Let $\lim_{n\to\infty}\frac{a_n}{n}=0$, the series $\sum\frac{a_n}{n^2}$ converges? Being $\lim_{n\to\infty}\frac{a_n}{n}=0$ then i can suppose that $a_n$ is bounded, otherwise there would not be a limit. Therefore the series $\sum\frac{a_n}{n^2}$ can be rewritten as $\sum{a_n\frac{1}{n^2}}$ and so as $a_n$ is bounded and $\frac{1}{n^2}$ convergent, then $\sum\frac{a_n}{n^2}$ is convergent!
That's nice? Am i wrong?
 A: There is a problem with the part 

Being $\lim_{n\to\infty}\frac{a_n}{n}=0$ then i can suppose that $a_n$ is bounded, otherwise there would not be a limit.

In this context, the limit of $a_n/n$ can be zero when $a_n$ goes to infinity slowly, for example $a_n=\log n$. 
Therefore, your approach only works if the sequence $(a_n)$ is bounded, but it is not guaranteed by the assumptions. 
The problem can be rephrased in the following way: if $\left(\epsilon_n\right)_{n\geqslant 1}  $ is a sequence of real numbers which converges to $0$, then the series $\sum_{n=1}^{ +\infty}\epsilon_n /n$ converges. 


*

*There are examples of sequence $\left(\epsilon_n\right)_{n\geqslant 1}$ converging to $0$ such that the series $\sum_{n=1}^{ +\infty}\epsilon_n /n$ converges, like $\epsilon_n:=1/n$. 

*There are examples of sequence $\left(\epsilon_n\right)_{n\geqslant 1}$ converging to $0$ such that the series $\sum_{n=1}^{ +\infty}\epsilon_n /n$ diverges, like $\epsilon_n:=1/\log n$ (the keyword is Bertrand series). 

A: Hint: Consider
$$
a_n=\frac{n}{\log(n)}
$$
and use the Cauchy Condensation Test to check the convergence of
$$
\sum_{n=2}^\infty\frac{a_n}{n^2}
$$
