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.)

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    $\begingroup$ (You had typos in your TeX that left mismatched braces in both equations) $\endgroup$ – Steven Stadnicki Apr 27 '18 at 0:52
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    $\begingroup$ Are you interested in all cube-free numbers, or just squares that are cube-free? (There's a bit of a mismatch between the title and the post) $\endgroup$ – Steven Stadnicki Apr 27 '18 at 0:53
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    $\begingroup$ Of course the sum exists (that is, converges), as it's a subsum of the sum of the reciprocals of all the squares, which converges. I think you are really asking whether there is a formula for it in terms of standard functions. If so, that's what the title should ask. And if you have found an answer, terrific – you should write it up, and post it as an answer. $\endgroup$ – Gerry Myerson Apr 27 '18 at 7:09
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    $\begingroup$ So, have you found an answer, or haven't you? It's not clear from your "Omg" comment. $\endgroup$ – Gerry Myerson Apr 27 '18 at 21:19
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    $\begingroup$ Anything you post here gets a timestamp which should be sufficient for establishing priority. $\endgroup$ – Gerry Myerson Apr 29 '18 at 4:00

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