for $|x|<1$. If $\lim_{x \rightarrow 1^-}\sum a_nx^n=A$, prove that $\sum a_n$ converges to $A$.
My attempt: given $\epsilon>0$, we can find $\delta$ such that if $x \in (1-\delta,1)$ then $|\sum a_nx^n-A| < \epsilon$. Now consider $|\sum a_n - A| \leq |\sum a_n - \sum a_nx^n| + |\sum a_nx^n-A|$ and since $x^n \rightarrow 1$ as $x \rightarrow 1^-$, we have $|\sum a_n - \sum a_nx^n| \rightarrow 0$ and so $\sum a_n$ converges to $A$. Is my proof valid? Thanks.
Edit: As suggested by @Clement C., my notes also say that by Tauberian theorem if we prove that $\lim_{n \rightarrow \infty}na_n=0$ then we can conclude the statement. I wonder if this is possible.
Suppose $\sum_{n=1}^\infty a_n = B \in [0, \infty]$. Of course $\sum_{n=1}^\infty a_n \ge \sum_{n=1}^\infty a_n x^n$, so we must have $B \ge A$.
On the other hand, suppose $B > A$. Take $C_1, C_2$ so $B > C_1 > C_2 > A$. There must be $N$ such that $\sum_{n=1}^N a_n \ge C_1$. Take $r \in (0,1)$ so $r^N > C_2/C_1$. Then if $r < x < 1$ we have $$A = \sum_{n=1}^\infty a_n x^n \ge \sum_{n=1}^N a_n (C_2/C_1) \ge C_2 > A$$ contradiction.