Answer
$1$
Work Step by Step
Consider $a_n=\dfrac{2n+1}{n^2(n+1)^2}=[\dfrac{1}{n^2}-\dfrac{1}{(n+1)^2}]$
The Nth partial sums are: $ s_n=(1-\dfrac{1}{4})+(\dfrac{1}{4}-\dfrac{1}{9})+...[\dfrac{1}{n^2}-\dfrac{1}{(n+1)^2}]=1-\dfrac{1}{(n+1)^2}$
Hence, we have
$\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} (1)-\lim\limits_{n \to \infty}\dfrac{1}{(n+1)^2}=1-0=1$
