## Thomas' Calculus 13th Edition

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.