Abscissa of absolute convergence of the product of two Dirichlet series 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.
 A: I think it is false:
Consider a simple series with a zero at $s=2$.  For example
$$E(s)=1-\frac{1}{2^s}-\frac{12}{4^s}=P(2^{-s}),\quad \text{with} \quad P(x)=1-x-12x^2.$$
We have $E(2)=0$ and $E(1)=-5/2$. 
Since
$$\frac{1}{1-x-12x^2}=\frac{1}{7}\sum_{k=0}^\infty (4^{k+1}+(-1)^k 3^{k+1})x^k,\qquad |x|<\frac14.$$
We have
$$E(s)^{-1}=1+\frac{1}{7}\sum_{k=1}^\infty (4^{k+1}+(-1)^k 3^{k+1})\frac{1}{2^{ks}},\qquad 
\sigma>2.$$
Consider 
$$A(s)=\zeta(s) E(s)$$
The Dirichlet series $A(s)$ is absolutely convergent for $\sigma>1$, because
it is the product
of two absolutely convergent Dirichlet series is absolutely convergent. The abscissa 
of absolute convergence of $A(s)$ is $a=1$ because the function $A(s)$ 
extends to a meromorphic
function and has a simple pole at $s=1$. 
Let now $B(s)=E(s)^{-1} \zeta(s)^{-1}$. The Dirichlet series $B(s)$ is absolutely convergent for $\sigma>2$  as product of two absolutely convergent Dirichlet series.
But the function $B(s)$ extends to a meromorphic function with a pole at $\sigma=2$. 
Therefore $b=2$ is is abscissa of absolute convergence.
The product $A(s)B(s)=1$ and its abscissa of absolute convergence is $-\infty$. 
