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.)

  • 1
    (You had typos in your TeX that left mismatched braces in both equations) – Steven Stadnicki Apr 27 at 0:52
  • 1
    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) – Steven Stadnicki Apr 27 at 0:53
  • 2
    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. – Gerry Myerson Apr 27 at 7:09
  • 2
    So, have you found an answer, or haven't you? It's not clear from your "Omg" comment. – Gerry Myerson Apr 27 at 21:19
  • 3
    Anything you post here gets a timestamp which should be sufficient for establishing priority. – Gerry Myerson Apr 29 at 4:00

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.