Chapter 12.5, Problem 12E

To determine

To Prove: $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\mathrm{.......}+\frac{1}{n\left(n+1\right)}=\frac{n}{n+1}$ is True for all natural numbers $n$ .

Explanation of Solution

Given information:

Use Mathematical induction to prove the above mentioned formula.

Concept used:

1.By using Mathematical induction technique we need to prove that this formula holds True for all natural numbers. The first step is to prove True for $n=1$ .

2.Second step by using induction hypothesis of mathematical induction we assume for $n=k$ given formula is True.

3.Third step is we need to prove that above formula is True for $n=k+1$ by using formula is True for $n=k$ .

Proof:

First, we need to prove the above formula is True for $n=1$ .

Given formula is

  $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\mathrm{.......}+\frac{1}{n\left(n+1\right)}=\frac{n}{n+1}$

Put $n=1$

We get

Left hand side = $\frac{1}{1\cdot 2}=\frac{1}{2}$

Right hand side = $\frac{1}{1+1}=\frac{1}{2}$

So Left hand side = Right hand side

It True for $n=1$ .

Let us assume it is True for $n=k$

So

  $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\mathrm{.......}+\frac{1}{k\left(k+1\right)}=\frac{k}{k+1}$ is True

We need to prove it is True for $n=k+1$

Left hand side = $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\mathrm{.......}+\frac{1}{k\left(k+1\right)}+\frac{1}{\left(k+1\right)\left(k+1+1\right)}$

  $\Rightarrow$ Left hand side = $\frac{k}{k+1}+\frac{1}{\left(k+1\right)\left(k+2\right)}$

Taking $\frac{1}{k+1}$ common from both sides

We get

  $\Rightarrow$ Left hand side = $\frac{1}{k+1}\left[k+\frac{1}{k+2}\right]$

  $\Rightarrow$ Left hand side = $\frac{1}{k+1}\left[\frac{k\left(k+2\right)+1}{k+2}\right]$

  $\Rightarrow$ Left hand side = $\frac{1}{k+1}\left[\frac{k^2+2k+1}{k+2}\right]$

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

Cancelling $\left(k+1\right)$ in numerator and denominator

We get

  $\Rightarrow$ Left hand side = $\frac{k+1}{k+2}$

Right hand side = $\frac{k+1}{k+1+1}$

  $\Rightarrow$ Right hand side = $\frac{k+1}{k+2}$

Left hand side = Right hand side

And hence the proof by Mathematical induction.