Closed form for $\sum\limits_{k=1}^{\infty}\zeta(4k-2)-\zeta(4k)$

Question

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.

Answer 1

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.