Question

Prove the following by using principle of mathematical induction for all $n\in N$ :
$1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+.......+\frac{1}{(1+2+3+\dots n)}=\frac{2n}{(n+1)}.$

Answer 1

Let the given statement be $P\left(n\right)$, i.e.

$P\left(n\right):1+\frac{1}{1+2}+\frac{1}{1+2+3}+.....+\frac{1}{1+2+3+....n}=\frac{2n}{n+1}$

For $n=1$, we have

$P\left(1\right):1=\frac{2.1}{1+1}=\frac{2}{2}=1$, which is true.

Let $P\left(k\right)$ be true for some positive integer $k$ i.e,

$1+\frac{1}{1+2}+....+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+....+k}=\frac{2k}{k+1}$ ..........(i)

We shall now prove that $P(k+1)$ is true.

Consider

$1+\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+....+k}+\frac{1}{1+2+3+....+k+(k+1)}$

$=(1+\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+....k})+\frac{1}{1+2+3+....+k+(k+1)}$

$=\frac{2k}{k+1}+\frac{1}{1+2+3+...+k+(k+1)}$ [ Using (i)]

$=\frac{2k}{k+1}+\frac{1}{\left(\frac{(k+1)(k+1+1)}{2}\right)}$ $[1+2+3+....+n=\frac{n(n+1)}{2}]$

$=\frac{2k}{(k+1)}+\frac{2}{(k+1)(k+2)}$

$=\frac{2}{(k+1)}(k+\frac{1}{k+2})$

$=\frac{2}{(k+1)}\left(\frac{k^2+2k+1}{k+2}\right)$

$=\frac{2}{(k+1)}\left[\frac{(k+1{)}^2}{k+2}\right]$

$=\frac{2(k+1)}{(k+2)}$

Thus, $P(k+1)$ is true whenever $P\left(k\right)$ is true.

Hence, by the principle of mathematical induction, statement $P\left(n\right)$ is true for all natural numbers i.e. $N$.