jmc

We can rewrite the summand as $$\begin{array}{rl}{\log }_2\left(1+\frac{1}{k}\right){\log }_{k}2{\log }_{k+1}2& =\frac{{\log }_2\left(\frac{k+1}{k}\right)}{{\log }_{2}k{\log }_2(k+1)}\\ & =\frac{{\log }_2(k+1)-{\log }_{2}k}{{\log }_{2}k{\log }_2(k+1)}\\ & =\frac{1}{{\log }_{2}k}-\frac{1}{{\log }_2(k+1)}.\end{array}$$Therefore, the sum telescopes: $$\begin{array}{rl}\sum _{k=2}^{63}{\log }_2\left(1+\frac{1}{k}\right){\log }_{k}2{\log }_{k+1}2& =\left(\frac{1}{{\log }_{2}2}-\frac{1}{{\log }_{2}3}\right)+\left(\frac{1}{{\log }_{2}3}-\frac{1}{{\log }_{2}4}\right)+\cdots +\left(\frac{1}{{\log }_{2}63}-\frac{1}{{\log }_{2}64}\right)\\ & =\frac{1}{{\log }_{2}2}-\frac{1}{{\log }_{2}64}\\ & =1-\frac{1}{6}\\ & =\boxed{\frac{5}{6}}.\end{array}$$