# Analytic continuation of ordinary Dirichlet series

Consider an ordinary Dirichlet series which is absolutely converge in some half plane Re s>c.

Question:Suppose it can be extended meromorphically to the whole complex plane with finite many poles.is it of finite order?If not,is it possible to construct a counterexample?

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Well, a simple counter example is $$A(s)=\sum_{n=1}^\infty a_n n^{-s}= e^{\eta(s)},$$ where $$\eta(s)=\sum_{n=1}^\infty (-1)^{n-1} n^{-s}=(1-2^{1-s})\zeta(s).$$ This Dirichlet series is obviously meromorphic since it is in fact entire and it is also absolutely convergent on some half plane Re$(s)>c$. This entire function is not of finite order by the functional equation of the Riemann zeta-function and Stirling's formula.
Update: An even simpler example along the same lines is $$B(s)=\sum_{k=0}^\infty \frac{2^{-ks}}{k!} =e^{2^{-s}}.$$ This Dirichlet series is an entire function that is absolutely convergent in the full complex plane, so it is absolutely convergent for any half plane Re$(s)>c$. However we have that $B(-x)= e^{2^x}$, and thus it does not fulfill $B(-x)\ll e^{x^c}$ for any $c>0$ and it is not an entire function of finite order. If we would like to have a meromorphic function with some pole that has some abscissa of convergence we can consider $C(s)=\zeta(s)+B(s)$. This function is an ordinary Dirichlet series that is absolutely convergent if and only if Re$(s)>1$, has a pole at Re$(s)=1$ and meromorphic continuation to the entire complex plane, but it does not have finite order.
well an entire function is of finite order if $f(s) \ll e^{|s|^c}$ for some $c>0$. I am not quite sure about the meromorphic analogue, but in case the meromorphic function is entire they should surely coincide. Of course even if $f(s) \ll e^{|s|^c}$ is of finite order the exponential $e^{f(s)}$ of this function should not in general be of finite order (unless $f(s)$ is a polynomial). –  Johan Andersson Nov 3 '11 at 18:41
@Paul: He considers $\exp(\eta(s))$, not $\eta(s)$. –  GH from MO Nov 3 '11 at 18:42