# Continuation up to zero of a Dirichlet series with bounded coefficients

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?

• Surely (meaning of course, that I am not completely sure) this is false. A strategy for constructing a counterexample would be to take an infinite sum of Riemann zeta-like functions, each one having a single pole in $0 < \Re(s) < 1$, in such a way so that the set of poles has an accumulation point in, say, $\Re(s) \geq \frac{1}{2}$. Jan 9, 2010 at 22:06
• @Pete. Then how do you establish continuation of the Dedekind zeta function up to zero? Jan 9, 2010 at 22:45

The following paper seems to (among other things) give a detailed construction roughly along the lines of my comment above:

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

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.

No. There is no particular reason (if the $a_n$ are arbitrary bounded positive) even to expect an analytic continuation across $\mathrm{Re}(s) = 1$. A natural boundary at this line would be the generic case, and analytic continuation across it a consequence of some special property, for example the symmetry underlying the functional equation of the Dedekind zeta function.

• But before applying the functional equation you have to continue it a little bit! Jan 9, 2010 at 22:56
• Well, the situation is the same as for the Riemann zeta function: The analytic continuability everywhere (except $s = 1$) is something you establish in the course of proving the functional equation. By the time you have finished proving the functional equation, you already know that there is an analytic continuation! Jan 9, 2010 at 23:25
• Yes, I agree with engelbrekt. Analytic continuation beyond the region of absolute convergence should be something special, but I don't know how to make the idea of a "generic Dirchlet series" precise. Jan 10, 2010 at 6:32
• @Idoneal: There are several papers on random Dirichlet series. I didn't reference these, because Anweshi was especially interested in the case of non-negative coefficients, whereas the typical context for random Dirichlet series is complex coefficients, e.g. with the $a_n$'s taken to be IID random variables with values on the unit circle. The typical theorem is: a random Dirichlet series has a natural boundary at its abscissa of convergence. Jan 10, 2010 at 11:57