# Abscissa of convergence $\sum_{n=0}^{\infty}\frac{\mu(n)}{n^s}$

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 $$\sigma_c$$?

• 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. – Dark Malthorp Nov 3 '18 at 5:30
• 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 – Shree Nov 3 '18 at 5:38
• @Shree, this sum is $1/\zeta(s)$ (@DarkMalthorp is right). – metamorphy Nov 3 '18 at 5:48
• 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. – Shree Nov 3 '18 at 5:54
• $\sigma_c > \frac12$ 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 $>\frac12$. – Dark Malthorp Nov 3 '18 at 15:19