The reciprocal of Abel's Theorem for Limits 
Let $\sum_{n=0}^{\infty} a_n x^n$ be a series of powers with a convergence radius $r \in (0, \infty)$, $a_n \ge 0$ for all $n \in \mathbb Z_ +$ and
$$f(x)=\sum_{n=0}^{\infty} a_n x^n, \quad |x| \lt r.$$
Supposing further that there is $$\lim_{x\to r^-} f(x) = s \in \mathbb R. $$ Show that $$f(x)=\sum_{n=0}^{\infty} a_nr^n = s. $$

Let's do it in two steps
1° $r= 1$ : Let's take $x=ry$.
$$\sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} a_n r^ny^n = \sum_{n=0}^{\infty} b_n y^n; \quad b_n=a_nr^n$$
This is a power series with a convergence ray
$$r' = \frac{1}{\lim_{n \to \infty}|b_n|^{\frac{1}{n}}} = \frac{1}{\lim_{n \to \infty}|a_nr^n|^{\frac{1}{n}}} = \frac{1}{\lim_{n \to \infty}|a_n|^{\frac{1}{n}}r}=1 $$
So, if we have given any power series with radius of convergence $r$, we can transform it into another power series with radius of convergence $1$. So we can take $r = 1$.
Let $\sum_{n=0}^{\infty} a_n x^n$ be a power series with radius of convergence 1.
$$\sum_{n=0}^{\infty} a_n \quad \text{then} \quad \lim_{x \to 1^-} f(x) = \sum_{n=0}^{\infty} a_n $$
Let $s_n= a_0 + a_1 + \cdots +a_n$, $ s_{-1} = 0$, and $\sum_{n=0}^{\infty} a_n = s $.
\begin{align*}
\sum_{n=0}^{m} a_n x^n
&= \sum_{n=0}^{m}(s_n -  s_{n-1}x^n) \\
&= \sum_{n=0}^{m} s_nx^n - \sum_{n=0}^{m} s_{n-1}x^n \\
&= \sum_{n=0}^{m-1} s_nx^n + s_mx^m - x \sum_{n=0}^{m} s_{n-1}x^{n-1} \\
&= \sum_{n=0}^{m-1} s_nx^n + s_mx^m - x\sum_{n=1}^{m} s_{n-1}x^{n-1} \\
&= \sum_{n=0}^{m-1} s_nx^n + s_mx^m - x \sum_{n=0}^{m-1} s_nx^n,
\end{align*}
and so,
$$\sum_{n=0}^{m} a_nx^n = (1-x)\sum_{n=0}^{m-1} s_nx^n + s_mx^m .$$
2° $|x| \lt 1$, $x^m \to 0 $ as $m \to \infty$ and $s_m \to s$. Therefore
$$\lim_{m \to \infty} \sum_{n=0}^{m} a_nx^n = \lim_{m \to \infty} (1-x) \sum_{n=0}^{m-1} s_nx^n + \lim_{m \to \infty} s_mx^m,$$
and
$$ f(x) = \sum_{n=0}^{\infty} (1-x) s_nx^n \quad\text{for} \quad 0 \lt x \lt 1.$$
Since $s_n \to s$, therefore for $\epsilon \gt 0$ then there is an integer  $N$ such that $|s_n -s| \lt \frac{\epsilon}{2}$ for all $n \ge N $
$$(1-x)\sum_{n=0}^{\infty} x^n = (1-x)\frac{1}{1-x}$$ for $ |x| \lt 1$
$$(1-x)\sum_{n=0}^{\infty} x^n = 1$$
Therefore, for $ n \ge N$ and $0 \lt x \lt 1 $,
\begin{align*}
|f(x) -s|
&= \left|(1-x)\sum_{n=0}^{\infty} s_nx^n - s \cdot 1\right| \\
&= \left|(1-x)\sum_{n=0}^{\infty} s_nx^n - (1-x)\sum_{n=0}^{\infty} sx^n \right| \\
&= \left|(1-x)\sum_{n=0}^{\infty} (s_n - s)x^n \right| \\
&\le  (1-x)\sum_{n=0}^{N} |s_n - s|x^n + \frac{\epsilon}{2}(1-x)\sum_{n=N+1}^{\infty} x^n \\
&\le (1-x)\sum_{n=0}^{N} |s_n -n|x^n + \frac{\epsilon}{2} \cdot 1
\end{align*}
Now, for a fixed $N$,
$$(1-x)\sum_{n=0}^{N}|s_n - s|x = g(x).$$
$g(x) = 0$ at $ x = 1 \to g(1) = 0$. For definition of continuity, given $\epsilon \gt 0 $ there exist $\delta \gt 0$ s.t $|g(x) - g(1)| \lt \frac{\epsilon}{2}$ whenever $1-\delta \lt x \lt 1$,
$$(1-x)\sum_{n=0}^N |s_n - s| x^n \lt \epsilon$$
So, $|f(x) - s| \lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$ whenever $1-\delta \lt x \lt 1$. Therefore
$$\lim_{x \to 1^-} f(x) = s = \sum_{n=0}^{\infty}a_n$$
That's all I could do, how bad is it?
Thanks in advance for any help.
 A: 1. The most significant gap in your proof is:

