I am looking for a closed form of the expression $$ \sum_{k=1}^{\infty}\zeta(4k-2)-\zeta(4k) $$ Closed form would be something in terms of constants such as $\pi$, $\gamma$, $e$, etc.


This is $$ \sum_{n=2}^\infty\frac1{n^2+1}=-\int_0^1\frac{x\,\sin(\log x)}{1-x}\,\mathrm dx=\ldots=\frac12\pi\coth\pi-1\approx0.576674. $$ More generally, $$ \sum_{n\in\mathbb Z}\frac{a}{a^2+n^2}=\pi\coth a\pi. $$ Edit: To reach the first series above, note that, for each $k\geqslant1$, $$ \zeta(4k-2)-\zeta(4k)=\sum_{n\geqslant2}\left(1-\frac1{n^2}\right)\frac1{n^{4k-2}}, $$ hence the sum to be computed is $$ \sum_{n\geqslant2}\left(1-\frac1{n^2}\right)\sum_{k\geqslant1}\frac1{n^{4k-2}}=\sum_{n\geqslant2}\left(1-\frac1{n^2}\right)\frac1{n^2}\frac1{1-\frac1{n^4}} =\sum_{n\geqslant2}\frac1{n^2+1}. $$ The change of the order of summations is valid since every term is nonnegative.

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    $\begingroup$ Could you elucidate for me, please ? I am not very familiar with the Zeta function. When I compute numerically the sum asked in this post I arrive to 0.576674 and I would like to know what is this number. Thanks, cheers and Merry Xmas. $\endgroup$ – Claude Leibovici Dec 25 '13 at 8:59
  • $\begingroup$ @ClaudeLeibovici I am afraid one should rather find 1.076674. You might be missing .5 somewhere. Two questions: Did you read the link? Did you downvote? $\endgroup$ – Did Dec 25 '13 at 9:21
  • $\begingroup$ I went to the link and I have been lost. What I typed in Mathematica is NSum[Zeta[4 k - 2] - Zeta[4 k], {k, 1, Infinity}]and I got this value of 0.576674. I id not downvote (it would be a shame for me to downvote in any area I am ignorant). Please let me know where I am wrong. Cheers $\endgroup$ – Claude Leibovici Dec 25 '13 at 9:28
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    $\begingroup$ I get your number if the summation starts at k=0 and not k=1 as I did according to the post. $\endgroup$ – Claude Leibovici Dec 25 '13 at 9:32
  • $\begingroup$ @ClaudeLeibovici I was wrong, thanks for asking (I will post some explanations about how to reach the series). $\endgroup$ – Did Dec 25 '13 at 9:35

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