Let $$f(x) = \sum_{n=0}^{\infty} (-1)^n \frac{\text{sech}\left((2n+1)x\right)}{2n+1} $$ which converges for all $x\in \mathbb R$. How can it be shown that $f\left(\frac{\pi}{2} \right) = \frac{\pi}{8}$ as WolframAlpha claims?
Here’s a way to write it as an alternative sum. Assume WLOG $x\gt 0$.
$$ f(x) = 2 \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \cdot \frac{e^{-(2n+1)x}}{1+e^{-2(2n+1)x}} \\ = 2\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \sum_{k=0}^{\infty} (-1)^k e^{-(2k+1)(2n+1)x} \\ = 2\sum_{k=0}^{\infty} (-1)^k \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} e^{-(2k+1)(2n+1)x} \\ = 2 \sum_{k=0}^{\infty} (-1)^k \tan^{-1} \left( e^{-(2k+1)x} \right)$$
