To prove extension of a holomorphic function to an entire function

Question

Suppose f is holomorphic in an open neighbourhood of $p\in \mathbb {C}$. Given that the series $\sum_{n=0}^\infty{f^{(n)} (p)}$ converges absolutely, prove that $f$ can be extended to an entire function.

What I did: By cauchy-taylor theorem $f (z) =\sum_{n=0}^\infty {f^{(n)} (p)(z-p)^n\over n!}$ Then I showed that $\frac{1}{R}=\lim\frac {f^{(n+1)}(p) n!}{f^{(n)} (p) (n+1)!} =0 $, $R$ is radius of convergence. My question is does the $\lim \frac {f^{(n+1)}(p)}{f^{(n)}(p)}$ exist? Otherwise my proof is wrong.

Answer 1

The series $$u_n=\cases{\frac1{n^2}& $n$ even\cr \frac1{n^3}& $n$ odd}$$ converges absolutely, yet the ratio $\frac{u_n}{u_{n+1}}$ has no limit, its $\limsup$ is $\infty$ and its $\liminf$ is $0$.

Answer 2

$f^{(n)} (p) \to 0$ so $\{f^{(n)} (p)\}$ is a bounded sequence. So the series $\sum \frac {f^{(n)}(p) (z-p)^{n}} {n!}$ is dominated by a constant times the series for $e^{|z-p|}$ which converges for all $z$.