Does every Dirichlet series admit an analytic continuation? If so, to what extent? The identity theorem for analytic continuation shows that uniqueness of analytic continuations of functions is very easy to characterize. This helps us a lot when we are trying to extend functions like the Riemann zeta function or the gamma function.
However, examples such as the one in this post show that existence of analytic continuations of functions is not so easy. My question is, given a Dirichlet series
$$ F(s) = \sum_{n=1}^{\infty} \frac{a_{n}}{n^{s}} $$
defined on a half-plane $\text{Re }s > \sigma_{0}$ for some $\sigma_{0}$, does $F$ necessarily admit a complex-analytic extension to a larger domain? If so, what is the largest domain to which we can extend $F$?
My dubious conjecture is that if $F$ is well-defined on a half-plane, then it admits a meromorphic extension to all of $\mathbb{C}$, but I don't know if this is really true. Is this reasonable or not? How can I determine this?
 A: First, meromorphic continuation is definitely too much to ask for, as the example $\log \zeta(s) = \sum_{p^r} \frac1r \frac1{(p^r)^s}$ shows: it has logarithmic singularities at $s=1$ and at every zero of $\zeta(s)$.
Even a continuation with branch cuts cannot necessarily extend to $\Bbb C$: the prime zeta function is an example of a Dirichlet series with the imaginary axis as its natural boundary.
I think that in a suitable sense, "most" Dirichlet series have its line of convergence as a natural boundary (in other words, no merormophic extension at all!). I couldn't find a clean statement to this effect, but you can look for example at some work of Bhowmik and Shlage-Puchta from 2007.
A: Look at $$F(s)=\prod_{n\ge 2} (1-n^{-ns+1})$$
It converges and it is analytic for $\Re(s) > 0$ and its zeros are at $$\Omega=\{  \frac{2i\pi k}{n\log n}+\frac{1}{n}, k\in \Bbb{Z}, n\ge 2\}$$
Every point of $i\Bbb{R}$ is an accumulation point of $\Omega$,
so $F$ can't extend analytically to any point of the imaginary axis.
