I been reading the paper Novel integral representations of the Riemann zeta-function and Dirichlet eta-function, close expressions for Laurent series expansions of powers of trigonometric functions and digamma function, and summation rules of Sergey K. Sekatskii and he gives $$\frac1{\sinh^2 x}=\frac1{x^2}-\frac13+\frac2{\pi^2}\sum_\limits{n=1}^\infty(-1)^{n+1}(2n+1)\left(\frac{x}{\pi}\right)^{2n}\zeta(2n+2)$$
I have recalculated the series and gives $$\frac{1}{\sinh ^2(x)}=\sum _{k=1}^{\infty } x^k \left(-2 (k+1) \pi ^{-k-2} \zeta (k+2) \cos \left(\frac{\pi k}{2}\right)\right)+\frac{1}{x^2}-\frac{1}{3}$$
What do you think?
By applying $-\frac{d^2}{dx^2}\log(\cdot)$ to both sides of $$ \sinh(x) = x \prod_{n\geq 1}\left(1+\frac{x^2}{\pi^2 n^2}\right) \tag{1}$$ we get $$ \frac{1}{\sinh^2 x}=\frac{1}{x^2}-2\sum_{n\geq 1}\frac{\pi^2 n^2-x^2}{(\pi^2 n^2+x^2)^2}=\frac{1}{x^2}-2\sum_{n\geq 1}\frac{1}{\pi^2 n^2}\sum_{m\geq 0}(-1)^m(2m+1)\left(\frac{x}{\pi n}\right)^{2m}\tag{2} $$ and $$ \frac{1}{\sinh^2 x}=\frac{1}{x^2}-2\sum_{m\geq 0}\frac{(-1)^m(2m+1)\zeta(2m+2)}{\pi^{2m+2}}\,x^{2m}\tag{3}$$ so Sergey K. Sekatskii is definitely right. On the other hand $\cos\frac{\pi k}{2}$ is zero for odd values of $k$, hence the two representations are equivalent.