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(Crossposted to Math Overflow) 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?

Edit: 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 series 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 $\epsilon > 0$?

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  • $\begingroup$ Your formula for $a_n$ is incorrect if the Euler factors are $1/(1 + a_p/p^s)$. Your factors should be $1/(1 - a_p/p^s)$. The notation $\sigma_a$ and $\sigma_c$ are standard for half-planes of convergence of Dirichlet series, not Euler products, so I had to reread your question to figure out what you were actually asking. It is not true that $\sigma_c < \sigma_a$, but rather $\sigma_c \leq \sigma_a$. Is there an actual purpose you have in mind for the Euler product in your motivational example, or is it just curiosity? $\endgroup$
    – KCd
    Commented Jul 18, 2020 at 2:40
  • $\begingroup$ Also posted to MO, mathoverflow.net/questions/365905/… $\endgroup$ Commented Jul 18, 2020 at 2:52
  • $\begingroup$ @KCd I fixed the factors. The question is specifically about the case when $\sigma_c < \sigma_a$ with a strict inequality, because I want to know if the corresponding Dirichlet series also converges in the vertical strip between $\sigma_c$ and $\sigma_a$. $\endgroup$ Commented Jul 23, 2020 at 23:53

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