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: 
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?
Let $\displaystyle a\_n = \frac{\log^2 n}{n^2}$.
Then
$$\lim\_{n\to\infty} n\left(\frac{a\_{n+1}}{a\_n}-1\right) = -2 < -1.$$
The series converges by [Raabe's test](http://en.wikipedia.org/wiki/Ratio_test#Raabe.27s_test).