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.
Note that, by the [integral test](https://math.stackexchange.com/questions/357954/does-the-following-series-converge)

$$ \int\_{1}^{\infty} \frac{(\ln(x))^2}{x^2}dx = -{\frac { \left( \ln \left( x \right) \right) ^{2}}{x}}-2\,{\frac {
\ln \left( x \right) }{x}}-\frac{2}{x}
\Big|\_{x=1}^{\infty} =2 . $$

the series converges. 

**Main result:** Suppose $f$ is continuous, positive, decreasing function on $[1,\infty)$ and let $a\_n=f(n)$. Then 

$(a)$ if $\int\_{1}^{\infty}f(x) dx$ is convergent, then $\sum\_{n=1}^{\infty} a\_n $ is convergent. 

$(b)$ if $\int\_{1}^{\infty}f(x) dx$ is divergent, then $\sum\_{n=1}^{\infty} a\_n $ is divergent.