The given series is $\large\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+..........$
Step 1
We have to find the $n^{th}$ term $t_n$ of the series.
The numerator of all the terms is $1$.
The denominator of each term consists of product of two terms
The first term of the denominator forms the sequence
$1,2,3,..........n$
$\therefore$ The first term in the denominator of $t_n$ is $n$
Step 2
The second term of the denominators form the series
$2,3,4..........(n+1)$
$\therefore$ The second term in the denominator of $t_n$ is $n+1$
$\Rightarrow\:$ The $n^{th}$ term of the series $t_n=\large\frac{1}{n(n+1)}$
Step 3
We can express each term of the series as
$\large\frac{1}{1\times 2}=\frac{1}{1}-\frac{1}{2}$
$\large\frac{1}{2\times 3}=\frac{1}{2}-\frac{1}{3}$
$\large\frac{1}{3\times 4}=\frac{1}{3}-\frac{1}{4}$
.............
$\large\frac{1}{n\times (n+1)}=\frac{1}{n}-\frac{1}{n+1}$
$\therefore\:$ The sum of $n$ terms of the series $S_n$ is
$S_n=\large\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+..........$
$=\bigg(\large\frac{1}{1}-\frac{1}{2}\bigg)+\bigg(\frac{1}{2}-\frac{1}{3}\bigg)+\bigg(\frac{1}{3}-\frac{1}{4}\bigg)+.......\bigg(\frac{1}{n}-\frac{1}{n+1}\bigg)$
$=\large\frac{1}{1}-\frac{1}{n+1}$
$=\large\frac{n+1-1}{n+1}=\frac{n}{n+1}$