Answer
Convergent
Work Step by Step
Given: $\lim\limits_{n \to \infty} n^2 a_n=0$
Let us consider that there exists a natural number $N$ such that $0 \lt n^2 a_n \leq 1$ for all $n \geq N$
This implies that when we compare it with: $\Sigma_{n=N}^\infty a_n \leq \Sigma_{n=N}^\infty \dfrac{1}{n^2}$, we find that the series $\Sigma_{n=N}^\infty a_n $ converges by the Limit comparison test.
Therefore, the series $\Sigma _{n=1}^\infty a_n$ converges.
