I first asked the following question on Mathematics StackExchange (a few weeks ago), since the content of MathOverflow is mostly above my academic level. I didn't want to bother people on this forum with a maybe trivial question. But since I got no answer, I'm trying my luck here.

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) I need help to prove the theorem in the case $\alpha_f \neq \alpha_g$.

After searching the literature, I found similar theorems stated in a few sources, but never saw any proof.

What I have managed to prove up to now:

• If $f$ and $g$ are positive real-valued functions, the result is obvious.
• Suppose $\alpha_f < \alpha_g$. Then, if $$\sum_{n=2}^{+\infty}\frac{|f(n)|}{n^{\alpha_g}} < |f(1)|$$ the result is also true.

I tried to experiment with particular functions not satisfying either of these strong conditions, but I fail to see why the result is true in the general case.

I first asked the following question on Mathematics StackExchange (a few weeks ago), since the content of MathOverflow is mostly above my academic level. I didn't want to bother people on this forum with a maybe trivial question. But since I got no answer, I'm trying my luck here.

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. I need help to prove the theorem in the case $\alpha_f \neq \alpha_g$.

After searching the literature, I found similar theorems stated in a few sources, but never saw any proof.

What I have managed to prove up to now:

• If $f$ and $g$ are positive real-valued functions, the result is obvious.
• Suppose $\alpha_f < \alpha_g$. Then, if $$\sum_{n=2}^{+\infty}\frac{|f(n)|}{n^{\alpha_g}} < |f(1)|$$ the result is also true.

I tried to experiment with particular functions not satisfying either of these strong conditions, but I fail to see why the result is true in the general case.