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.
The answer is NO.

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 $F(1)=1$.
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 $F(z_0)=1 \ne 0$.