Question

If $S_{\infty }=\frac{1}{1\cdot 3\cdot 5}+\frac{1}{3\cdot 5\cdot 7}+\frac{1}{5\cdot 7\cdot 9}+\cdots$ up to infinite term, then the value of $24S_{\infty }$ is

Answer 1

$T_r=\frac{1}{(2r-1)(2r+1)(2r+3)}=\frac{1}{4}[\frac{1}{(2r-1)(2r+1)}-\frac{1}{(2r+1)(2r+3)}]\Rightarrow S_n=\frac{1}{4}\sum _{r=1}^n[\frac{1}{(2r-1)(2r+1)}-\frac{1}{(2r+1)(2r+3)}]\Rightarrow 4S_n=(\frac{1}{1\cdot 3}-\frac{1}{3\cdot 5})+(\frac{1}{3\cdot 5}-\frac{1}{5\cdot 7})+(\frac{1}{5\cdot 7}-\frac{1}{7\cdot 9})⋮+(\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)})\Rightarrow 4S_n=\frac{1}{3}-\frac{1}{(2n+1)(2n+3)}\Rightarrow 4S_{\infty }=\frac{1}{3}\therefore 24S_{\infty }=2$