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It is known that if $\sum_{k=1}^{\infty} \frac{\mu(k)}{k^s} = \frac{1}{\zeta(s)}$ for $\Re(s) > 1/2$ then RH holds. My question is:

  1. Under RH why is it not $\sum_{k=1}^{\infty} \frac{\mu(k)}{k^s} = \frac{1}{\zeta(s)}$ for $\Re(s) > 0$?

  2. Is it known that $\sum_{k=1}^{\infty} \frac{\mu(k)}{k^s}$ diverges for $0 < \Re(s) \leq 1/2$ (whether under RH or not)?

Any reference material or relevant information will be much appreciated.

Thanks,

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    $\begingroup$ By general properties of Dirichlet series, if $\sum_k\mu(k)k^{-s}$ converged for some $s$ with $\sigma:=\operatorname{Re}s<1/2$, then it would converge to an analytic function in the half-plane $\operatorname{Re}s>\sigma$. However, $1/\zeta(s)$ is not analytic in such a half-plane, as there are zeros of $\zeta$ on the critical line. $\endgroup$ Jul 6, 2014 at 16:08

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We can not hope to the estimate of $M(n):=\sum_{k\leq n} \mu(k)$ better than $M(n)=O(n^{1/2+o(1)})$. Take $s\in (0,1/2)$. Rewrite partial sum of our series as $$\sum_{k=n}^m (M(k)-M(k-1))/k^s=-M(n-1)/n^s+M(m)/m^s+\sum_{k=n+1}^{m-1} M(k)(k^{-s}-(k+1)^{-s}).$$ Assume that both $n$, $m$ are large, $M(m)$ is much greater than $m^s$, $M(n-1)=0$ and $M(k)$ is non-negative for $n\leq k\leq m$. We see that partial sum is large, so the series diverges.

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