Suppose $\{a(n)\}_{n\ge 1}$ is a bounded complex sequence. Let $\phi(s)=\sum_{n\ge 1} \frac{a(n)}{n^s}$. Obviously, the Dirichlet series $\phi(s)$ is absolutely convergent for $\mathcal{R}(s)>1$. I would like to know the minimal value $\alpha$ such that the Dirichlet series $\phi(s)$ can have a meromorphic extension on the half plane $\mathcal{R}(s)>\alpha$? The converse question is whether there is some condition that ensures the Dirichlet series $\phi(s)$ can have a meromorphic extension on the half plane $\mathcal{R}(s)>1/2$?
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2$\begingroup$ This question is way too broad; one cannot do better than $\alpha = 1$ without some conditions on $\{a(n)\}$. $\endgroup$– Peter HumphriesCommented Feb 18, 2019 at 13:55
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The maximal domain of meromorphic continuation of a Dirichlet series can be anything.
More precisely, for every connected open subset $O$ of $\mathbb{C}$, which contains the half plane $\{\Re s>1\}$, there exists a Dirichlet series $D$ with bounded coefficients, which is holomorphic in $O$, cannot be extended continuously beyond $O$, and all isolated points of $\mathbb{C}\setminus O$ are simple poles of $D$.
If $O$ is simply connected, and is contained in the half plane $\{\Re s>\frac{1}{3}\}$, you can arrange $D$ to have an Euler product.
answered Feb 18, 2019 at 15:38
Jan-Christoph Schlage-PuchtaJan-Christoph Schlage-Puchta
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$\begingroup$ Citations along with these statements would be very welcome! $\endgroup$ Commented Feb 18, 2019 at 23:49
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1$\begingroup$ @Greg Martin: Bhowmik, Schlage-Puchta: The maximal domain of meromorphic continuation of a Dirichlet series, Analysis 26 (2010) $\endgroup$ Commented Feb 19, 2019 at 16:01