I'd like some help to prove the following theorem :
Let $\sum_{n \geq 1}\frac{f(n)}{n^s}$ and $\sum_{n \geq 1}\frac{g(n)}{n^s}$ be two Dirichlet series with respective abscissas of absolute convergence $\alpha_f$ and $\alpha_g$ ($\alpha_f, \alpha_g \neq -\infty$). Then the abscissa of convergence of $\sum_{n \geq 1}\frac{f*g(n)}{n^s}$ is :
- $\max(\alpha_f, \alpha_g)$ if $\alpha_f \neq \alpha_g$ ;
- less than or equal to $\alpha$ if $\alpha_f = \alpha_g = \alpha$
where $f*g$ refers to the Dirichlet convolution of $f$ and $g$.
The result is to be proved in an exercise from this book (exercise 9 page 259)
The statement of the book is probably buggy, I had to modify a few details for it to be plausible (for instance, replacing $\min$ by $\max$ and "greater than" by "less than").
I need a proof of the first point (case $\alpha_f \neq \alpha_g$) of the theorem. I'd prefer proofs not making use of complex analysis, since I don't know about it.
My ultimate goal is to prove that if $f$ is invertible (for Dirichlet convolution) and $\alpha_f \neq -\infty$, then $\alpha_{f^{-1}} \leq \alpha_f$.