If x,y,z are three consecutive positive integers, then $\frac{1}{2}(\log \mathrm{x}+\log \mathrm{z})+\frac{1}{2\mathrm{xz}+1}+\frac{1}{3.(2\mathrm{xz}+1)}+\dots \infty =$

$\begin{array}{rcl}\mathrm{x}& =& \mathrm{y}-1, \mathrm{z}=\mathrm{y}+1\\ \mathrm{xz}& =& {\mathrm{y}}^2-\mathrm{y}\end{array}$

Given eqn

$\begin{array}{rcl} & =& \frac{1}{2} \log \left(\mathrm{xz}\right)+\frac{1}{2}\log \left|\frac{1+2\mathrm{xz}+1}{1-2\mathrm{xz}-1}\right|\\ & & \\ & =& \frac{1}{2} \log \left(\mathrm{xz}\right)+\frac{1}{2}\log \left|\frac{2(1+\mathrm{xz})}{-2\mathrm{xz}}\right|\\ & =& \frac{1}{2} \log \left|\mathrm{xz} \left(\frac{1+\mathrm{xz}}{\mathrm{xz}}\right)\right|\\ & =& \frac{1}{2} \left|\log \right| \left|{\mathrm{y}}^2\right|=\log \mathrm{y}\end{array}$