I have seen statements like $\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}$ is convergent for $\Re(s)>1$, and I have seen proof of it being zero (and therefore convergent) when $s = 1$ but haven’t seen specific claim that the abscissa of convergence $\sigma_c=1$ for this Dirichlet series. And I’ve seen proof of $\sigma_a = 1$

Does anyone know of a reference to a proof for the value of $\sigma_c$?

Thanks in advance.

  • 4
    $\begingroup$ Finding value of $\sigma_c$ would be equivalent to solving the Riemann zeta hypothesis. $\sigma_c = \frac12$ would imply that RZH is true, and $\sigma_c > \frac12$ would imply RZH is false. $\endgroup$ – Dark Malthorp Nov 3 '18 at 5:30
  • $\begingroup$ It is possible $\sigma_c =1$ and the sum oscillates without exploding. (Like $\eta(0)$ is Grandi series) - in which case it won’t prove or disprove RH $\endgroup$ – Shree Nov 3 '18 at 5:38
  • $\begingroup$ @Shree, this sum is $1/\zeta(s)$ (@DarkMalthorp is right). $\endgroup$ – metamorphy Nov 3 '18 at 5:48
  • $\begingroup$ I do agree it is $\frac{1}{\zeta(s)}$ and $\sigma_c=\frac{1}{2}$ proves RH, but I do think $\sigma_c>\frac{1}{2}$ does not disprove RH. $\endgroup$ – Shree Nov 3 '18 at 5:54
  • 1
    $\begingroup$ $\sigma_c > \frac12$ would imply non-analyticity of $\frac1{\zeta}$ at some point with real part $>\frac12$, but since $\zeta$ is holomorphic except for a single pole at $s = 1$, this is only possible if $\zeta(s)$ has a zero with real part $>\frac12$. $\endgroup$ – Dark Malthorp Nov 3 '18 at 15:19

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