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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?

Thanks in advance!

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    $\begingroup$ The proof that under RH it converges for $\Re(s) > 1/2$ is very similar to the PNT $\endgroup$
    – reuns
    Commented Oct 17, 2018 at 6:08

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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 $\Re(s)>1/2$.

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  • $\begingroup$ I was on the verge to posting the same response as GH when I saw your question. Perhaps the upper bound obtained so far for the de Bruijn-Newman constant (less or equal than 0.22 so far according to Polymath 15) can lead to some abcissa of convergence strictly less than 1. $\endgroup$
    – Sylvain JULIEN
    Commented Oct 17, 2018 at 11:21
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    $\begingroup$ @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). $\endgroup$
    – Peter Humphries
    Commented Oct 17, 2018 at 12:51

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