Sum of the series $\frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}\right)-\frac{1}{4}\left(\frac{1}{2^2}+\frac{1}{3^2}\right)+\frac{1}{6}\left(\frac{1}{2^3}+\frac{1}{3^3}\right)\dots \dots$is

$\begin{array}{rcl} & =& \frac{1}{2} \left[\frac{1}{2}-\frac{{\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^2}{2}+\frac{{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^3}{3}-......\right]\\ & & +\frac{1}{2}\left(\frac{1}{3}-\frac{{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}^2}{2}+\frac{{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}^3}{3}-.....\right) \\ & =& \frac{1}{2} \left[\log \left(1+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\right]+\frac{1}{2}\left[\log \left(1+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)\right]\\ & =& \frac{1}{2} \left[\log \frac{3}{2} \frac{4}{3}\right]=\frac{1}{2} \log 2\end{array}$