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How to compute $\sum_{n=0}^{\infty} \frac{(-1)^n}{\cosh((n+\frac{1}{2}) \pi)}$?

My attempt was trying to consider the Mellin transform of $\frac{1}{\cosh(x)}$ and use the inverse Mellin transform as a representation of $\frac{1}{\cosh(x)}$

$$\frac{1}{\cosh(x)}=\mathcal{M}^{-1}\Big(2\Gamma(s)\beta(s)\Big)(x) $$ Which implies $$\sum_{n=0}^{\infty} \frac{(-1)^n}{\cosh((n+\frac{1}{2}) \pi)}=\frac{1}{2\pi i}\int_{\sigma - i\infty}^{\sigma +i\infty}2^{s+1} \frac{\Gamma(s)\beta^2(s)}{\pi^s}\, ds$$

This integral didn't lead me anywhere.

I did realize the series may be similar to a particular value of theta functions, but I couldn't find any that arrives to this sum.

Any help is appreciated

Gary
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Dqrksun
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    Is $\beta$ the Dirichlet beta function? – Gary Jan 28 '24 at 12:25
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    Without the alternating signs (ie factor $(-1)^n$) we do have a closed form in terms of theta and elliptic functions. But the alternating signs are creating problem in getting a closed form. – Paramanand Singh Jan 29 '24 at 06:25

1 Answers1

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From this website (See "Hyperbolic series involving the lemniscate functions") we have:

$$\sum _{n=0}^{\infty } \frac{(-1)^n}{\cosh \left(\left(n+\frac{1}{2}\right) \pi \right)}=\\\sum _{m=0}^{\infty } (-1)^m \text{sech}\left(\left(\frac{1}{2}+m\right) \pi \right)=\\\frac{i \log (1+\text{sech}(\pi ))}{\pi }-\frac{i \psi _{e^{-4 \pi }}^{(0)}\left(\frac{1}{4}-\frac{i}{4}\right)}{\pi }+\frac{2 i \psi _{e^{-2 \pi }}^{(0)}\left(\frac{1}{4}-\frac{i}{4}\right)}{\pi }+\frac{i \psi _{e^{-2 \pi }}^{(0)}\left(\frac{1}{2}-\frac{i}{2}\right)}{\pi }-\frac{2 i \psi _{e^{-\pi }}^{(0)}\left(\frac{1}{2}-\frac{i}{2}\right)}{\pi }$$

Mathematica code:

(I*Log[1 + Sech[Pi]])/Pi - (I*QPolyGamma[0, 1/4 - I/4, E^(-4*Pi)])/Pi + ((2*I)*QPolyGamma[0, 1/4 - I/4, E^(-2*Pi)])/Pi + (I*QPolyGamma[0, 1/2 - I/2, E^(-2*Pi)])/Pi - ((2*I)*QPolyGamma[0, 1/2 - I/2, E^(-Pi)])/Pi

Mariusz Iwaniuk
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