To sum 1/[k(k+d)]: decompose as $\frac{1}{d}$[1/k - $\frac{1}{k+d}$]. Examples: 1/[k(k+1)] = 1/k - $\frac{1}{k+1}$. 1/[k(k+2)] = $\frac{1}{2}$[1/k - $\frac{1}{k+2}$]. 1/[k(k+3)] = $\frac{1}{3}$[1/k - $\frac{1}{k+3}$]. For products of 3 consecutive: 1/[k(k+1)(k+2)] = $\frac{1}{2}$[$\frac{1}{k(k+1}$) - $\frac{1}{(k+1}$(k+2))], which telescopes after one more layer.