Let $f_n$ be a sequence of holomorphic functions defined on an open set $\Omega\subset \mathbb{C}$.
We say that the series $\sum_n f_n$ is uniformly absolutely-convergent if $\sum_n |f_n|$ converges uniformly on any compact subset $K\subset \Omega$.
We say that the series $\sum_n f_n$ converges normally if for every compact subset $K\subset \Omega, \sum_n \sup_K |f_n|$ converges.
We know if the series converge normally, then it must be uniformly absolutely-convergent. But how about the converse?
Does there exist a counterexample which shows that a series of holomorphic functions may be uniformly absolutely convergent but not normally.
Also, there exist a counterexample if we just consider single compact subset [1]
[1]. Does absolute and uniform convergence imply normal convergence?
Theorem Notation as above, TFAE:
(1) The series $\sum_n f_n$ is uniformly absolutely-convergent.
(2) The series $\sum_n f_n$ converges normally.
Proof We just need to prove (1) implies (2). By the continuity, for any $n\in \mathbb{N}^{+}$, we can choose $z_n\in K$ such that $$|f_n(z_n)|=\sup_K|f_n|$$ Consider two cases:
(i) If $\partial \Omega\neq \emptyset$, let $$\delta:=d(K,\partial\Omega)>0,\ L:=\{z\in\Omega|\ d(z,K)\leq \frac{\delta}{2}\},$$ then $L\supset K$ is a compact subset of $\Omega$, and the balls $B(z_n,\delta/2) \subset L$. By [Hor, Thm 1.6.7], $|f_n|$ is subharmonic for any $n$. By [Hor,Thm 1.6.3], we have $$|f_n(z_n)|\int_0^{\delta/2} rdr\leq \frac{1}{2\pi}\int_L |f_n(z)|d\lambda(z)\Longrightarrow |f_n(z_n)\leq \frac{4}{\pi\delta^2}\int_L|f_n(z)|d\lambda(z).$$ where $\lambda$ denotes the Lebesgue measure.
By the convergence, we know the series $\sum_n |f_n|$ is uniformly bounded over $L$, hence $$\sum_n \sup_n |f_n(z)|=\sum_n |f_n(z_n)|\leq \frac{4}{\pi\delta^2}\int_L \sum_n |f_n(z)|d\lambda(z)<\infty.$$ (ii) If $\partial \Omega=\emptyset$, then $\Omega=\mathbb{C}$, this can be proved similary.
[Hor] Lars Hormander. An introduction to several complex variables. $\square$