Let $f(s)=\sum\_n a\_n n^{-s}$ be a Dirichlet series whose coefficients satisfy $\lvert a\_n\rvert\leq n^{C}$. Then $f(s)$ converges absolutely in some halfplanes, and is conditionally convergent in (potentially different) halfplane. It might happen sometimes that $f(s)$ admits meromorphic continuation to a larger domain. Consider maximal such domain. Is this domain always a halfplane? For purposes of this question, I take it to mean that $\mathbb{C}$ is a halfplane.


The question is motivated by known results about analytic continuation of $L$-functions. It has vexed me: to what extent the analytic continuation is special to the algebraic world of $L$-functions, and how much its properties are common to the analytic world of Dirichlet series.



Is the maximum domain to which a Dirichlet series can be continued always a halfplane?
It should be possible to make a Dirichlet series whose domain of meromorphicity is as screwy as you want. Notice that $\zeta(s - 1 - \alpha) = \sum n^{1+\alpha}/n^s$, so $\zeta(s - 1-\alpha)$ is a Dirichlet series, with pole at $\alpha$. Let $\gamma$ be a curve dividing the complex plane into two pieces, one of which contains all $z$ with $\Re(z)$ sufficiently large; and choose $\alpha\_i$ a sequence of points of $\gamma$ which is dense in $\gamma$. Consider 
$$\sum c\_i \zeta(s -1- \alpha\_i)$$
where $c\_i$ is some sequence which goes to zero very fast.


As long as the $c\_i$ go to zero fast enough, this should converge absolutely on away from $\gamma$; and will be represented by a Dirichlet series on the half plane to the right of the rightmost point of $\gamma$. The dense set of poles will ensure that you can't continue past $\gamma$. Details are left to the reader. :)


