[tex]\displaystyle \frac{1}{k(k+1)}= \frac{(k+1)-k}{k(k+1)}= \frac{k+1}{k(k+1)}- \frac{k}{k(k+1)}= \frac{1}{k}- \frac{1}{k+1}. \\ \\ \\ Deci~ \boxed{ \frac{1}{k(k+1)} = \frac{1}{k}- \frac{1}{k+1} }~.~(k \notin \{-1;0 \}).\\ \\ \\ S= \Big ( \frac{1}{1}- \frac{1}{2} \Big)+ \Big( \frac{1}{2} - \frac{1}{3} \Big) + \Big( \frac{1}{3}+ \frac{1}{4} \Big)+...+ \Big( \frac{1}{n}- \frac{1}{n+1} \Big)= \\ \\ \\ =1- \frac{1}{n+1}= \\ \\ \\ = \frac{n}{n+1} ~.[/tex]
