Does $\sum_{n=2}^\infty \frac{1}{(\ln(n))^2}$ converge or diverge?

Question

I've tried, the limit comparison test with several values and have tried finding some values for the direct comparison test but nothing really concrete has come out of it.

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

I don't think the integral test would help, and I'm not really sure how to go about this.

Answer 1

Observe that $x > (\ln x)^2 $ and so

$$ \frac{1}{(\ln x)^2} > \frac{1}{x} $$

for $x>1$. Applying this inequality to your case, you can see that the given numerical series is divergent

Answer 2

Since for $n>1, n>\log n$ then: $n\log{n}>\log^2{n}$ and thus: $$\sum_{n>1}{\frac{1}{\log^2{n}}}\ge\sum_{n>1}{\frac{1}{n\log n}}=\infty$$

Answer 3

One can prove that any sum of the form $$\sum_{k=2}^\infty\frac1{\ln^n{(k)}}$$ diverges where $n\in\mathbb{R}$. Firstly, it can be proven by induction and L'Hôpital's rule that $$\lim_{x\to\infty}\frac{x^n}{e^x}=0$$ for all $n\in\mathbb{R}$. Then by using this fact we must have $$x^n\lt e^x$$ for all $x\gt x_0$ as $x^n$ is increasing and using the definition of a limit. Taking $x=\ln{(t)}$ for some $t\in\mathbb{R}$ we get $$\ln^n{(t)}\lt t$$ for all $t\gt t_0$ where $t_0=e^{x_0}$. Thus for all $k\ge \lceil t_0\rceil$ we have that $$\ln^n{(k)}\lt k$$ and hence $$\frac1{\ln^n{(k)}}\gt \frac1k$$ So the sum $$\sum_{k=\lceil t_0\rceil}^\infty \frac1{\ln^n{(k)}}\gt\sum_{k=\lceil t_0\rceil}^\infty\frac1k\to\infty$$ Hence $$\sum_{k=2}^\infty\frac1{\ln^n{(k)}}\gt\sum_{k=\lceil t_0\rceil}^\infty \frac1{\ln^n{(k)}}\to\infty$$