# Convergence of Euler product implies convergence of Dirichlet series?

(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 $$\epsilon > 0$$?

• 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?
– KCd
Jul 18, 2020 at 2:40
• Also posted to MO, mathoverflow.net/questions/365905/… Jul 18, 2020 at 2:52
• @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 $\sigma_a$. Jul 23, 2020 at 23:53