Prove that if lim sup$_{n} n^{2}a_{n} = 0 \Rightarrow \sum_{k = 1}^{\infty} a_{k}$ converges. Also assume that the sequence is positive.
lim sup$_{n} n^{2}a_{n} = 0$ means that for every $\epsilon$, there exists an $N \in \mathbb{N}$ such that $n > N$ implies ... then I'm kind of lost after this part. I know that $s = sup A$ iff for every $\epsilon > 0$, there exists an $a \in A$ such that $a > s - \epsilon$. But I'm not sure how I can apply that to this.
There is some $N_0$ such that $n > N_0 \Rightarrow n^2a_n < 1 \Rightarrow a_n < \dfrac{1}{n^2} \Rightarrow \displaystyle \sum_{n=N_0}^\infty a_n \leq \displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2} \leq \dfrac{\pi^2}{6}$