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

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?

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  • 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
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$$\frac1{(2n+1)^2(2n-1)^2}=\frac14\left(\frac1{2n-1}-\frac1{2n+1}\right)^2=\frac14\left(\frac1{(2n-1)^2}-\frac2{4n^2-1}+\frac1{(2n+1)^2}\right)$$

Now, we get

$$\frac{\pi^2}6=\sum_{n=1}^\infty\frac1{n^2}=\frac14\sum_{n=1}^\infty\frac1{n^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}\implies$$

$$\sum_{n=1}^\infty\frac1{(2n-1)^2}=\frac{\pi^2}8\;,\;\;\sum_{n=1}^\infty\frac1{(2n+1)^2}=\frac{\pi^2}8-1$$

Telescoping, we also get

$$\sum_{n=1}^\infty\frac1{4n^2-1}=-\frac12$$

so all in all we get that your series equals

$$\frac14\left(\frac{\pi^2}8-1+\frac{\pi^2}8-1\right)=\frac14\left(\frac{\pi^2}4-2\right)=\frac1{16}\left(\pi^2-8\right)$$

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