Determine whether the infinite series $\sum _{n=1}^{\infty }\frac{1}{n^2+3n+2}$ converges, and if it does, find its sum.

Step-by-step solution

1. Factor the denominator: $n^2+3n+2=(n+1)(n+2)$.
2. Rewrite the general term using partial fractions: $\frac{1}{(n+1)(n+2)}=\frac{1}{n+1}-\frac{1}{n+2}$.
3. Hence the series becomes ${\sum }_{n=1}^{\infty }(\frac{1}{n+1}-\frac{1}{n+2})$.
4. Write the first few terms to see cancellation: $(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\cdots +(\frac{1}{N+1}-\frac{1}{N+2})$.
5. All intermediate terms cancel, leaving $S_N=\frac{1}{2}-\frac{1}{N+2}$.
6. Take the limit as $N\to \infty$: $displaystyleli{m}_{N\to \infty }(\frac{1}{2}-\frac{1}{N+2})=\frac{1}{2}$.
7. Since the limit exists, the series converges and its sum is $\frac{1}{2}$.

Concept tested: Telescoping series

Published 16 October 2026 · How this solution was prepared and checked