$$\sum_{c=1}^\infty \frac{1}{(2c+1)^2(2c-1)^2} = \frac{1}{16}(\pi^2 -8)$$

I got the result using wolfram alpha but I don't know how to calculate it. I tried breaking it into telescopic sums but it can't be separated like that. Any hints?

  • 2
    $\begingroup$ A partial fraction decomposition looks like a good idea. $\endgroup$ – Daniel Fischer Mar 6 '16 at 22:33
  • $\begingroup$ See also Basel problem. $\endgroup$ – Lucian Mar 7 '16 at 6:48


Now, we get



Telescoping, we also get


so all in all we get that your series equals



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