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

The series converges to $3$.
Here, $s_n=(\dfrac{3}{1}-\dfrac{3}{4}) +(\dfrac{3}{4}-\dfrac{3}{9})+.....+(\dfrac{3}{n^2}-\dfrac{3}{(n+1)^2})=3-\dfrac{3}{(n+1)^2}$ Thus, we have $\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} 3- \lim\limits_{n \to \infty}\dfrac{3}{(n+1)^2}=3$ Now, $s_n \to 3$ as $n \to \infty$, so, the given series will converge to $3$.