Evaluate the sum $\sum _{n=1}^{100}\frac{1}{n(n+1)}$.

Step-by-step solution

1. Rewrite the general term using partial fractions: $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$.
2. Hence the series becomes a telescoping sum: ${\sum }_{n=1}^{100}\left(\frac{1}{n}-\frac{1}{n+1}\right)$.
3. Write out first few terms: $(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+\cdots +(\frac{1}{100}-\frac{1}{101})$.
4. All intermediate terms cancel, leaving $1-\frac{1}{101}$.
5. Thus the sum equals $1-\frac{1}{101}=\frac{101-1}{101}=\frac{100}{101}$.

Concept tested: Telescoping series

Published 4 October 2026 · How this solution was prepared and checked