Let $u(x,y)$ be the solution of the Laplace equation $\Delta u=0$ on the unit square $(0,1)\times (0,1)$ with boundary condition:
$$ u(x,1)=1, u(x,0)=0, u(0,y)=0, u(1,y)=0$$
The series solution is $$\sum\_{n=1}^\infty \frac{4\sin((2n-1)\pi x)\mbox{sinh}((2n-1)\pi y)}{(2n-1)\pi\mbox{sinh}((2n-1)\pi)}$$
The value of $u(1/2,1/2)$ is then (use identity for hyperbolic sine) the series: $$\sum\_{n=1}^\infty \frac{(-1)^{n+1}2}{(2n-1)\pi\cosh((2n-1)\pi/2)}$$
Numerical evidence suggests that this series sums to 1/4.


But this is not obvious at all. Does anyone know how to sum this series? Or use other methods to get this answer?


This problem was posted in [math.stackexchange](https://math.stackexchange.com/questions/358957/the-series-sum-n-1-infty-frac-1n12n-1-mboxsech-left-frac2n). 



A series question related to solution of Laplace equation
Let $v,w,z$ be the functions obtained from $u$ by composing with a rotation of angle $\frac\pi4,\frac\pi2,\frac{3\pi}4$ about $(\frac12,\frac12)$. The sum $u+v+w+z$ is a harmonic function, whose value at the boundary is constant equal to $1$. Thus $u+v+w+z\equiv1$. This gives you $4u\left(\frac12,\frac12\right)=1$.