Q. How do you know that $s_m \to s$?

Recall that $s$ is defined as $\displaystyle \lim_{x \to 1^-} \sum_{n=0}^{\infty} b_n x^n$. So, whether $s$ is equal to $\displaystyle \sum_{n=0}^{\infty} b_n = \lim_{n\to\infty} s_n$ is not known in advance. And most of all, this is precisely what we want to prove. So your argument is suffering from a circular reasoning.
Edit: Upon a second reading, it seems that you redefined $s$ to be the $\sum_{n=0}^{\infty} a_n$ in the intermediate step of your proof. Then there is only one minor issue remaining: How do you know that the sum $\sum_{n=0}^{\infty} a_n$ is finite?

2. So, how one would prove the claim? I will present two proofs.
1st Proof. Fix $N$, then for each $x \in (0, r)$, the non-negativity of $a_n$'s implies that
$$ \sum_{n=0}^{N} a_n x^n \leq f(x) \leq \sum_{n=0}^{\infty} a_n r^n. $$
Letting $x \to r^-$, we get
$$ \sum_{n=0}^{N} a_n r^n \leq s \leq \sum_{n=0}^{\infty} a_n r^n. $$
Now letting $N \to \infty$ proves the desired equality.
2nd Proof. For any function $f : A \times B \to \mathbb{R}$, we can prove that
$$ \sup_{x \in A} \biggl( \sup_{y \in B} f(x, y) \biggr) = \sup_{(x,y)\in A\times B} f(x, y) = \sup_{y \in B} \biggl( \sup_{x \in A} f(x, y) \biggr) $$
in $\overline{\mathbb{R}} = \mathbb{R}\cup\{-\infty,+\infty\}$. Then by using the monotonicity of $\sum_{n=0}^{N} a_n x^n$ both in $N \in \mathbb{Z}_{\geq 0}$ and in $x \in [0, \infty)$, we get
\begin{align*}
s
&= \lim_{x \to r^-} \lim_{N\to\infty} \sum_{n=0}^{N} a_n x^n \\
&= \sup_{x \in (0, r)} \biggl( \sup_{N\in\mathbb{Z}_{\geq 0}} \sum_{n=0}^{N} a_n x^n \biggr) \\
&= \sup_{N\in\mathbb{Z}_{\geq 0}} \biggl( \sup_{x \in (0, r)} \sum_{n=0}^{N} a_n x^n \biggr) \\
&= \lim_{N\to\infty} \lim_{x \to r^-} \sum_{n=0}^{N} a_n x^n \\
&= \sum_{n=0}^{\infty} a_n r^n.
\end{align*}
Note that this in fact holds regardless of whether both sides are finite or $+\infty$.
