For all $n \ge 1$, prove that $\dfrac{1}{1.2} + \dfrac{1}{2.3} + \dfrac{1}{3.4} + \ldots + \dfrac{1}{n(n+1)} = \dfrac{n}{n+1}$.

Question

Accepted Answer

Let $P(n)$ be the statement
$$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\ldots+\frac{1}{n(n+1)}=\frac{n}{n+1}.$$

**Base case:** For $n=1$,
$$	ext{LHS}=\frac{1}{1\cdot2}=\frac12,\qquad 	ext{RHS}=\frac{1}{1+1}=\frac12.$$
So $P(1)$ is true.

**Inductive step:** Assume $P(k)$ is true for some $k\ge1$:
$$\frac{1}{1\cdot2}+\ldots+\frac{1}{k(k+1)}=\frac{k}{k+1}. \qquad 	ext{(Induction Hypothesis)}$$

We must show
$$\frac{1}{1\cdot2}+\ldots+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k+1}{k+2}.$$

Using the induction hypothesis on the LHS:
$$\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k(k+2)+1}{(k+1)(k+2)}=\frac{k^2+2k+1}{(k+1)(k+2)}=\frac{(k+1)^2}{(k+1)(k+2)}=\frac{k+1}{k+2}.$$

This is exactly $P(k+1)$.

> [!ANSWER] Since $P(1)$ is true and $P(k)\Rightarrow P(k+1)$ for every $k\ge1$, by PMI, $\dfrac{1}{1\cdot2}+\ldots+\dfrac{1}{n(n+1)}=\dfrac{n}{n+1}$ for all $n\ge1$.

Q.For all $n \ge 1$ , prove that $\dfrac{1}{1.2} + \dfrac{1}{2.3} + \dfrac{1}{3.4} + \ldots + \dfrac{1}{n(n+1)} = \dfrac{n}{n+1}$ .

Concept understanding — Induction for Telescoping Sums

Q.For all n≥1n \ge 1n≥1, prove that 11.2+12.3+13.4+…+1n(n+1)=nn+1\dfrac{1}{1.2} + \dfrac{1}{2.3} + \dfrac{1}{3.4} + \ldots + \dfrac{1}{n(n+1)} = \dfrac{n}{n+1}1.21​+2.31​+3.41​+…+n(n+1)1​=n+1n​.

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Q.For all $n \ge 1$ , prove that $\dfrac{1}{1.2} + \dfrac{1}{2.3} + \dfrac{1}{3.4} + \ldots + \dfrac{1}{n(n+1)} = \dfrac{n}{n+1}$ .

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