Prove that convergent sum of analytic functions in a neighborhood of $z_0$ is an entire function Let $f(z)$ be an analytic function in a neighborhood of $z_0$. Prove that if series $$\sum_{n=1}^{\infty}f^{(n)}(z_0)$$ converges (not absolutely necessarily), then $f(z)$ is an entire function.
I don't really know how to approach this one. I just see that $|f^{(n)}(z_0)|\to0$ for a large $n$, but I don't know how can you prove from there that $f$ is an entire function itself.
Thanks for the time and suggestions.

In a neighbourhood of $z_0$ one has
\begin{equation}
f(z)= \sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
\end{equation}
If the $f^{(n)}(z_0)$ are bounded by $C$, we have
\begin{equation}
\left|\frac{f^{(n)}(z_0)}{n!}\right|\le \frac{C}{n!}
\end{equation}
hence the radius of convergence of the series is $\infty$.
