Let $a,b\in\mathbb{R}$. I want to investigate the convergence/divergence of the sequence $(a_n)$ defined by $a_n=1/(n^a\ln^b(n))$ for $n\geq 2$.
Using WolframAlpha, I suspect that this sequence converges to $0$ when $a>0$ and $b\in \mathbb{R}$, while it tends to infinity when $a<0$ and $b\in \mathbb{R}$.
I will try proving the convergence part. We start with the case $a>0,b<0$.
First, we have $a_n=[\ln(n)/n^{-a/b}]^{-b}$. Here is $$ \lim_{n\to\infty}\frac{\ln(n)}{n^{-a/b}}\stackrel{L.H.}=\lim_{n\to\infty}\frac{1/n}{(-a/b)n^{-a/b-1}}=\lim_{n\to\infty}\frac{(-b/a)}{n^{-a/b}}=0, $$ since $-a/b>0$, and L.H. stands for L'Hospital's rule. This implies that $a_n\to 0^{-b}=0$ as $n\to\infty$.
Now, let $\beta\in\mathbb{R}$ be such that $b<\beta$. We have $$ \frac{\ln^{-\beta}(n)}{n^a}=\frac{1}{\ln^b(n)}\frac{\ln^{b-\beta}(n)}{n^a}. $$ The sequence $(1/\ln^b(n))$ tends to infinity, while $(\ln^{b-\beta}(n)/n^a)$ converges to $0$, since $b-\beta<0$ by the above argument. So $\frac{\ln^{-\beta}(n)}{n^a}\to 0$ as $n\to\infty$. Since $\beta$ can be negative or positive, we conclude that the sequence $(\frac{\ln^{-\beta}(n)}{n^a})$ converges to $0$ when $a>0$ and $\beta\in \mathbb{R}$.
Is this proof valid, or/and is there a better proof? If there are some mistakes in the proof, please let me know.
Alternatively, since $e^x > x \implies \ln x < x$ we have for $a >0$ and $b < 0$,
$$0 <\frac{1}{n^a (\ln n)^b} = \frac{(\ln n )^{|b|}}{n^a} = \frac{(\frac{2|b|}{a}\ln n^{\frac{a}{2|b|}} )^{|b|}}{n^a} < \left(\frac{2|b|}{a}\right)^{|b|}\frac{n^{a/2}}{n^a} \longrightarrow_{n \to \infty} 0$$