Find the sum of the series:

Solution:

In the given question we need to find the sum of the series.

For that, first, we need to find the $\mathrm{n}^{\text {th }}$ term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is $\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+. .$ to $\mathrm{n}$ terms.

The series can be written as, $\frac{1}{1 \times 2}, \frac{1}{2 \times 3}, \frac{1}{3 \times 4}, \ldots, \frac{1}{n \times(n+1)}$.

So, $\mathrm{n}^{\text {th }}$ term of the series,

$a_{n}=\frac{1}{n(n+1)}$

By the method of partial fractions, we can factorize the above term.

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

$a_{1}=1-\frac{1}{2} \rightarrow(1)$

$a_{2}=\frac{1}{2}-\frac{1}{3} \rightarrow(2)$

$a_{n-1}=\frac{1}{n-1}-\frac{1}{n} \rightarrow(n-1)^{\text {th }}$ equation

$\mathrm{a}_{\mathrm{n}}=\frac{1}{\mathrm{n}(\mathrm{n}+1)}=\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1} \rightarrow \mathrm{n}^{\text {th }}$ equation

Now, we need to find the sum of this series, Sn.

This can be found out by adding the equation (1), (2)...up to $\mathrm{n}^{\text {th }}$ term.

$\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{n}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{n}}$

$\mathrm{S}_{\mathrm{n}}=1-\frac{1}{\mathrm{n}+1}=\frac{\mathrm{n}}{\mathrm{n}+1}$

So, Sum of the series, $\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{\mathrm{n}+1}$