infinite sum of holomorphic functions Let $H$ be the open upper half plane. Let $\{f_n\}$ be a sequence of holomorphic functions in $H$ and continuous in $\bar{H}$. Suppose $\lim_{|z| \rightarrow \infty} f_n(z)$ exists and $|f_n(x)| \leq a_n$ for real $x$, where $\sum a_n$ converges.
I would like to prove that $\sum f_n$ converges to a holomorphic function in $H$. If I show $\sum f_n$ converges in a real line, then how can I show it converges in $H$?
 A: EDIT: This is for the original question, which had "$\lim_{n \to \infty} f_n(z)$ exists" rather than 
"$\lim_{|z|\to \infty} f_n(z)$ exists".
It's not true.
Let $p_n(z)$ be a polynomial (existence guaranteed by Runge's theorem) such that $|p_n(z) - \ln(n^2)| < 1/n$ for $z = x+iy$ with
$-n \le x \le n$, $1/n \le y \le n$, and $|p_n(z)| < 1/n$ for $-n \le x \le n$, $-n \le y \le -1/n$. Let $$f_n(z) = n^{-2} \exp\left(p_n(z) - \overline{p_n(\overline{z})}\right)$$
which is analytic.  Note that for real $x$, $p_n(x) - \overline{p_n(x)}$ is imaginary, so $|f_n(x)| = n^{-2}$.  Thus the condition on real $x$ is satisfied with $a_n = n^{-2}$.
On the other hand, for $z = x + i y$ with $-n \le x \le n$, $1/n \le y \le n$, $p_n(z) - \overline{p_n(\overline{z})} = \ln(n^2) + h_n(z)$
with $|h_n(z)| \le 2/n$, so $$|f_n(z) - 1| =  |\exp(h_n)-1| \le \exp(2/n)-1 \to 0 \ \text{as} \ n \to \infty$$
and since every $z \in H$ is in the rectangle $-n\le x \le n, 1/n \le y \le n$ for sufficiently large $n$, $f_n(z) \to 1$ pointwise on $H$ as $n \to \infty$.
However, $\sum_{n=1}^\infty f_n(z)$ does not converge for any $z \in H$, as $f_n(z)$ does not converge to $0$.
