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: 
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.
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).