Suppose $\lim_{n \rightarrow \infty } n^2 a_n =1$ then $ \sum _{n=1} ^{\infty} a_n$ is convergent or divergent?
We use the definition of the limit, then for $\varepsilon =1$ there must exist an $n_0\geq n$ such that $$ |n^2 a_n -1|<1$$ so we can say that: $$ 0 < n^2 a_n < 2 $$ We divide by: $$ 0 < a_n < \frac{2}{n^2} $$ We now know if we take the infinite sum we get: $$ 0 < \sum _{n=1} ^{\infty} a_n < \sum _{n=1} ^{\infty} \frac{2}{n^2} $$ We know that the right side converges, so the sum is bounded, but how would I prove that is converges? Or can we find a counterexample that diverges, bur stays within these bounds.