Question

Find the sum to n terms

$\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\dots \dots$

Answer 1

The given series is

$\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\dots \dots \text{to n terms}$

Let '$a_n$' be nth term of the given series and '$S_n$' be the sum of nth terms of the given series.

$\therefore a_n=\frac{1}{\left(\text{nth term of 1, 2, 3,....}\right)\left(\text{nth term of 2, 3, 4....}\right)}$

$=\frac{1}{[1+(n-1)\times 1][2+(n-1)\times 1]}$

$=\frac{1}{n(n+1)}=\frac{1}{n}+\left(\frac{-1}{n+1}\right)$ [By partial fraction]

Putting n = 1, 2, 3, .... n, we have

$a_1=\frac{1}{1}-\frac{1}{2}$;

$a_2=\frac{1}{2}-\frac{1}{3}$;

$a_3=\frac{1}{3}-\frac{1}{4}$; .....

$a_n=\frac{1}{n}-\frac{1}{n+1}$; .....

Adding vertically, we have:

$S_n=a_1+a_2+a_3+\dots \dots +a_n$

$=\frac{1}{1}-\frac{1}{n+1}=\frac{n+1-1}{n+1}=\frac{n}{n+1}$

$\therefore S_n=\frac{n}{n+1}$