Answer to Question #198619 in Calculus for Moe

Question

3) Prove by induction that for all natural numbers n:

(c) 1/(1)(2) + 1/(2)(3) + · · · +1/(n)(n + 1) = n/(n + 1)

Answer 1

Let $P(n)$ be the statement that $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+…+\frac{1}{n\cdot(n+1)}=\frac{n}{n+1}$ .

1) For $n=1$ we have that $\frac{1}{1\cdot 2}=\frac{1}{2}=\frac{1}{1+1}$ . So, $P(1)$ is true.

2) Suppose that $P(k)$ holds, that is, $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+…+\frac{1}{k\cdot(k+1)}=\frac{k}{k+1}$ .

$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+…+\frac{1}{k\cdot(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k(k+2)+1}{(k+1)(k+2)}=\frac{k^2+2k+1}{(k+1)(k+2)}=\frac{(k+1)^2}{(k+1)(k+2)}=\frac{k+1}{k+2}$

We have that $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+…+\frac{1}{(k+1)\cdot(k+2)}=\frac{k+1}{(k+1)+1}$ .

Therefore, $P(k+1)$ holds.

By the Principle of Mathematical Induction, for all $n\in\mathbb{N}:\$ $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+…+\frac{1}{n\cdot(n+1)}=\frac{n}{n+1}$