\begin{align*} \sum_{k=1}^{n-1} \frac{1}{\cosh^2\left(\frac{\pi k}{n}\right)}\end{align*}
I tried to solve with mathematica that shows Does anyone know how to derive this and does it is possible for Human to derive this(speaking seriously)
\begin{align*} \sum_{k=1}^{n-1} \frac{1}{\cosh^2\left(\frac{\pi k}{n}\right)}\end{align*}
I tried to solve with mathematica that shows Does anyone know how to derive this and does it is possible for Human to derive this(speaking seriously)
First of all, you could have simplified the formula $$\frac{\pi ^2}{n^2}\sum_{k=1}^{n-1} \frac{1}{\cosh^2\left(\frac{\pi k}{n}\right)}=$$ $$\psi _{e^{\frac{\pi }{n}}}^{(1)}\left(\left(1-\frac{i}{2}\right) n\right)+\psi _{e^{\frac{\pi }{n}}}^{(1)}\left(\left(1+\frac{i}{2}\right) n\right)-\psi _{e^{\frac{\pi }{n}}}^{(1)}\left(1-\frac{i n}{2}\right)-\psi _{e^{\frac{\pi }{n}}}^{(1)}\left(1+\frac{i n}{2}\right)$$
Next, you could bound the summation by the two integrals $$I_1=\int_{1}^{n-1} \frac{dk}{\cosh^2\left(\frac{\pi k}{n}\right)}=\frac n \pi \left(\tanh \left(\frac{\pi (n-1)}{n}\right)-\tanh \left(\frac{\pi }{n}\right)\right)$$
$$I_2=\int_{0}^{n-1} \frac{dk}{\cosh^2\left(\frac{\pi k}{n}\right)}=\frac n \pi \tanh \left(\frac{\pi (n-1)}{n}\right) $$ which, compared to your summation, differ at most by $\pm \frac 12$.
All of that leads to the asymptotics
$$\sum_{k=1}^{n-1} \frac{1}{\cosh^2\left(\frac{\pi k}{n}\right)}\sim -\frac 12+\frac{ \tanh (\pi )}{\pi }n-\frac{\pi \tanh (\pi ) \text{sech}^2(\pi )}{n}+O\left(\frac{1}{n^2}\right)$$