Let $(f_n)_{n=1}^\infty$ be a sequence of functions $f_n:U\to\mathbb{C}$ with $U$ open and each $f_n$ analytic. If $f_n\to f$ locally uniformly on $U$, is $f$ necessarily analytic?
I believe this is true. Since every point $z$ of $U$ has a $\delta$-neighborhood $\mathcal{N}_\delta(z)$ on which $f_n\to f$ uniformly, $f$ is analytic on each neighborhood $\mathcal{N}_\delta(z)$. As this holds for each $z\in U$, $f$ is analytic on $U$.
Does this suffice? Is there any counterexample to this statement? For context, I want to show that the following function is analytic on $S\equiv\mathbb{C}\setminus\{\sqrt{n}:n\in\mathbb{N}\}$:
$$f(z)=\sum_{n=1}^\infty\frac{z^4}{n\sqrt n(z-\sqrt n)}$$
If the above holds, then it suffices to show that the sequence of partial sums of the above series converges uniformly to $f$ on compact subsets of the form $\mathcal{N}_R(0)\setminus\bigcup_{n\in\mathbb{N}}\mathcal{N}_{\varepsilon}(\sqrt{n})$, as each $z\in S$ is contained in a set of this form for some $(R,\varepsilon)$. This implies locally uniform convergence on $S$, and hence analyticity.
Best Answer
Your assertion is correct. See this.
For a more expanded exposition, I'd recommend Gamelin V.2. The entire chapter is about power series of analytic functions. The next chapter deals with Laurent series of functions.