Proving $\frac{200}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)\cosh\left(\frac{\pi}{2}(2n+1)\right)}=25$ I solve a partial differential equation (Laplace equation) with specific boundary conditions and I finally found the answer:
$$U(x,y)=\frac{400}{\pi}\sum_{n=0}^{\infty}\frac{\sin\left((2n+1)\pi x\right)\sinh\left((2n+1)\pi y\right)}{(2n+1)\sinh\left((2n+1)\pi\right)}$$
If I put $x=\frac{1}{2}$ and $y=\frac{1}{2}$, then the equation becomes as follow:
$$U\left(\frac{1}{2},\frac{1}{2}\right)=\frac{200}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)\cosh\left(\frac{\pi}{2}(2n+1)\right)}$$
At the end we can draw the answer by MS. Excel software and see that the $U\left(\frac{1}{2},\frac{1}{2}\right)= 25$.
Honestly, my professor asked me to prove $U\left(\frac{1}{2},\frac{1}{2}\right)= 25$. However, it is not included in my course and I do not have any idea.
Could you help me? Thanks.
 A: We will evaluate
$$S=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)\cosh\left(\frac{\pi}{2}(2n+1)\right)}$$
Since the series is alternating, use $f(z)=\pi\csc(\pi z)$ and we have $$\oint\frac{\pi \csc(\pi z)}{(2z+1)\cosh\left(\frac{\pi}{2}(2z+1)\right)}\,dz$$
The poles are at $z=-\frac{1}{2}, \;\ z=n, \;\ z=\frac{(2n+1)\,i}{2}-\frac{1}{2}$, so
$$\begin{align}&\text{Res}\left(f(z)\,;\,z=-\frac{1}{2}\right)=-\frac{\pi}{2}\\
&\text{Res}(f(z)\,;\,z=n)=\frac{(-1)^{n}}{(2n+1)\cosh\left(\frac{\pi}{2}(2n+1)\right)}\\
&\text{Res}\left(f(z)\,;\,z=\frac{(2n+1)i}{2}-\frac{1}{2}\right)=\frac{(-1)^{n}}{(2n+1)\cosh\left(\frac{\pi}{2}(2n+1)\right)}\end{align}$$
Hence:
$$\oint\frac{\pi \csc(\pi z)}{(2z+1)\cosh(\frac{\pi}{2}(2z+1))}dz=-\frac{\pi}{2}+4\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)\cosh\left(\frac{\pi}{2}(2n+1)\right)}$$
As $N\to \infty$, then the integral on the left tends to $0$. So, we have $0=-\frac{\pi}{2}+4S$. Solving for $S$ gives $\,\dfrac{\pi}{8}$. Hence
$$\frac{200}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)\cosh\left(\frac{\pi}{2}(2n+1)\right)}=25$$
