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

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

• This is not the same series as you have mentioned but it may be useful none-the-less. Warren D. Smith, "Cruel and unusual behavior of the Riemann zeta function" secamlocal.ex.ac.uk/people/staff/mrwatkin/zeta/cruel.ps
May 3, 2010 at 21:49
• To compute $c(n)$ use the identities $1/zeta(s) = \sum_{n=1}^{\infty} \mu(n)/n^{s}$ (this follows from Mobius inversion) and $\zeta(s+1) = \sum_{n=1}^{\infty} 1/n^{s+1}$, which hold in an appropriate half-plane. May 3, 2010 at 21:57
• The computation is discussed in G. P\'olya and G.~Szeg\"o, Problems and theorems in analysis, Vol.~II, Grundlehren Math. Wiss. 216, Springer-Verlag, Berlin et al. (1976), Division 8, Chapter 1, Sections 5--7. May 3, 2010 at 22:14
• The only Franel-Riemann connection I've been able to find concerns Farey series. How have you come to believe that Franel did what you say he did? May 4, 2010 at 0:35
• It is this equivalence that Franel uses in connecting Farey and Riemann. May 4, 2010 at 10:34

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.
• The conditional convergence of $\sum_{n=1}^{\infty} \mu(n) n^{-s}$ in $\Re s >1/2$ is equivalent to the Riemann Hypothesis. We know that $\sum_{n=1}^{\infty} n^{-s-1}$ converges for $\Re s >0$. From these two fact, you can deduce that the convergence of $\zeta(s+1)/\zeta(s)$ in $\Re s >1/2$ is actually equivalent to RH. Nov 25, 2010 at 15:42