Answer
$3$
Work Step by Step
We have: $a_n=\dfrac{6}{(2n-1)(2n+1)}=3(\dfrac{1}{2n-1}-\dfrac{1}{2n+1})$
Here, $ s_n=3(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}+.....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1})=3(1-\dfrac{1}{2n+1})$
Thus, we have the sum: $\lim\limits_{n \to \infty} s_n=3(1-\dfrac{1}{2n+1})=3$
