Is this series convergent? $\sum_{i=1}^{\infty} \frac{(\log n)^2}{n^2}$ $\sum_{i=1}^{\infty} \frac{(\log n)^2}{n^2}$
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 $n$? 
 A: 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.
A: 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?
A: What you did is wrong since you have to compare with the general term of a converging series.
You can try Cauchy's Condensation Test
$$\frac{2^nn^2\log^22}{2^{2n}}=\log^2\frac{n^2}{2^n}$$
and now it's easy to check that $\,\displaystyle{\sum_{n=1}^\infty\frac{n^2}{2^n}}\;$ converges (for example, the n-th root test)
A: Note that, by the integral test
$$ \int_{1}^{\infty} \frac{(\ln(x))^2}{x^2}dx = -{\frac { \left( \ln  \left( x \right)  \right) ^{2}}{x}}-2\,{\frac {
\ln  \left( x \right) }{x}}-\frac{2}{x}
\Big|_{x=1}^{\infty} =2 . $$
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.
A: 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.
A: 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. 
