If $\displaystyle\lim_{n\to\infty}(n^2a_n)$ exist then show that $\sum a_n$ converges. If $\displaystyle\lim_{n\to\infty}(n^2a_n)$ exist then show that $\sum a_n$ converges.
my try:We have to somehow use the fact that $\sum\frac1{n^2}$ converges but don't know how.
 A: Let $L = \lim_{n\to\infty} n^2 |a_n|$.  We know this exists because $\lim_{n\to\infty} n^2 a_n$ exists.  Also $L \geq 0$.
Let $b_n = \frac{1}{n^2}$.  We know that $\sum_{n=1}^\infty b_n$ converges.  Moreover,
$$
    \lim_{n\to\infty} \frac{|a_n|}{b_n} = \lim_{n\to\infty} \frac{|a_n|}{1/n^2}
    = \lim_{n\to\infty} n^2 |a_n| = L
$$
So by the Limit Comparison Test $\sum_{n=1}^\infty |a_n|$ converges.
(The usual Limit Comparison Test requires $L>0$.  But it can be extended to: if $\frac{a_n}{b_n} \to 0$, and $\sum b_n$ converges, then $\sum a_n$ converges.)
A: Every convergent sequence is bounded. Hence,
$$
\sum|a_n|=\sum\frac{|n^2a_n|}{n^2}\le M\sum\frac1{n^2},
$$
where $M>0$ is such that $|n^2a_n|\le M$ for each $n\ge1$. Every absolutely convergent sequence converges. We conclude that $\sum a_n$ converges.
A: Here you want to use the limit comparison test https://en.wikipedia.org/wiki/Limit_comparison_test. This is a pretty straightforward example and yes you want to use that series $\sum \frac{1}{n^2}$ converges
