I'm crossposting this from math.stackexchange because I think it might be inappropriately research-level for the community over there.

Suppose we have an Euler product over the primes

$$F(s) = \prod_{p} \left( 1 - \frac{a_p}{p^s} \right)^{-1},$$

where each $a_p \in \mathbb{C}$. The Euler product is convergent in the range $Re(s) > \sigma_c$, and absolutely convergent in the range $Re(s) > \sigma_a$, for some $\sigma_c < \sigma_a \in \mathbb{R}$. If we multiply out the Euler product, we get a Dirichlet series

$$F(s) = \sum_{n=1}^\infty \frac{a_n}{n^s},$$

where $a_n := \prod_{p^k || n} a_p^k$ is completely multiplicative as a function of $n$.

Question: We know that the Dirichlet series for $F(s)$ must converge absolutely in the half-plane $Re(s) > \sigma_a$. Must the Dirichlet series for $F(s)$ also converge in the half-plane $Re(s) > \sigma_c$? If not, what is a counterexample?

My question is motivated by considering a product like

$$F(s) = \left(1 - \frac{1}{2^s}\right)^{-1}\left(1 + \frac{1}{3^s}\right)^{-1}\left(1 - \frac{1}{5^s}\right)^{-1}\left(1 + \frac{1}{7^s}\right)^{-1} ... = \prod_{n=1}^\infty \left( 1 + \frac{(-1)^n}{p_n^s} \right)^{-1},$$

where a classical result on infinite products demonstrates convergence for $Re(s) > 1/2$ [although absolute convergence only happens in the half-plane $Re(s) > 1$]. This product for $F(s)$ will have no zeroes in the half-plane $Re(s) > 1/2$, so if we multiply it out to get the Dirichlet series

$$F(s) = \sum_{n=1}^\infty \frac{a_n}{n^s} = 1 + \frac{1}{2^s} - \frac{1}{3^s} + \frac{1}{4^s} + \frac{1}{5^s} - \frac{1}{6^s} - \frac{1}{7^s}...,$$

does the Dirichlet series converge too? Can we then conclude that the coefficients $a_n$ satisfy

$$\sum_{j = 1}^n a_j = O(n^{1/2 + \epsilon}),$$

for all $\epsilon > 0$?

  • 3
    $\begingroup$ "I'm crossposting this from math.stackexchange because I think it might be inappropriately research-level for the community over there." There is nothing the least bit inappropriate about posting research-level questions to math.stackexchange. What is inappropriate is crossposting here without leaving a link there to the question here. $\endgroup$ Jul 18, 2020 at 2:52
  • $\begingroup$ Duly noted, thank you! $\endgroup$ Jul 18, 2020 at 13:44

2 Answers 2


First of all, I think in your first formula you want $F(s)=\prod_p\left(1-\frac{a_p}{p^s}\right)^{-1}$ so that when it multiplies out, the coefficients at the primes match. We should also have all $|a_p|\le1$ since otherwise the Euler product still makes sense as a product, but does not expand to an absolutely convergent Dirichlet series.

Secondly, there are several classical results on Dirichlet series that are no longer well known, but are nicely documented in chapter 9 (Dirichlet Series) of The Theory of Functions, by Titchmarsh. Many of these come down to summation by parts, bounds on vertical growth, and some kernels.

In particular, it is non-trivial that the locus of conditional convergence is always a half-plane (9.12). Further $\sigma_a - \sigma_c \le 1$ (9.13). The region of conditional convergence has polynomial growth on vertical lines $|F(\sigma+i T)|\ll|t|^{1-(\sigma-\sigma_c)+\epsilon}$ (9.33). A partial converse is that if the function is regular and $O(t^\epsilon)$ then the Dirichlet series is at least conditionally convergent. There are similar results for the half-plane where the mean value exists.

For dealing with Euler products, the primary trick is that $F(s)$ is convergent as an Euler product iff $\log F(s)$ is absolutely convergent as a sum. Expanding this out, since the square and higher terms will absolutely converge further, this is essentially saying that $\sum a_p p^{-s}$ converges.

This suggests a way to answer your question in the negative. If $F(s)$ converges conditionally then $F(s)$ grows polynomially in $\Im s$, so $\log F(s)$ grows subpolynomially. Rig up the $a_p$ so that $\sum a_p p^{-s}$ is conditionally convergent for some $1/2 < \sigma$ (achieved if $\sum_{p<N}a_p\ll \sqrt N$ but grows too quickly vertically (make long subsequences look like $p^{-i T}$). In fact, this may be the generic case, and a probabilistic argument could work, since the $p^{i T}$ can be modeled as independent random variables. Here is one potential construction that I think will work. Take the $a_p$ to be in {1,0,-1} such that the non-zero terms are alternating, to ensure convergence. For $2^n < p < 2^{n+1}$ also require $a_p$ to either be 0 or sign$(\Re p^{-i 2^n})$, which you can do a fixed proportion of the time. Then at $\sigma + i 2^n$ the function will be as large as $2^{n/4}$.

[Edit] Correction, I was completely wrong about this being generic. In fact, Kowalski mentions a result from Bagchi's thesis, that almost surely a random Euler product will converge between 1/2 and 1 (and have polynomial growth)

  • $\begingroup$ I fixed the sign of $a_p$ in my Euler product, but I don’t necessarily want $|a_p| \leq 1$ for all $p$, because I’m not assuming $\sigma_c < \sigma_a \leq 1$. Your answer is very persuasive, but I still have misgivings. Since the Euler product shows $F(s)$ is nonzero to the right of $\sigma_c$, why doesn’t such a zero-free region imply that $F(s)$ converges as a Dirichlet series here, too? And the same for $1/F(s)$ in that half-plane to the right of $\sigma_c$? My understanding was that zero-free regions imply convergence (unless ruled out for trivial reasons e.g. $n$th term test). $\endgroup$ Jul 18, 2020 at 13:42

Maybe not exactly what you're looking for, but you may be interested in this preprint by Kaczorowski and Perelli: arXiv:1506.07630 where the authors study the links between several kinds of convergence abscissae for the Selberg class and the extended Selberg class as well.

  • $\begingroup$ Neat! That reminds me of a paper by Brevig & Heap showing an even stronger result (which I see K & P give a hat-tip to at the start of their paper): if the coefficients of the Dirichlet series are multiplicative, then the abscissas of uniform & absolute convergence always coincide. doi.org/10.1016/j.exmath.2015.10.005 $\endgroup$ Jul 24, 2020 at 0:26

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