infinite vs finite sum of holomorphic functions is holomorphic?

We know that if $f(z)$ and $g(z)$ are holomorphic on a domain $D$, then so is the sum $f(z)+g(z)$.

We can then say the finite sum of holomorphic functions is also holomorphic:

$$ h(z) = \sum_{n=1}^N f_n(z)$$

Question: If an infinite sum of holomorphic functions (such as a power series) doesn't diverge, it is also holomorphic. Why?

$$ h(z) = \sum_{n=1}^\infty f_n(z)$$

My uncertainty and caution is from the fact that "the limit of the sum" is not always the "sum of the limit" from elsewhere, and holomorphicity is about the existence of the derivate limit.

Solution 1:

Elaborating my comment, as requested:

Here's the standard way of showing the existence of a sequence of polynomials $g_n(z)$ that converges to $0$ for each $z$ with $\text{Re}(z) < 0$ and $1$ for each $z$ with $\text{Re}(z) \ge 0$.

Let $A_n = \{z: |z| \le n,\ \text{Re}(z) \le -1/n\}$ and $B_n = \{z: |z| \le n,\ \text{Re}(z) \ge 0\}$. These are disjoint compact sets, and the complement of $A_n \cup B_n$ is connected. By Runge's theorem, there is a polynomial $g_n$ such that $|g_n(z)| < 1/n$ for $z \in A_n$ and $|g_n(z) - 1| < 1/n$ for $z \in B_n$. Now any complex number $z$ is in $A_n \cup B_n$ for all sufficiently large $n$: if $\text{Re} (z ) \ge 0$ you just need $n > |z|$, while if $\text{Re}(z) < 0$ it's true if $n > \max(|z|, -1/\text{Re}(z))$. And then it's easy to see that $g_n(z)$ converges to $0$ if $\text{Re}(z) < 0$ and $1$ if $\text{Re}(z) \ge 0$.

To make the sequence into a series, just take $f_1 = g_1$ and $f_n = g_n - g_{n-1}$ otherwise: we have $\sum_{k=1}^n f_k(z) = g_n(z)$.