Answer
Converges
Work Step by Step
As we are given that $\lim\limits_{n \to \infty} n^2 a_n=0$
Consider, there will exist a natural number $N$ such that $0 \lt n^2 a_n \leq 1$; $n \geq N$
Thus, we have $\Sigma_{n=N}^\infty a_n \leq \Sigma_{n=N}^\infty \dfrac{1}{n^2}$
By the Limit comparison test, $\Sigma_{n=N}^\infty a_n $ converges.
Hence, the given series $\Sigma _{n=1}^\infty a_n$ converges.
