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:



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> 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)?
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Any reference material or relevant information will be much appreciated.


Thanks,



On the convergence of Dirichlet series over the Mobius Mu function
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.


