## University Calculus: Early Transcendentals (3rd Edition)

$1$
We have: $a_n=\dfrac{2n+1}{n^2(n+1)^2}=\dfrac{1}{n^2}-\dfrac{1}{(n+1)^2}$ Here, $s_n=(1-\dfrac{1}{4})+(\dfrac{1}{4}-\dfrac{1}{9})+...[\dfrac{1}{n^2}-\dfrac{1}{(n+1)^2}]=1-\dfrac{1}{(n+1)^2}$ Thus, we have the sum: $\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} [1-\dfrac{1}{(n+1)^2}]=1$