Find the sum to n terms of the series

The given series is $\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\ldots$

$n^{\text {th }}$ term, $a_{n}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ (By partial fractions)

$a_{1}=\frac{1}{1}-\frac{1}{2}$

$a_{2}=\frac{1}{2}-\frac{1}{3}$

$a_{3}=\frac{1}{3}-\frac{1}{4} \ldots$

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

Adding the above terms column wise, we obtain

$a_{1}+a_{2}+\ldots+a_{n}=\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots \frac{1}{n}\right]-\left[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots \frac{1}{n+1}\right]$

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