Sum of sech....

Let, $S=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)\cosh[(n+\frac{1}{2})\pi]}$

$\begin{aligned} \implies 2S=\displaystyle\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(2n+1)\cosh[(n+\frac{1}{2})\pi]}\\ =\frac{1}{2}\displaystyle\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(n+\frac{1}{2})\cosh[(n+\frac{1}{2})\pi]}\end{aligned}$

$\implies S=\frac{1}{4}\displaystyle\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(n+\frac{1}{2})\cosh[(n+\frac{1}{2})\pi]}$

Now, let us assume, $S=\displaystyle\sum_{n=-\infty}^{\infty} (-1)^n f\Big(n+\frac{1}{2}\Big) \implies f(z)=\frac{1}{4z\cosh (z\pi)}$

$f(z)$ has simple poles at $z=0, i(m+\frac{1}{2})$ where $m$ is an integer $\color{#3D99F6} [\text{as }\cosh(ia)=\cos a]$ .

Let us use the contour-integral-based formula for summation $\forall g(z)$ , $\color{#3D99F6}\displaystyle\sum_{n=-\infty}^{\infty} (-1)^n g\Big(n+\frac{1}{2}\Big)=\sum\Big(\text{residues of } [g(z)\pi \sec (\pi z)] \text{ at singularities of } g(z)\Big)$ In our case, residue of $f(z)\pi \sec (\pi z)$ at $z=0$ is $\displaystyle\lim_{z\rightarrow 0}\Big[z\cdot\frac{1}{4z\cosh (z\pi)}\cdot\pi \sec (\pi z)\Big]=\frac{\pi}{4}$ .

Similarly, residue of $f(z)\pi \sec (\pi z)$ at $z= i(m+\frac{1}{2})$ is $\begin{aligned}&\lim_{z\rightarrow i(m+\frac{1}{2})}\Big[\big\{z- i(m+\frac{1}{2})\big\}\cdot\frac{1}{4z\cosh (z\pi)}\cdot\pi \sec (\pi z)\Big]&\\=&\frac{\pi\sec[i(m+\frac{1}{2})\pi]}{4i(m+\frac{1}{2})}\cdot\lim_{z\rightarrow i(m+\frac{1}{2})}\Big[\frac{z- i(m+\frac{1}{2})}{\cosh (z\pi)}\Big]&\\=&\frac{\pi}{4i(m+\frac{1}{2})\cosh[(m+\frac{1}{2})\pi]}\cdot\lim_{z\rightarrow i(m+\frac{1}{2})}\Big[\frac{1}{\pi\sinh(z\pi)}\Big]\quad\small \color{#3D99F6} [\text{using L'Hospital's Rule}] &\\=&-\frac{1}{4}\cdot\frac{(-1)^m}{(m+\frac{1}{2})\cosh[(m+\frac{1}{2})\pi]}\quad\small \color{#3D99F6} [\text{as } \sinh[i(m+\frac{1}{2})\pi]=i\sin[(m+\frac{1}{2})\pi]=i(-1)^m] \end{aligned}$ So, using the above formula, $\begin{aligned}& S=\frac{\pi}{4}-\frac{1}{4}\sum_{m=-\infty}^{\infty}\frac{(-1)^m}{(m+\frac{1}{2})\cosh[(m+\frac{1}{2})\pi]}=\frac{\pi}{4}-S&\\&\implies 2S=\frac{\pi}{4}&\\& \implies\boxed{S=\frac{\pi}{8}=0.392699}\end{aligned}$