Telescoping series — find sum of 1/(1·2) + 1/(2·3) + ... + 1/(n(n+1))

Question

Find the sum: $S = \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4} + \cdots + \dfrac{1}{n(n+1)}$

(NCERT Class 11 — Sequences and Series)

Solution — Step by Step

Why This Works

This is called a telescoping series because, like a collapsible telescope, the intermediate terms fold into each other and vanish. The trick is the partial fraction decomposition that writes each term as a difference of two consecutive terms.

As $n \to \infty$, $S \to 1$. So the infinite series $\sum_{k=1}^{\infty} \frac{1}{k(k+1)} = 1$.

Verification for small $n$: when $n = 1$, $S = \frac{1}{2} = \frac{1}{2}$ (and $\frac{1}{1+1} = \frac{1}{2}$). When $n = 2$, $S = \frac{1}{2} + \frac{1}{6} = \frac{2}{3}$ (and $\frac{2}{3}$). Checks out.

Alternative Method

You can prove the formula by mathematical induction. Base case: $n = 1$ gives $\frac{1}{2} = \frac{1}{2}$. Inductive step: assume $S_n = \frac{n}{n+1}$, then $S_{n+1} = S_n + \frac{1}{(n+1)(n+2)} = \frac{n}{n+1} + \frac{1}{(n+1)(n+2)} = \frac{n(n+2) + 1}{(n+1)(n+2)} = \frac{n^2 + 2n + 1}{(n+1)(n+2)} = \frac{(n+1)^2}{(n+1)(n+2)} = \frac{n+1}{n+2}$. Done.

Common Mistake