# Does $\sum_{n=1}^\infty \frac{\mu(n)}{n^s}$ converge for $\sigma > \frac{1}{2}$?

Looking at @Lucia's answer to this question 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?

• The proof that under RH it converges for $\Re(s) > 1/2$ is very similar to the PNT Oct 17 '18 at 6:08
The convergence of $$\sum\mu(n)/n^s$$ for $$\Re(s)>1/2$$ is equivalent to the 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 $$\Re(s)>1/2$$.
• @SylvainJULIEN, no, an abscissa of convergence $\Theta < 1$ implies that $\zeta(s)$ has no zeroes for $\Re(s) > \Theta$. This is not implied by improved bounds for the de Bruiijn-Newman constant (it is not a substitute for RH in this sense). Oct 17 '18 at 12:51