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 $\epsilon > 0$?

isinappropriate is crossposting here without leaving a link there to the question here. $\endgroup$ – Gerry Myerson Jul 18 '20 at 2:52