Double alternating sum

Evaluate the series

$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} (-1)^{m+n} \frac{m}{m+n}$$

Solution

We have successively:


\begin{align*}
\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} (-1)^{m+n}  \frac{m}{m+n} &=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} (-1)^{m+n} m \int_{0}^{\infty} e^{-(m+n)x}\, {\rm d}x \\
 &= \int_{0}^{\infty} \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} (-1)^{m+n} m e^{-(m+n)x} \, {\rm d}x\\
 &=\int_{0}^{\infty} \frac{e^x}{\left ( 1+e^x \right )^3}\, {\rm d}x \\
 &\overset{u=e^x}{=\! =\! =\!} \int_{1}^{\infty} \frac{{\rm d}u}{(1+u)^3} \\
 &= \left [ -\frac{1}{2\left ( u+1 \right )^2} \right ]_1^{\infty} \\
 &=\frac{1}{8}
\end{align*}