Answer
converge to $3$
Work Step by Step
The nth partial sums are: $ 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}$
Now, apply limits, we have $\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} 3- \lim\limits_{n \to \infty}[\dfrac{3}{(n+1)^2}]=3$
and when $n \to \infty$, then $s_n \to 3$.
Thus, the series converges to $3$
