Landau's Theorem for Dirichlet series with real coefficients ($c_n$) states that if the coefficients are of fixed sign for all sufficiently large $n$, then the point $\sigma_0$ on the abscissa of convergence of the series is itself a singularity of the function represented by the series in the half plane $\sigma>\sigma_0$.

One can also show that the conclusion of Landau's Theorem applies in a broader context. For example, if the partial sums $\sum_{n\leq x} c_n\geq 0$, then the real point $\sigma_0$ is a singularity of the function.

My question is this: To what extent does the converse implication hold, that is, if $\sigma_0$ is a singularity of the function represented for $\sigma>\sigma_0$ by some Dirichlet series with real coefficients, then under what additional conditions may we conclude that the ($c_n$) are of fixed sign for sufficiently large $n$?

  • $\begingroup$ As Daniel explained, you can rig counterexamples to any simple statement you might hope to prove. However, theorems of this flavor tend to be provable under conditions on the $c_n$ which are strong, but typical for the kinds of Dirichlet series that show up in number theory. What example are you interested in? $\endgroup$ – Frank Thorne Oct 30 '11 at 20:24
  • $\begingroup$ Thanks for the input, Frank. I am interested in the sequence $c_n=2\mu{n}M(n)$- its partial sum is $M^2(x)$ - and so satisfies the statement I made in the question. I don't expect it to be of constant sign for sufficiently large $n$ (that would be asking a lot!), yet I am wondering if the pole on the real axis (which occurs at $2\Omega$, where $\Omega$ is the largest real part of a zero of $\zeta(s)$) + some other hypotheses would imply the positivity of the partial sum. $\endgroup$ – Kevin Smith Oct 30 '11 at 23:21
  • $\begingroup$ Error: should read $c_n=2\mu(n)M(n)$. $\endgroup$ – Kevin Smith Oct 30 '11 at 23:23
  • $\begingroup$ One can see that it is not of constant because $M(x)$ is known to be zero infinitely often. $\endgroup$ – Kevin Smith Oct 30 '11 at 23:34
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    $\begingroup$ Sorry, what is $M(n)$? $\endgroup$ – Frank Thorne Oct 31 '11 at 0:20

I'm not sure you can hope for much. For example consider the case $c_n=1$ if $n$ is not a square, and $c_n=-1$ otherwise. The associated Dirichlet series has a pole at $s=1$, but of course the terms are not of fixed sign for sufficiently large $n$.

In general, knowing analytic properties of a Dirichlet series (such as convergence) cannot tell you much about any of the individual terms $c_n$, since you can always change infinitely many of the $c_n$ for $n$ in a "sparse" set.

  • $\begingroup$ Thanks Daniel. I should perhaps consider the sign on "dense" sets, maybe the squarefree integers, yet, the function I am interested in is not multiplicative so that wouldn't really help. $\endgroup$ – Kevin Smith Oct 30 '11 at 20:03

Let $n_1<n_2<\cdots$ be any increasing sequence of positive integers, and define an arithmetic function $f$ by declaring that $f(n_k) = n_k^{\sigma_0}/k$ for each $k\ge1$, while $f(n)=-2^{-n}$ if $n$ is not equal to any of the $n_k$. Then $F(s) = \sum_{n=1}^\infty f(n)n^{-s}$ converges for $\sigma>\sigma_0$, but $\lim_{\sigma\to\sigma_0+} F(s) = +\infty$ and so $\sigma_0$ is a singularity. This class of examples shows that the set of positive integers at which $f(n)$ is positive can literally be any infinite set of numbers. (I say "positive" instead of "fixed sign" here to emphasize the fact that this can happen even when the one-sided limit is $+\infty$ rather than $-\infty$.)

One concrete example that's similar is: let $j$ be a positive integer, and set $G(s) = \zeta(js) - \zeta(s+1)$ (where $\zeta$ is the Riemann zeta-function). Then $G(s)$ has a singularity at $s=1/j$, indeed a simple pole with positive residue; however, its Dirichlet series coefficients are almost all negative, as the coefficient of $n^{-s}$ equals $-1/n$ unless $n$ is a $j$th power, in which case it equals $1-1/n$.

  • $\begingroup$ Thank you Greg - Your answer is very illuminating. In hindsight I think that its the positivity of the partial sum for sufficiently large $n$ that I want to get at, so I will re-think and post a new question. $\endgroup$ – Kevin Smith Oct 31 '11 at 7:54

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