Why is a absolutely and uniformly convergent series in the complex plane holomorphic?

Question

Suppose I have some series $f(z) = \sum_{k = 0}^\infty a_n(z)$, with $a_i$ holomorphic on $\mathbf{H}$, that is absolutely convergent for all $z \in \mathbf{H}$ and uniformly convergent on compact subsets of the upper half plane $\mathbf{H}$. My lecture notes state that $f(z)$ then is holomorphic as a function on $\mathbf{H}$. But why is that so?

Thanks!

Answer 1

Here is the general result sometimes known as "Weierstrass theorem".

Let $\Omega$ be an open set of $\mathbb{C}$ and $f_n$ a sequence of holomorphic functions on $\Omega$ such that $f_n \to f$ uniformly on each compact $K \subset \Omega$. Then $f$ is holomorphic on $\Omega$.

Once you have prove that, you can apply this theorem to $\Omega =\textbf{H}$ and $f_n = \sum_{k=0}^n a_k$ to get the desired result.

The proof of Weierstrass theorem is classic and should be in your classbook. Here is an other approach : Uniform limit of holomorphic functions.