Question

If $S_n=\frac{1}{1\times 3}+\frac{1}{3\times 5}+\frac{1}{5\times 7}+\cdots n\text{terms}$, then $S_{\infty }$ is

• A. $3$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

The correct option is C $\frac{1}{2}$

Let $T_r$ be the $r^{th}$ term of the given series. Then,
$T_r=\frac{1}{(2r-1)(2r+1)},r=1,2,3,\dots \dots ,n$
$=\frac{1}{2}(\frac{1}{2r-1}-\frac{1}{2r+1})$
Hence, the required sum is
$S_n=T_1+T_2+T_3+\cdots \cdots +T_n=\frac{1}{2}(\frac{1}{1}-\frac{1}{3})+\frac{1}{2}(\frac{1}{3}-\frac{1}{5})+\frac{1}{2}(\frac{1}{5}-\frac{1}{7})+⋮⋮+⋮⋮+\frac{1}{2}(\frac{1}{2n-1}-\frac{1}{2n+1})\Rightarrow S_n=\frac{1}{2}(1-\frac{1}{2n+1})$
For very large value of $n$, $\frac{1}{2n+1}\cong 0$
$\therefore S_{\infty }=\frac{1}{2}(1-0)=\frac{1}{2}$