The series of $\frac{1}{\cosh(z)}$ How to show that
$$\frac{1}{\cosh(z)} =\sum _{n=0}^{\infty }{\frac {\left( 
-1 \right)^{n}\left(\psi \left( 2\,n,\frac{3}{4}\right)-\psi \left( 2\,n,\frac{1}{4} \right) \right) {z}^{2\,n}}{ {4}^{n}{\pi }^{2\,n+1}\left( 2\,n \right) !}},$$
where $\psi$ is a polygamma function?
Is there maybe some link with an answer, or a book in which the above is shown?
 A: Commonly known, the expansion
$$\frac{1}{\cosh(x)}=\sum_{n=0}^\infty E_n \frac{t^n}{n!}$$
with $E_n$ the Euler numbers (see here or definition here).
In order to grasp the foundation of the calculation from here, I have to refer you first directly to the following article from Kölbig that discusses deeply the relation between the Euler numbers and the polygamma function $\psi^{(k)}(x)$ for $x=\frac{1}{4}$ and $x=\frac{3}{4}$. This is a standard article that is often cited.
However, then you will need another reference since the calculation not trivial: there are deep fundamental relations between the Dirichlet $\beta$-function, Euler numbers, and Riemann $\zeta$-function for positive integers; this is shown by Idowu in another standard article where the author based on the work of Kölbig and a prior article shows that the Euler numbers $(E_{2k})$ relate in the above scheme, which you have, to the polygamma function $\psi^{(2k)}(x)$ for $x=\frac{1}{4}$ and $x=\frac{3}{4}$, see final result on last page equation 3.3.
So as you can see the answer to your question is not trivial and it needs some further elaboration and study by yourself in the details. But the articles above should deliver the foundation.
