Series convergence ques

Question

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 :)

Answer 1

$\sum_{n=1}^{\infty}\frac{(ln(n))^2}{n^2}< \sum_{n=1}^{\infty}\frac{ln(n) \sqrt{n}}{n^2}<\sum_{n=1}^{\infty}\frac{ln(n)}{n^{\frac{3}{2}}}<\sum_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}$

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.

Answer 2

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 $f''(2) = \sum_{n=1}^{\infty} \frac{(\ln(n))^{2}}{n^2}$.

Answer 3

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}} $.