Converse to a theorem of Landau on Dirichlet series - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T10:05:13Z http://mathoverflow.net/feeds/question/79536 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79536/converse-to-a-theorem-of-landau-on-dirichlet-series Converse to a theorem of Landau on Dirichlet series Kevin Smith 2011-10-30T18:51:53Z 2011-10-30T23:41:25Z <p>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 $\sigma>\sigma_0$.</p> <p>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. </p> <p>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 $n$?</p> http://mathoverflow.net/questions/79536/converse-to-a-theorem-of-landau-on-dirichlet-series/79540#79540 Answer by Daniel Loughran for Converse to a theorem of Landau on Dirichlet series Daniel Loughran 2011-10-30T19:33:27Z 2011-10-30T20:25:08Z <p>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 $n$.</p> <p>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.</p> http://mathoverflow.net/questions/79536/converse-to-a-theorem-of-landau-on-dirichlet-series/79561#79561 Answer by Greg Martin for Converse to a theorem of Landau on Dirichlet series Greg Martin 2011-10-30T23:41:25Z 2011-10-30T23:41:25Z <p>Let <code>$n_1&lt;n_2&lt;\cdots$</code> 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 $-\infty$.)</p> <p>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 $1-1/n$.</p>