Question

If $S_n=\frac{1}{1\cdot 3\cdot 5}+\frac{1}{3\cdot 5\cdot 7}+\frac{1}{5\cdot 7\cdot 9}\cdots$, then $24S_{\infty }=$

Answer 1

$T_n=\frac{1}{(2n-1)(2n+1)(2n+3)}=\frac{1}{4}[\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)}]$
Which is the form of $\frac{1}{4}(V_n-V_{n+1})$
$\therefore S_n=\frac{1}{4}(V_1-V_{n+1})\Rightarrow S_n=\frac{1}{4}[\frac{1}{3}-\frac{1}{(2n+1)(2n+3)}]$
So when denominator become vary large then fraction will be $\cong 0$
Hence $S_{\infty }=\frac{1}{4}[\frac{1}{3}-0]=\frac{1}{12}$