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Let $f$ and $g$ be an arithmetic functions, and let $f*g$ be the Dirichlet convolution of $f$ and $g$.

As known from fundamental analytic number theory, the Dirichlet series generating function is: $DG(f;s)=\sum_{n=1}^\infty \frac{f(n)}{n^s}$.

Hence, The multiplication of Dirichlet series can be written as: $DG(f;s)DG(g;s)=DG(f*g;s)$, but as we know, Riemann series theorem says that if the series is conditionally convergence, then any permutation of the series may generate another sum.

My question is, if $DG(f;s)$ $DG(g;s)$ are conditionally convergence, how can we definde $DG(f;s)DG(g;s)$?

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    $\begingroup$ if a Dirichlet series converges anywhere, it will also converge absolutely at an abscissa $+1$ from the conditional convergence one (at least of course), so one defines the Dirichlet product where $f,g$ are both absolutely convergent and then extend it as much as possible by analytic continuation which is unique $\endgroup$
    – Conrad
    Oct 28 '20 at 13:28
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    $\begingroup$ Also @Conrad $\sum_{n=1}^\infty (-1)^nn^{-s}$ converges for $\Re(s) > 0$ while $\sum_{n=1}^\infty (\sum_{d| n}(-1)^{d}(-1)^{n/d})n^{-s}$ does not. Anywhere $DG(f,s),DG(g,s),DG(f\ast g,s)$ converge we have $DG(f,s)DG(g,s)=DG(f\ast g,s)$ (analytic continuation stuff plus $DG(f,s)=\lim_{h\to 0^+} DG(f,s+h)$) $\endgroup$
    – reuns
    Oct 28 '20 at 13:49
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This question can be completely answered by the following surprising theorem.

Theorem: Let $D(f,s)$, $D(g,s)$ be the Dirichlet series corresponding to $f$ and $g$. If $D(f,s)$ and $D(g,s)$ conditionally converge at $s=0$, then the convolution Dirichlet series $D(f*g,s)$ must be conditionally converges at $s=\frac{1}{2}+it$ for each $t\in \mathbb{R}$. And the constant "$\frac{1}{2}$" is optimal as concerns $\text{Re}\hspace{0.05cm}s$.


The proof can be found in the book Diophtine apprximation and Dirichlet series,pp114 theorem 4.3.4 for second Edition.

I really like its proof, which used hyperbola method of Dirichlet to prove the convergence, and used Banach-Steinhuas theorem and Kronecker's lemma to prove the optimality.


Other interesting consequences can also be found in section 4.3 of this book.

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    $\begingroup$ Your username just reminded me of Landau's lemma, another helpful result in this area :D $\endgroup$
    – TravorLZH
    Nov 13 '21 at 6:20

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