Let $a\_n$ be a bounded sequence of positive real numbers. Is it the case that the Dirichlet series $\sum \frac{a\_n}{n^s}$ can be meromorphically continued up to the right of zero, with at the most a pole at $1$?


This question occurred to me while trying to extend the Dedekind zeta function up to zero. If the above is true, then we get the result.


Related question: What if we assume $a\_n$ to be only a bounded sequence of complex numbers?



Continuation up to zero of a Dirichlet series with bounded coefficients
The following paper seems to (among other things) give a detailed construction roughly along the lines of my comment above:




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Bhowmik, Gautami, Schlage-Puchta, Jan-Christoph
Natural boundaries of Dirichlet series. (English summary)
Funct. Approx. Comment. Math. 37 (2007), part 1, 17--29.


In this paper, the authors prove some conditions for the existence of natural boundaries of Dirichlet series, and give applications to the determination of asymptotic results.


Let $n\_\nu$ be rational integers, assume the series $\sum\frac{n\_\nu}{2^{\epsilon\nu}}$ converges absolutely for every $\epsilon>0$, and let $\mathcal{P}$ be the set of prime numbers $p$ such that $n\_p>0$. Assume that the Riemann $\zeta$-function has infinitely many zeros on the line $\frac{1}{2}+it$, and suppose that $f$ is a function of the form $$ f(s)=\prod\_{\nu\geq1}\zeta\left(\mu\left(s-\frac{1}{2}\right)+\frac{1}{2}\right)^{n\_\nu}. $$ Then $f$ is holomorphic in the half-plane $\Re s>1$ and has a meromorphic continuation to the half-plane $\Re s>\frac{1}{2}$. If, for all $\epsilon>0$, $ \mathcal{P}((1+\epsilon)x)-\mathcal{P}(x)\gg x^{\frac{\sqrt{5}-1}{2}}\log^2x, $ then the line $\Im s =\frac{1}{2}$ is the natural boundary of $f$; more precisely, every point of this line is an accumulation point of zeros of $f$.


As an example on the existence of a natural boundary, $\Omega$-results for Dirichlet series associated to counting functions are obtained. It is proved that if $D(s)=\sum\frac{a(n)}{{n^s}}$ has a natural boundary at $\Re s=\sigma$, then there does not exist an explicit formula of the form $ A(s) := \sum\_{n\leq x}a\_n=\sum\_{\rho}c\_\rho x^\rho+O(x^\sigma), $ where $\rho$ is a zero of the Riemann zeta-function, and hence it is possible to obtain a term $\Omega(x^{\sigma-\epsilon})$ in the asymptotic expression for $A(x)$.


Reviewed by Roma Kačinskaitė




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In the above review, where $\Im(s) = \frac{1}{2}$ appears, I'm sure $\Re(s) = \frac{1}{2}$ is intended. Also the "assume" is a bit strange, since it is an old, famous theorem of G.H. Hardy that $\zeta(s)$ has infinitely many zeros on the critical line. 


