Answer
$5$
Work Step by Step
Consider $a_n=\dfrac{40 n}{(2n-1)^2(2n+1)^2}=5[\dfrac{1}{(2n-1)^2}-\dfrac{1}{(2n+1)^2}]$
The nth partial sums are: $ s_n=5(\dfrac{1}{1^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{5^2}+\dfrac{1}{5^2}+.....+\dfrac{1}{(2n-1)^2}-\dfrac{1}{(2n+1)^2})=[5(1-\dfrac{1}{(2n-1)^2})]$
Hence, we have
$\lim\limits_{n \to \infty} s_n=5(\lim\limits_{n \to \infty}1-\lim\limits_{n \to \infty}\dfrac{1}{(2n-1)^2})=5$
