Suppose $\lim\limits_{n \rightarrow \infty } n^2 a_n =1$, then $ \sum\limits _{n=1} ^{\infty} a_n$ is convergent? Or is it divergent? 
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.
 A: A series converges iff its remainder term $\sum\limits_N^\infty a_n \rightarrow 0$ as $N\rightarrow \infty$. Your argument shows that you can, for large enough $N$, bound $|\sum\limits_N^\infty a_n|$ by $\sum\limits_N^\infty \frac{1}{n^2}$. The latter term approaches zero as $N\rightarrow \infty$, since $\sum \frac{1}{n^2}$ converges.
A: The hypothesis means first that, if $n$  is large enough, $a_n>0$, and also that $a_n$ is asymptotically equivalent to $\dfrac1{n^2}$.
Now two series with (eventually positive) equivalent terms both converge or both diverge.
A: Option.
$\lim_{n \rightarrow \infty}n^2a_n=1.$
This Implies for  $n \in \mathbb{Z^+}$ :
$n^2|a_n| \lt M$, positive real number.
$|a_n| \lt M/n^2.$
By comparison test $\sum |a_n|$ is convergent,
hence $\sum a_n$ is convergent.
A: Your argument is correct, more simply we can refer directly to limit comparison test with $$\sum \frac1{n^2}$$
indeed
$$\frac{a_n}{ \frac1{n^2}}=n^2a_n\to 1$$
therefore $\sum a_n$ converges.
It is important to recognize that since, more in general, when we solve a problem we don't need everytime to prove all the results that we have already proved in a general way.
Therefore if you are not requested to use esplicitely the $\epsilon-\delta$ definition a solution by limit comparison test is fine.
