Notation. This question deals with series of real terms $\{a_n\}_{n\in\Bbb N}$ such that $\limsup_{n\to\infty}\sqrt[n]{|a_n|}=1$: the following terminology is used.
- If $\lim_{N\to\infty}\sum_0^N a_n=\lim_{N\to\infty} s_N=\sum_0^\infty a_n=\pm\infty$ the series is divergent.
- If $-\infty<\lim_{N\to\infty} s_N=\sum_0^\infty a_n<+\infty$ the series is convergent.
- If $\nexists\lim_{N\to\infty} s_N=\sum_0^\infty a_n$ i.e. $\liminf_{N\to\infty} s_N\neq \limsup_{N\to\infty} s_N$ the series is not convergent or oscillating.
The heart of the matter. This question was inspired by this one and related answers. Simply stated, the result proved there says that if $\{a_n\}_{n\in\Bbb N}$ is a real sequence such that $\lim_{N\to\infty} s_N=\sum_0^\infty a_n$ exists, then
$$
\text{if }\lim_{x\to 1^-}\sum_{n=0}^\infty a_n x^n\text{ is finite}\implies \sum_{n=0}^\infty a_n\neq\pm\infty\label{1}\tag{1}
$$
i.e. Abel summability of the series implies that it cannot diverge. As noted by p4sch, if we even assume that $s_N$ is not converging, it can nevertheless happen that its Abel summability implies its boundedness, as the elementary example
$$
\{a_n\}_{n\in\Bbb N}=\{(-1)^n\}_{n\in\Bbb N}
$$ shows. This leads to the following
Question: does Abel summability of the series $\sum_0^N a_n$ always implies the boundedness of the sequence $\{s_N\}_{n\in\Bbb N}$ of its partial sums, or more precisely does the following implication holds?
$$
\text{if }\lim_{x\to 1^-}\sum_{n=0}^\infty a_n x^n\text{ is finite}\implies -\infty<\liminf_{N\to\infty}\sum_{n=0}^N a_n\le\limsup_{N\to\infty}\sum_{n=0}^N a_n<+\infty\label{2}\tag{2}
$$
My attempt. I tried to extend the method of proof used by zwh, since it seemed suited to obtain the sought for result. I started attacking the problem by trying to prove that
$$
\lim_{x\to 1^-}\sum_{n=0}^\infty a_n x^n\neq\infty \implies \limsup_{N\to\infty}\sum_{n=0}^N a_n<+\infty\label{1a}\tag{1a}
$$
provided $\liminf_{N\to\infty} s_N>-\infty$. Now it is only the right side of \eqref{1a} which holds therefore, for a general positive $M$, there exist an integer $N'$ for which there exists an infity of partial sums $s_{n_i}$ (but not all) such that
$$
s_{n_i}\ge M\qquad\forall n_{i}>N'\label{3}\tag{3}
$$
Thus we have
$$
\begin{split}
\sum_{n=0}^\infty a_n x^n &= (1-x)\sum_{n=0}^\infty s_nx^n\\
&= (1-x)\left(\sum_{n=0}^{N'} s_nx^n + \sum_{n_i\ge N'+1}^\infty s_{n_i}x^{n_i} +\sum_{n_j\ge N'+1}^\infty s_{n_j}x^{n_j}\right)
\end{split}
$$
where the partial sums $\{s_{n_j}\}_{j\in\Bbb N}$ are the ones which do not satisfy \eqref{3}. At this point, however, I am not able to find an arbitrary large lower bound for $\sum_{n=0}^\infty a_n x^n$ and obtain the "reductio ad absurdum".
Best Answer
I think the statement is in general false. Consider $$f(x):=\sum_{n\geq1}(-1)^{n+1}nx^n=\frac{x}{(1+x)^2}$$ $f(x)$ converges to $1/4$ as $x\rightarrow1-$ but partial sums of the series $\sum_{n\geq1}(-1)^{n+1}n$ oscillate with increasing amplitude: $-1,1,-2,2,...,-n,n,...$