Let be $f(x)=\sum\limits_{n=0}^{\infty}a_nx^n$ in $(-r,r)$ and $\lim\limits_{x\to r^{-}}f(x)=L$. Prove that $\sum\limits_{n=0}^{\infty}a_nr^n=L$ Question: Let be $a_n\geq0$ for all $n\in\mathbb{N}$ and suppose that $f(x)=\sum\limits_{n=0}^{\infty}a_nx^n$ in the interval $(-r,r)$ and $\lim\limits_{x\to r^{-}}f(x)=L$. Prove that $\sum\limits_{n=0}^{\infty}a_nr^n=L$.
My attempt: I was trying to prove that $s_k=\sum\limits_{n=0}^{k}a_nr^n$ is convergent because I think that I could use the Abel's Theorem:

Let $\sum a_nx^n$ be a power series which the radius of convergence $r$ is finite and positive. If $\sum a_nr^n$ converges, then $\sum a_nx^n$ converges uniformly in $[0,r]$ and $\lim\limits_{x\to r^{-}}\sum a_nx^n=\sum a_nr^n$.

I don't know if I'm on the right way, because I got stuck with my ideia.
 A: The radius of convergence is at least $r$. If it is greater than $r$ then the result follows by continuity. If it is equal to $r$ the result follows by Abel's Theorem.
Convergence of $\sum a_nr^{n}$: Consider $\sum\limits_{n=0}^{N} a_n(r'^{n})$ where $0< r'<r$. Note that $\sum\limits_{n=0}^{N} a_n r'^{n}\leq f(r') <L+1$ if $r'$ is close enough to $r$. Hence, $\sum\limits_{n=0}^{N} a_nr^{n}\leq L+1$. Now let $N \to \infty$.
A: We go to show that $\sum_{n=0}^{\infty}a_{n}r^{n}$ converges by contradiction.
Suppose the contrary that $\sum_{n=0}^{\infty}a_{n}r^{n}$ diverges,
then $\sum_{n=0}^{\infty}a_{n}r^{n}=\infty$ because $a_{n}\geq0$.
Choose $N$ such that $\sum_{n=0}^{N}a_{n}r^{n}>L+1$. Observe that
$x\mapsto\sum_{n=0}^{N}a_{n}x^{n}$ is a continuous function, so there
exists $\delta\in(0,\frac{1}{2}r)$ such that $\sum_{n=0}^{N}a_{n}x^{n}>L+1$
whenever $x\in(r-\delta,r]$. For any $x\in(r-\delta,r)$, we have
that $L+1<\sum_{n=0}^{N}a_{n}x^{n}\leq\sum_{n=0}^{\infty}a_{n}x^{n}=f(x)$.
This contradicts to the fact that $\lim_{x\rightarrow r-}f(x)=L$.

Remark: Without the assumption that $a_{n}\geq0$, the proposition
is false. Counter-example: Let $r=1$. For $x\in(-1,1)$, we have
that $1-x+x^{2}-x^{3}+\ldots=\frac{1}{1+x}:=f(x)$. Clearly, $f(x)\rightarrow\frac{1}{2}$
if $x\rightarrow1-$. However, the infinite series $\sum_{n=0}^{\infty}a_{n}(1)^{n}$
diverges (where $a_{n}=(-1)^{n}$) because the general term does not
converge to zero.
