Question

${lim}_{n\to \infty }[\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\cdots \frac{1}{(2n+1)(2n+3)}]$ is equa to

• A. 0
• B. $\frac{1}{2}$
• C. $\frac{1}{9}$
• D. 2

Answer 1

The correct option is B

$\frac{1}{2}$

${lim}_{n\to \infty }[\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\cdots \frac{1}{(2n+1)(2n+3)}]$
Here, $T_n=\frac{1}{(2n+1)(2n+3)}$
$\Rightarrow T_n=\frac{A}{(2n-1)}+\frac{B}{2n+1}$
On equating $A=\frac{1}{2}$ and $B=-\frac{1}{2}$:
$T_n=\frac{1}{2(2n-1)}-\frac{1}{2(2n+1)}$
$\Rightarrow T_1=\frac{1}{2}[1-\frac{1}{3}]$
$\Rightarrow T_2=\frac{1}{2}[\frac{1}{3}-\frac{1}{5}]$
$\Rightarrow T_{n-1}=\frac{1}{2}[\frac{1}{2n-1}-\frac{1}{2n-1}]$
$\Rightarrow T_n=\frac{1}{2}[\frac{1}{2n-1}-\frac{1}{2n+1}]$
$\Rightarrow T_1+T_2+T_3\cdots T_n=\frac{1}{2}[1-\frac{1}{2n+1}]$
$\Rightarrow T_1+T_2+T_3\cdots T_n=\frac{1}{2}\left[\frac{2n}{2n+1}\right]$
$\Rightarrow T_1+T_2+T_3\cdots T_n=\frac{n}{2n+1}$
$\therefore {lim}_{n\to \infty }[\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\cdots \frac{1}{(2n+1)(2n+3)}]$
$={lim}_{n\to \infty }\left[{\sum }_{n=1}^n\frac{1}{(2n+1)(2n+3)}\right]$
$={lim}_{n\to \infty }\left(\frac{n}{2n+1}\right)$
$=li{m}_{n\to \infty }\left(\frac{1}{2+\frac{1}{n}}\right)$
$=\frac{1}{2}$