# Analytic continuation commuting with series

Suppose $f_1,f_2,...$ are entire functions, and there is an open subset $U \subseteq \mathbb{C}$ such that the series $F(z) = \sum_{n=1}^{\infty} f_n(z)$ converges normally on $U$. Also suppose that $F$ can be analytically continued to an entire function.

I have a situation where all $f_n$ vanish at some point $z_0$, but unfortunately $z_0 \notin U.$ Can we still say that $F(z_0) = 0$?

I would guess not, since it feels like bending the rules of analytic continuation in a way that shouldn't be allowed. But I didn't think of a counterexample.

Example. Let $U$ be the unit disk, and $$f_n(z)=(1-z)z^n.$$ Then $F(z)=\sum f_n(z)=z$, and while $f_n(1)=0$, we have that $F(1)=1$.
• Or $f_n(z) =(z-1) (z-10) (z^n + (z/10)^n)$ to make the point far away – reuns Jul 6 '17 at 7:13
No. Let $f_n(z)=\frac{z^n}{n!}$ , $z_0=0$ and $U= \mathbb C \setminus \{0\}$. Then $\sum_{n=0}^{\infty} f_n(z)$ converges normally on $U$ to $F(z)=e^z$. But $F(z_0)=1 \ne 0$.
• No, it converges to $e^z - 1$ which takes the value $0$. It's important that $\sum f_n$ doesn't converge in $z_0$ since otherwise you can just plug it in. – user457218 Jul 6 '17 at 6:26
• Ooops. It should be read $\sum_{n=0}^{\infty} f_n(z)$ ! – Fred Jul 6 '17 at 6:32
• Then $f_0(0) \ne 0$. You can't get a counterexample this way. – user457218 Jul 6 '17 at 6:38