If a Dirichlet series has coefficients +1 and -1 and an analytic continuation without poles (or zeros) to the right of Re(s) = 1/2, what can we say about it's abscissa of convergence?
Is it always at most (at or to the left of) 1/2? Would it help if the coefficients are multiplicative? *And this is a normal dirichlet series, so each term is n^-x
A motivation is the reciprocal zeta series, where the abscissa of convergence can be determined by knowing up to where the analytic continuation doesn't have poles (Riemann hypothesis bounds the Mertens function to o(sqrt(n)n^ε)).
Actually my original functions must have an Euler product of (1+p^-x)(1-p^-x)(1+0p^-x) for the primes divided into two subgroups, the remaining primes ignored. I think my statement can be proven if my functions' equivalent of a prime zeta function converges (at least to 1/2) which can be proven if my functions' number of zeros satisfy a similar asymptomatic bound as the zeta function. Namely the sum of (1/r+1/r) converges (I think it must be absolutely) where this sum goes over the reciprocal of each root/zero of my function and with multiplicacy -1 over each pole, and each is paired with its complex conjugate. Can this be proven?
It is possible for the abscissa of convergence to be anything between $0$ and $1$.
To see that it can be as small as $0$, note that the Dirichlet series $$ \sum_{n \geq 1} \frac{(-1)^n}{n^s} $$ has abscissa of convergence equal to $0$.
It is clear that $1$ is possible, from $\zeta(s)$, and standard coefficient bounds guarantee that the abscissa cannot be greater than $1$.
The fact any abscissa of convergence between $0$ and $1$ is possible follows from the fact that for any $\theta \in (0, 1)$, one can choose signs $\epsilon_i \in \{ -1, +1 \}$ such that $$ \sum_{n \leq N} \epsilon_i \asymp N^\theta. $$ Abel summation then shows that $\theta$ is the abscissa of convergence.