Q:

Explanations about some Mittag-leffler partial fraction expansions

Is it possible to show where the following series come from?
$$\sum _{k=1}^{\infty } \left(\frac{1}{\pi ^2 k^2}-\frac{2}{(x-2 \pi  k)^2}-\frac{2}{(2 \pi  k+x)^2}\right)+\left(-\frac{2}{x^2}-\frac{1}{6}\right)=\frac{1}{\cos (z)-1}$$
$$\sum _{k=1}^{\infty } \left(-\frac{2 \pi ^2 k^2+3}{6 \pi ^4 k^4}+\frac{2}{3 (x-2 \pi  k)^2}+\frac{4}{(x-2 \pi  k)^4}+\frac{2}{3 (2 \pi  k+x)^2}+\frac{4}{(2 \pi  k+x)^4}\right)+\frac{4}{x^4}+\frac{2}{3 x^2}+\frac{11}{180}=\frac{1}{(\cos (z)-1)^2}$$
$$\sum _{k=1}^{\infty } \left(\frac{8 \pi ^4 k^4+15 \pi ^2 k^2+15}{60 \pi ^6 k^6}-\frac{4}{15 (x-2 \pi  k)^2}-\frac{2}{(x-2 \pi  k)^4}-\frac{8}{(x-2 \pi  k)^6}-\frac{4}{15 (2 \pi  k+x)^2}-\frac{2}{(2 \pi  k+x)^4}-\frac{8}{(2 \pi  k+x)^6}\right)+\left(-\frac{8}{x^6}-\frac{2}{x^4}-\frac{4}{15 x^2}-\frac{191}{7560}\right)=\frac{1}{(\cos (z)-1)^3}$$ sorry for the inconvenient I forget to add a part of the formula.

A:

As I commented earlier, I have a problem with the first expression. So, since Maple said that it is correct, I suppose I am wrong but I would like to know where.
Let me consider $$S_1=\sum _{k=1}^{\infty } \frac{1}{\pi ^2 k^2}\qquad S_2=\sum _{k=1}^{\infty }\frac{1}{(x-2 \pi  k)^2}\qquad S_3=\sum _{k=1}^{\infty }\frac{1}{(x+2 \pi  k)^2}$$ So $$S_1=\frac 16\qquad S_2=\frac{\psi ^{(1)}\left(1-\frac{x}{2 \pi }\right)}{4 \pi ^2}\qquad S_3=\frac{\psi ^{(1)}\left(1+\frac{x}{2 \pi }\right)}{4 \pi ^2}$$ where appears the first derivative of the digamma function.$$S_1-2S_2-2S_3=\frac{1}{6}-\frac{\psi ^{(1)}\left(1-\frac{x}{2 \pi }\right)+\psi ^{(1)}\left(1+\frac{x}{2 \pi }\right)}{2 \pi ^2}=\frac{1}{6}+\frac{2}{x^2}+\frac{1}{\cos (x)-1}$$ the last simplification being obtained using a CAS.
Using Taylor around $x=0$, what I find is $$S_1-2S_2-2S_3=-\frac{x^2}{120}-\frac{x^4}{3024}-\frac{x^6}{86400}+O\left(x^{8}\right)$$ while $$\frac{1}{\cos (x)-1}=-\frac{2}{x^2}-\frac{1}{6}-\frac{x^2}{120}-\frac{x^4}{3024}-\frac{x^6}{86400}+O\left(x^{8}\right)$$

