I know that the Converse of Abel's Theorem is not true, but there are partial converses with added hypotheses. I have not been able to find whether this one is true or not online, but I strongly suspect that it is. The statement is:
If $f(x)=\sum_{k=0}^\infty a_kx^k$ converges on $|x|<1$, where $a_k\ge0\forall k$, and $\lim_{x\rightarrow1^-} f(x) = l \in\mathbb{R}$, then $\sum_{k=0}^\infty a_k=l$.
I highly suspect that it is true, but my attempts at a proof have not succeeded. I have tried using the Cauchy Criterion to utilise the convergence of $\sum_{k=0}^\infty a_kx^k$ for $x \in (0,1)$ but since the convergence is not known to be uniform I cannot take the sup over $x \in (0,1)$. I have also tried to bound $a_k$ directly which would also complete the proof, since if $\sum_{k\ge0} a_k$ converges then it must converge to $l$ by Abel's Theorem.
This question has been asked here before Converse of Abel's theorem, but the accepted answer doesn't answer the question, it mentions a paper but that paper doesn't mention this partial Converse anywhere.
I'm looking for either a proof, or an explicit counterexample in case it is False.
I have managed to prove it by using the second approach mentioned in my question and getting a hint from this question. (The part where he writes $\lim_{x \to 1} \sum_{n = 0}^\infty a_nx^n \geq \sum_{n = 0}^N a_n$ for all N)
Let $f(x)=\sum_{k=0}^\infty a_kx^k$ for $x\in(0,1)$, then $f$ is increasing. So $l=\lim_{x\rightarrow1^-} f(x)$ is the sup of $f$ over $(0,1)$.
So,\begin{aligned} \sum_{k=0}^\infty a_kx^k \le l\ \forall x\in(0,1)\\ \Longrightarrow \sum_{k=0}^N a_kx^k \le l\ \forall x\in(0,1)\ \forall N\in \mathbb N\ \\\Longrightarrow \sum_{k=0}^N a_k \le l\ \forall N\in \mathbb N \end{aligned} By taking sup over $x\in(0,1)$.
This means that the sequence of partial sums is bounded, so by Monotone Convergence Theorem $\sum_{k=0}^\infty a_k$ converges. By Abel's theorem it must converge to $l$.