Chapter 11.7, Problem 14PPS

To determine

To prove that for each natural number

  $\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\mathrm{....}+\frac{1}{n\left(n+1\right)}=\frac{n}{n+1}$

Explanation of Solution

Given:

  $\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\mathrm{....}+\frac{1}{n\left(n+1\right)}=\frac{n}{n+1}$

Concept Used:

An identity can be proved using mathematical induction in two steps:

1. First prove the identity is true for $n=1$ .
2. Prove that if it is true for $n=k$ then it is also true for $n=k+1$ .

Calculation:

In order to prove

  $\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\mathrm{....}+\frac{1}{n\left(n+1\right)}=\frac{n}{n+1}$

Use mathematical induction.

First, let’s proof that the identity is true for $n=1$ .

For, the left hand side of given identity is:

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

And right hand side is:

  $\begin{array}{c}\frac{n}{n+1}=\frac{1}{1+1}\left(\text{Putting }n=1\right)\\ =\frac{1}{2}\\ =\text{Left side}\end{array}$

Thus, the given identity is true for $n=1$ .

Now, let it is true for $n=k;\text{ for some }k>1$ . Thus,

  $\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\mathrm{....}+\frac{1}{k\left(k+1\right)}=\frac{k}{k+1}$

Now, let’s show that it is also true for $n=k+1$ , i.e. prove

  $\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\mathrm{....}+\frac{1}{k\left(k+1\right)}+\frac{1}{\left(k+1\right)\left(\left(k+1\right)+1\right)}=\frac{k+1}{\left(k+1\right)+1}$

To prove above simplify the left side as,

  $\begin{array}{c}\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\mathrm{....}+\frac{1}{k\left(k+1\right)}+\frac{1}{\left(k+1\right)\left(\left(k+1\right)+1\right)}=\frac{k}{k+1}+\frac{1}{\left(k+1\right)\left(k+2\right)}\\ \because \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\mathrm{....}+\frac{1}{k\left(k+1\right)}=\frac{k}{k+1}\\ \frac{k}{k+1}+\frac{1}{\left(k+1\right)\left(k+2\right)}=\frac{k\left(k+2\right)+1}{\left(k+1\right)\left(k+2\right)}\\ =\frac{k^2+2k+1}{\left(k+1\right)\left(k+2\right)}\\ =\frac{{\left(k+1\right)}^2}{\left(k+1\right)\left(k+2\right)}\because k^2+2k+1={\left(k+1\right)}^2\\ =\frac{\left(k+1\right)}{\left(k+2\right)}\\ =\frac{k+1}{k+1+1}\\ =\text{Right side}\end{array}$

Thus, the identity is also true for $n=k+1$ when it is true for $n=k$ .

Thus, by mathematical induction it is proved that

  $\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\mathrm{....}+\frac{1}{n\left(n+1\right)}=\frac{n}{n+1}$