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.
No, you can't do it that way. You have shown that the terms are bounded by $1$, but that's also true of the series $1+1+1+\dots$; does that series converge?