7
$\begingroup$

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

$\endgroup$
7
  • 3
    $\begingroup$ 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 $\endgroup$
    – muad
    May 3, 2010 at 21:49
  • $\begingroup$ 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. $\endgroup$
    – Mark Lewko
    May 3, 2010 at 21:57
  • $\begingroup$ 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. $\endgroup$ May 3, 2010 at 22:14
  • 1
    $\begingroup$ 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? $\endgroup$ May 4, 2010 at 0:35
  • $\begingroup$ It is this equivalence that Franel uses in connecting Farey and Riemann. $\endgroup$ May 4, 2010 at 10:34

2 Answers 2

11
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – Marc Palm
    Nov 25, 2010 at 15:42
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Not quite. What I'd like is a citation for the equivalence that Franel uses in "Les suites...". $\endgroup$ May 4, 2010 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.