Question

The sum of the series $\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\cdots$ upto $n$ terms is $S_n.$ Then $4S_{\infty }$ equals

Answer 1

$T_1=\frac{1}{3}=\frac{1}{2^2-1}{T}_2=\frac{1}{15}=\frac{1}{4^2-1}{T}_3=\frac{1}{35}=\frac{1}{6^2-1}{T}_4=\frac{1}{63}=\frac{1}{8^2-1}$
$\therefore T_n=\frac{1}{(2n{)}^2-1}=\frac{1}{(2n+1)(2n-1)}\Rightarrow T_n=\frac{1}{2}[\frac{1}{(2n-1)}-\frac{1}{(2n+1)}]$

So, $S_n=\sum T_n$
$=\frac{1}{2}[\frac{1}{1}-\frac{1}{3}]+\frac{1}{2}[\frac{1}{3}-\frac{1}{5}]+\frac{1}{2}[\frac{1}{5}-\frac{1}{7}]$
$⋮$ $⋮$
$+\frac{1}{2}[\frac{1}{(2n-1)}-\frac{1}{(2n+1)}]$

$S_n=\frac{1}{2}[1-\frac{1}{(2n+1)}]$
For very large value of denominator, value of the fraction will be $\cong 0$
$\therefore S_{\infty }=\frac{1}{2}$