Question: $\sum\_{i=1}^{\infty} \frac{(\log n)^2}{n^2}$

I guess it is convergent, so I apply comparsion test for this.

$\frac{log^2}{n^2} < \frac{n^2}{n^2} = 1$ 

So it is bounded by 1 and hence it is convergent. 

Can I do in this way? It kind of make sense, but I have never deal with constant in this case

Usually we do bounding using another term depend on $n$?

Answer: 
You can't do that way as Gerry mentioned. However, you know that $\log^2n\leq\sqrt n$ for $n$ big enough and that's why for big $n$ you have $\frac{(\log n)^2}{n^2}\leq \frac{1}{n\sqrt{n}}$. For the latter series do converge.
Since $\log n\le n-1$ for all $n$, we get $\frac{1}{4}\log n \le \sqrt[4]{n}-1\le \sqrt[4]{n}$ for all $n$. So
$$\frac{\log^2 n}{n^2}\le \frac{16\sqrt{n}}{n^2}=\frac{16}{n\sqrt{n}}$$
and we know that $\sum \frac{16}{n\sqrt{n}}$ converges. So by comparison test, $\sum \frac{\log^2 n}{n^2}$ converges.