Let $(a_n)$ be an enumeration of all complex numbers with rational real and imaginary parts which are not contained in the closed unit disk (i.e., $\{z\in\mathbb{Q}[i] \colon |z|>1\}$).
Let $(c_n)$ be a decreasing sequence of real numbers converging “rapidly” toward zero. (I'll say more on “rapidly” in a moment.)
Define $F(z) = \sum_{n=0}^{+\infty} \frac{c_n}{z-a_n}$ for all $z$ such that this series converges. Since by “rapidly” I mean at the very least that $\sum_{n=0}^{+\infty} c_n$ converges, $F$ is defined at least on the open unit disk $U := \{z \colon |z|<1\}$. Let $f(z) = F(z)$ on $U$. Because the series converges uniformly on every compact domain in $U$, the function $f$ is holomorphic on $U$.
‣The question: possibly discussing on the meaning of “rapidly”, is it possible that $f$ should have a holomorphic extension on some larger (connected) open set than $U$? (And, if it is possible, is there any relation between the values of that extension with those of $F$ for points outside the unit disk where it is defined?)
Possible assumptions for “rapidly” might be: •the series for $F$ (or possibly even all its derivatives) converges uniformly on the closed unit disk; •or perhaps: $c_n = o(n^{-\alpha})$ for $\alpha > \frac{3}{2}$ say (if my quick computation is correct, this implies that the series for $F$ converges outside a set of Lebesgue measure zero).
[The reason this question came up is that a student asked me for an example of a holomorphic function on the open unit disk which extends continuously to the closed unit disk but not to a larger domain of holomorphy, and my rather stupid reaction was to try to construct it in this way. (Instead, I should really have invoked the Ostrowski–Hadamard gap theorem.)
I asked this question years ago on sci.math.research and I suggested this problem to a number of people, none of whom was able to provide a satisfactory answer.]
