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