Let $f(x)=\sum f_n (x)$ be a series, uniformly converging on some open set $D$.
For that matter I don't really care if $f_n$ are real or complex, so $D$ could be $(0,1)$ or $B_1(0)$ for example, depending on the type of series. I don't mind restricting to "simple" open sets (like intervals, balls, rings, half-planes), although I'm doubtful that the answer would depend on the type of open domain.
We may also assume that $f_n$ are nice functions, say analytic.
Assume that $f(z)=\infty$ for $z\in\partial D$. Does it mean that $\lim_{x\to z} f(x)=\infty$?
By $f(z)=\infty$ I mean $\sum f_n(z)=\infty$.
It sounds very logical to me, and the examples I have in mind for convering series do give a positive answer to my question. Yet, no proof comes to mind.
In the complex case, I can say that $f$ is analytic in $D$, and I would like to say that $z$ is some sort of pole - and we know that analytical functions aproach infinity near poles. Yet, this is problematic as $f$ is not defined in a a (punctured) neighborhood of $z$.
Following is a proposed solution.
Take $z_n \in D$ such that $\lim z_n = z$. For $n\in\mathbb{N}$ define $g_n:\mathbb{N}\to\mathbb R$ by $g_n(m)=f_m(z_n)$
Let us assume $g_n\geq0$. Then by Fatou's lemma (with respect to the counting measure on $\mathbb N$), $$\sum_m \liminf_{n}g_n(m)\leq\liminf_{n}\sum_m g_n(m)$$
If we assume $f_n$ is continuous, then $\liminf_{n}g_n(m)=f_m(z)$. By our assumptions, $\sum_m f_m(z)=\infty$ and so we obtain $$\infty\leq\liminf_{n}\sum_m f_m(z_n)$$
which is what we wanted. Q.E.D