Man finde eine explizite Darstellung für die Partialsummen der Reihe
und berechne damit - wenn möglich - die Summe.
(Anleitung: Man beachte, dass 1n(n+1)=1n−1n+1{\displaystyle {\frac {1}{n(n+1)}}={\frac {1}{n}}-{\frac {1}{n+1}}} gilt.)
∑k=1n1k(k+1)=∑k=1n(1k−1k+1)=11−12+12−13+13−14+14⋯−1n−1+1n−1−1n+1n−1n+1=1−1n+1{\displaystyle {\begin{aligned}\sum _{k=1}^{n}{\frac {1}{k(k+1)}}&=\sum _{k=1}^{n}\left({\frac {1}{k}}-{\frac {1}{k+1}}\right)\\&={\frac {1}{1}}-{\frac {1}{2}}+{\frac {1}{2}}-{\frac {1}{3}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{4}}\dots -{\frac {1}{n-1}}+{\frac {1}{n-1}}-{\frac {1}{n}}+{\frac {1}{n}}-{\frac {1}{n+1}}\\&=1-{\frac {1}{n+1}}\end{aligned}}}
limn→∞(1−1n+1)=1{\displaystyle \lim _{n\to \infty }\left(1-{\frac {1}{n+1}}\right)=1}