Looking at @Lucia's answer to [this question](https://mathoverflow.net/questions/164874/is-it-possible-to-show-that-sum-n-1-infty-frac-mun-sqrtn-diverg) it appears $\sum\_{n=1}^\infty \frac{\mu(n)}{n^s}$ converges for $\sigma > \frac{1}{2}$. Can someone point me to a proof or provide proof for this? If I misunderstood and if this is not true, is it known for what minimum value of $\sigma$ will it converge?


Thanks in advance!



Does $\sum_{n=1}^\infty \frac{\mu(n)}{n^s}$ converge for $\sigma > \frac{1}{2}$?
The convergence of $\sum\mu(n)/n^s$ for $\Re(s)>1/2$ is equivalent to the [Riemann hypothesis](https://en.wikipedia.org/wiki/Riemann_hypothesis). First, by Theorems 1.1 and 1.3 in Montgomery-Vaughan: Multiplicative number theory I, this convergence implies that $1/\zeta(s)$ is holomorphic in $\Re(s)>1/2$ (which is clearly equivalent to the Riemann Hypothesis), and also that $$M(x):=\sum\_{n\leq x}\mu(n)\ll\_\varepsilon x^{1/2+\varepsilon}\tag{$\*$}$$
for any $\varepsilon>0$. Conversely, by Theorems 13.24 and 1.3 in the same book, the Riemann Hypothesis implies $(\*)$, and also the convergence of $\sum\mu(n)/n^s$ for $\Re(s)>1/2$.


