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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.

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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 $F(1)=1$.

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    $\begingroup$ Or $f_n(z) =(z-1) (z-10) (z^n + (z/10)^n)$ to make the point far away $\endgroup$
    – reuns
    Commented Jul 6, 2017 at 7:13
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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 $F(z_0)=1 \ne 0$.

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    $\begingroup$ 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. $\endgroup$
    – user457218
    Commented Jul 6, 2017 at 6:26
  • $\begingroup$ Ooops. It should be read $ \sum_{n=0}^{\infty} f_n(z)$ ! $\endgroup$
    – Fred
    Commented Jul 6, 2017 at 6:32
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    $\begingroup$ Then $f_0(0) \ne 0$. You can't get a counterexample this way. $\endgroup$
    – user457218
    Commented Jul 6, 2017 at 6:38

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