$\zeta(s+1)/\zeta(s)$ - MathOverflow

$\zeta(s+1)/\zeta(s)$

2010-05-03T21:35:35Z

Franel uses the convergence of

$ \frac{\zeta(s+1)}{\zeta(s)} = \sum \frac{c(n)}{n^s}$

as an equivalent to the Riemann hypothesis.

Does anybody have a citation for this result and/or hints for computing $c(n)$?

Thanks for any insight.

Cheers, Scott

Answer by François G. Dorais for $\zeta(s+1)/\zeta(s)$

2010-05-04T01:41:36Z

Since $$\zeta(s+1) = \sum_{n=1}^\infty \frac{1/n}{n^s}$$ and $$\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \frac{\mu(n)}{n^s}$$ where $\mu$ is the Möbius function, we have $$c(n) = \sum_{d \mid n} \frac{d}{n}\mu(d) = \frac{1}{n}\prod_{p \mid n} (1-p)$$ using Dirichlet convolution.

Answer by Gerry Myerson for $\zeta(s+1)/\zeta(s)$

2010-05-04T13:00:43Z

If I understand correctly, what Scott wants is a citation for Franel's paper on (Farey series and) the Riemann Hypothesis. That would be Les suites de Farey et le problème des nombres premiers, Göttinger Nachr. (1924) 198–201.

Answer by Jakob Katz for $\zeta(s+1)/\zeta(s)$

2010-05-04T19:27:54Z

This would be a comment, but I don't have the points for it.

Also not quite what you're asking for, but for a different look at what I believe is the same relationship between Farey series and RH (not having read Franel's paper), check H.M. Edwards' book on the Riemann zeta function, paragraph 12.2. He references Franel, but proves the equivalence in question by other means.