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We know that given a multiplicative function $f$ for which the series $\sum_{n=1}^\infty f(n)$ converges absolutely then so does the Euler product $\prod_{p}\sum_{k=0}^\infty f(p^k)$, but does the reverse hold (at least up to conditional convergence)?

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  • $\begingroup$ Are you referring to a Dirichlet's series? Or to general series involving multiplicative functions? I remind you that the Euler product is a consequence of choosing a multiplicative function, it is an identity, whoch of course makes sense if you have convergence! $\endgroup$ – PITTALUGA Oct 2 '13 at 15:01
  • $\begingroup$ If the series $\sum_{n=1}^\infty f(n)$ converges absolutely, then the product converges absolutely as well (and we get identity), my question is the concerned with the opposite direction (being an identity is irrelevant here). $\endgroup$ – user98246 Oct 2 '13 at 15:38
  • $\begingroup$ OK, I'll be more direct: the answer is yes, if the Euler product converges also the initial series do. The point was that this just comes from the nature of the identity, which of course holds where you have convergence. It's like $1/(x^2-1)=1/((x+1)(x-1))$ which is true but makes no sense for $x = \pm 1$. $\endgroup$ – PITTALUGA Oct 2 '13 at 15:56
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Suppose $f$ is multiplicative and $\prod_p \sum_k |f(p^k)|$ converges, i.e. there is $L$ such that for every $\epsilon > 0$ there exist $K,P$ such that $\left|L - \prod_{p \le P'} \sum_{k \le K'} |f(p^k)|\right| < \epsilon$ whenever $K' > K$ and $P' > P$. Now $$\prod_{p \le P_1} \sum_{k \le K_1} |f(p^k)| \le \sum_{n \le N} |f(n)| \le \prod_{p \le P_2} \sum_{k \le K_2} |f(p^k)|$$ where $P_1$ and $K_1$ are such that all positive integers $\prod_{p \le P_1} p^{k(p)}$ with all $k(p) < K_1$ are at most $N$, while $P_2$ and $K_2$ are such that all $n \le N$ are of the form $\prod_{p \le P_2} p^{k(p)}$ with all $k(p) \le K_2$. We conclude that the sum converges absolutely.

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