Question

Sum the following series to n terms is $\sum _{r=1}^n\frac{1}{4r^2-1}$ be $S_n=\frac{1}{k}[1-\frac{1}{kn+1}]$

$S_{\infty }=\frac{1}{k}$. Find $k$ ?

Answer 1

$T_n=\frac{1}{4n^2-1}=\frac{1}{(2n+1)(2n-1)}=\frac{1}{2}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]$

$S_n=\frac{1}{2}[\frac{1}{2n-1}-\frac{1}{2n+1}+\frac{1}{2n-3}-\frac{1}{2n-1}+\frac{1}{2n-5}-\frac{1}{2n-3}+...........\frac{1}{3}-\frac{1}{5}+1-\frac{1}{3}]$

$S_n=\frac{1}{2}[1-\frac{1}{2n+1}]$

$S_{\infty }=\frac{1}{2}$
$\Rightarrow k=2$