I know there is a formula for $\displaystyle\sum_{k=1}^{\infty}\frac{\mu(k)}{k^s}=\zeta(s)^{-1}$, which in other words mean the sum over the square-free numbers, due to the present Mobius function.
A different result is $\displaystyle\sum_{k=1}^{\infty}\frac{|\mu(k)|}{k^s}=\frac{\zeta(s)}{\zeta(2s)}$
How about the cube-frees? Looking for: $\displaystyle\sum_{k \text{ } cube-free}\frac{1}{k^2}$.
(Site is not loading the equations I posted in math mode. Oh well.)