Let $(a_n)_{n \in \mathbb N} \subset \mathbb R$ be a series so that $\sum_{n=0}^\infty a_n x^n$ has convergence radius R. Suppose
$$ \sum_{n=0}^\infty a_n R^n = \infty $$ holds. Does it follow that
$$ \lim_{(-R, R) \ni x \rightarrow R} \sum_{n=0}^\infty a_n x^n = \infty? $$
My intuition tells me this should be the case, but sadly I have no idea how to prove this. (It took me quite a while to get that result for a certain series $(a_n)_{n \in \mathbb N}$ and now I wonder, whether I could have used a general result instead.)
We can modify the proof of Abel's theorem to deal with this situation. First, if necessary replace $a_n$ with $a_n R^n$ to get $R = 1$, which simplifies the notation but doesn't change anything important. Then denote
$$f(x) = \sum_{n = 0}^\infty a_n x^n\quad \text{for } x \in (-1,1)\quad\text{and}\quad s_n = \sum_{k = 0}^n a_k.$$
For $x \in (-1,1)$ we have
$$F(x) := \frac{f(x)}{1-x} = \sum_{n = 0}^\infty s_n x^n.$$
Now, given an arbitrary $M \in (0,+\infty)$, choose $N \in \mathbb{N}$ so that $s_n > M$ for all $n \geqslant N$. Then we can split the power series of $F$ at $N$ to obtain
\begin{align} f(x) &= (1-x)F(x) \\ &= (1-x)\sum_{n = 0}^{N-1} s_n x^n + (1 - x)\sum_{n = N}^\infty s_n x^n \\ &\geqslant (1 - x)\sum_{n = N}^\infty s_n x^n - (1 - x)\sum_{n = 0}^{N-1} \lvert a_n x^n\rvert \\ &> (1 - x)\sum_{n = N}^\infty M x^n - (1- x)\sum_{n = 0}^{N-1} \lvert a_n\rvert \\ &= M\cdot x^N - (1 - x) \sum_{n = 0}^{N-1} \lvert a_n\rvert. \end{align}
From this inequality, we obtain
$$\liminf_{x \to 1^-} f(x) \geqslant M,$$
since $x^N \to 1$, and $\sum_{n = 0}^{N-1} \lvert a_n\rvert$ is a fixed quantity. Since $M$ was arbitrary, it follows that
$$\lim_{x \to 1^-} f(x) = +\infty = \sum_{n = 0}^\infty a_n.$$