What conclusion can we make if $f$ is a holomorphic function?

Question

Suppose $f$ is holomorphic in an open neighbourhood of $z_0 \in \Bbb C$. Given that the series $$\sum\limits_{n=1}^{\infty} f^{(n)} (z_0)$$ converges absolutely, we can conclude that

$(1)$$\ \ \ \ f$ is constant

$(2)$$\ \ \ \ f$ is a polynomial

$(3)$$\ \ \ \ f$ can be extended to an entire function

$(4)$$\ \ \ \ f(x) \in \Bbb R$ for all $x \in \Bbb R$

If we take the function $f(z)=e^{\frac {iz} {2}},$ $z \in \Bbb C$ and let us take $z_0=0$. Then clearly $f$ is entire and hence holomorphic on any open neighbourhood of $0$ and $\sum\limits_{n=1}^{\infty} f^{(n)} (0)$ converges absolutely to $2$. But $f$ is neither a constant nor a polynomial. Also $f(1) \notin \Bbb R$. Which clearly indicates neither $(1)$ nor $(2)$ nor even $(4)$ is a correct option. So I think the correct option is $(3)$. But I have failed to prove it.

How should I proceed to prove that $(3)$ is indeed the correct option? Please help me in this regard.

Thank you very much.

Answer 1

Since $\sum_{n=1}^{\infty} f^{(n)} (z_0)$ converges absolutely, we have, by the root test:

$ \lim \sup |f^{(n)} (z_0)|^{\frac {1} {n}} \le 1$, hence $ \lim \sup \left |\frac{f^{(n)} (z_0)}{n!} \right |^{\frac {1} {n}}\ =0$.

Therefore the power series $ \sum_{n=0}^{\infty}\frac{f^{(n)} (z_0)}{n!}(z-z_0)^n$ has radius of convergence $= \infty$.

For $z \in \mathbb C$ put $g(z):=\sum_{n=0}^{\infty}\frac{f^{(n)} (z_0)}{n!}(z-z_0)^n$ . Then $g$ is entire and $f=g$ in an open neighborhood of $z_0$.

Answer 2

Fix $N\in \mathbb{N}$ and consider $z\in \mathbb{C}$ such that $|z-z_0|<N/e$. Now, for every $n\geq N$ we have that $$|z-z_0|^n<(N/e)^n\leq(n/e)^n<n!$$ and hence $|\frac{(z-z_0)^n}{n!}|<1$, therefore $$g(z):=\sum_{n\geq 1}\frac{f^{(n)}(z_0)}{n!}(z-z_0)$$ converges absolutely in the compact $D(z_0,N/e)$ for any $N\in \mathbb{N}$ and hence holomorphic in each $D(z_0,N/e)$, therefore entire.