Series convergence ques I was just solving another question regarding this topic and just wanna check if I am doing it right.
So the question is whether the given series converges or not, and justify.
$$\sum_{n = 1}^{\infty} \frac{(\ln{n})^2}{ n^2}$$
I have tried using ratio and comparison test, but that leads me nowhere.
For comparison test I compared the term with $\frac{1}{n^2}$. Am I doing that wrong?
Thank you so much :)
 A: Recall that $f(s) = \sum_{n=1}^{\infty} n^{-s}$ converges when $s>1$.
Then we observe that $f''(s)= \sum_{n=1}^{\infty} \frac{(\ln(n))^{2}}{n^s}$, with the same region of convergence.
Hence $f''(2) = \sum_{n=1}^{\infty} \frac{(\ln(n))^{2}}{n^2}$.
A: Here's a nice way
to show that,
for any $ a > 0$,
$\dfrac{\ln(x)}{x^a}
\to 0$
as
$x \to \infty$.
It actually shows that
$\dfrac{\ln(x)}{x^a}
\lt\dfrac{2}{ax^{a/2}}
$
for $x \gt 1$.
$\begin{array}\\
\ln(x)
&=\int_1^x \dfrac{dt}{t}\\
&<\int_1^x \dfrac{dt}{t^{1-c}}
\qquad\text{since } t^{1-c} < t
\text{ for }0 < c \\
&=\int_1^x t^{c-1}dt\\
&=\dfrac{t^c}{c}|_1^x\\
&=\dfrac{x^c-1}{c}\\
&<\dfrac{x^c}{c}\\
\end{array}
$
Let $c = \frac{a}{2}$,
so that
$\ln(x)
\lt \dfrac{x^{a/2}}{a/2}
= \dfrac{2x^{a/2}}{a}
$.
Dividing by $x^a$,
$\dfrac{\ln(x)}{x^a}
\lt\dfrac{2}{ax^{a/2}}
$.
A: $\sum_{n=1}^{\infty}\frac{(ln(n))^2}{n^2}< \sum_{n=1}^{\infty}\frac{ln(n) \sqrt{n}}{n^2}<\sum_{n=1}^{\infty}\frac{ln(n)}{n^{\frac{3}{2}}}<\sum_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}$ 
This is a p-series with p>1 so it converges. Thus $\sum_{n=1}^{\infty}\frac{(ln(n))^2}{n^2}$ converges as well.
