Let $p,q \in \mathbb{R}$. Show using the comparison test (or limit comparison test) that
$$ \sum\limits_{n=2}^{\infty} \dfrac{1}{n^p \ln(n)^q} $$
converges for $p>1$ and any value of $q$ and that it diverges for $p<1$ and any value of $q$.
My attempt:
I see that for $q \geq 0$ that $\sum\limits_{n=2}^{\infty} \dfrac{1}{n^p \ln(n)^q} \leq \sum\limits_{n=2}^{\infty} \dfrac{1}{n^p}$ which converges for $p>1$.
My trouble is with the case of $q<0$.
Use the fact that $\ln$ is (eventually) smaller than any positive power. More precisely, for every $\epsilon>0$ there exists $n_0 \in \mathbb{N}$ such that for $n \geq n_0$ we have $\ln(n)\leq n^{\epsilon}$. In order to prove this, it suffices to show that the limit
$$\lim_{x \to \infty} \frac{\ln(x)}{x^\epsilon} $$ exists and equal to zero, which can be done using l'Hôpital's rule.
EDIT: To specifically handle the case where $q<0$, write $q=-|q|$ and observe that for any $\epsilon>0$ eventually $\ln(n) \leq n^{\epsilon/|q|}$, and thus $\ln(n)^{|q|} \leq n^{\epsilon}$. Hence, for sufficiently large $n$ we get
$$\frac{1}{n^p \ln(n)^q}=\frac{\ln(n)^{|q|}}{n^p}\leq \frac{n^\epsilon}{n^p}=\frac{1}{n^{p-\epsilon}} .$$
Can you take it from here?
