Does the analytic continuation of $f$ always exist? Let $f(z)$ be a holomorphic funtion on region $\Omega$. Then, does the analytic continuation of $f$ always exist? Note that $f$ is always the analytic continuation of itself, so I exclude this case. 
If the answer is yes, then what is the analytic continuation of $\sum_{n=0}^\infty z^n$, defined on the unit disk?

The function
$$f(z)=\sum_{n=0}^\infty z^{n!}$$
has no analytic continuation beyond $|z|<1$. The circle $|z|=1$ is a natural boundary.
