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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?

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  • $\begingroup$ It's certainly true that $\sigma_a(P) \le \sigma_a(D)$ -- this is immediate from the fact that the divisor function $d(n)$ is $O(n^{\epsilon})$ for any $\epsilon > 0$. $\endgroup$ Sep 8 '13 at 15:11
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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 $\sigma_c(P) = \sigma_a(P) = -\infty$.

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  • $\begingroup$ Great example, it's simplicity is striking. Thank you. Now, do you think there are further hypotheses that would work? The case of interest to me has completely multiplicative coefficients. $\endgroup$ Sep 9 '13 at 14:14

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