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 $\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4} \ldots$ to $\mathrm{n}$ terms.

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

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

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

$a_{n}=\frac{1}{(2 n-1)(2 n+1)}=\frac{A}{2 n-1}+\frac{B}{2 n+1}$

$1=A(2 n-1)+B(2 n+1)$

On equating the like term on RHS and LHS,

$2 \mathrm{~A}+2 \mathrm{~B}=0 \rightarrow(\mathrm{a})$

$\mathrm{A}+\mathrm{B}=1 \rightarrow(\mathrm{b})$

On solving, we will get; $A=\frac{1}{2} ; B=-\frac{1}{2}$

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

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

$a_{2}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right) \rightarrow(2)$

$a_{n-1}=\frac{1}{2}\left(\frac{1}{2 n-3}-\frac{1}{2 n-1}\right) \rightarrow(n-1)^{\text {th }}$ equation

$a_{n}=\frac{1}{2}\left(\frac{1}{2 n-1}-\frac{1}{2 n+1}\right) \rightarrow n^{\text {th equation }}$

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.

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

$\mathrm{S}_{\mathrm{n}}=\frac{1}{2}\left[1-\frac{1}{2 \mathrm{n}+1}\right]=\frac{\mathrm{n}}{2 \mathrm{n}+1}$

So, Sum of the series, $\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2 \mathrm{n}+1}$