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?

  • $\begingroup$ 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, $\Re(s) \geq \frac{1}{2}$. $\endgroup$ Jan 9, 2010 at 22:06
  • $\begingroup$ @Pete. Then how do you establish continuation of the Dedekind zeta function up to zero? $\endgroup$
    – Anweshi
    Jan 9, 2010 at 22:45

2 Answers 2


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.

  • $\begingroup$ But before applying the functional equation you have to continue it a little bit! $\endgroup$
    – Anweshi
    Jan 9, 2010 at 22:56
  • 2
    $\begingroup$ 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! $\endgroup$
    – engelbrekt
    Jan 9, 2010 at 23:25
  • $\begingroup$ 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. $\endgroup$
    – Idoneal
    Jan 10, 2010 at 6:32
  • 1
    $\begingroup$ @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. $\endgroup$ Jan 10, 2010 at 11:57

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