Let $D(s)$ be a Dirichlet series with abscissa of convergence $\sigma\_c=\sigma\_a$. Does it follow that the Dirichlet series defined by $P(s)=D(s)\bar D(\bar s)$ has the same abscissa of convergence?


In light of Noam D. Elkies answer below, I would like to know about conditions on the coefficients that also lead to the divergence of the square? What if the coefficients are completely multiplicative? 



Abscissa of convergence of Dirichlet series
That's not true even for power series with real coefficients.
Let
$$
D(s) = \sqrt{1-2^{-s}} 
 = 1 - \frac12 2^{-s} - \frac18 4^{-s} - \frac1{16} 8^{-s} - \frac5{128} 16^{-s}
- \cdots .
$$
Then $P(s) = 1 - 2^{-s}$, so 
$\sigma\_c(D) = \sigma\_a(D) = 0$ but 
$\sigma\_c(P) = \sigma\_a(P) = -\infty$.


