series convergence with comparison test $\frac{1}{n\log(n)^p}$

Question

I came across a question about convergence tests and I'm not sure how to show that

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

converges for $p>1$. I found that the integral test works but I haven't learnt it yet. I have learnt the comparison test, so I tried that but couldn't find any series that converges to which I can compare.
So my question is both how can I show convergence here with comparison test and also: Is there some list of convergent series that I can compare to?

PS: I also found out that this is called a ln series

Also, this is not a duplicate of this question. (The power is on the $\log$ here)

Answer 1

Use the Cauchy Condensation Test:

$\sum a_n$ converge iff $\sum 2^na_{2n}$ converge.

So $\sum \frac{1}{n \log(n)^p}$ converge iff $$\sum 2^n\frac{1}{2^n \log(2^n)^p} \text{ converges}$$

And $$\sum 2^n\frac{1}{2^n \log(2^n)^p}=\sum \frac{1}{(n\log 2)^p}=\frac{1}{(\log 2)^p}\sum\frac{1}{n^p}$$

And from this it follows that the given integral converges for $$p>1$$

Answer 2

$$ \sum_{n\ge 2}\frac{1}{n\log^p n}\sim \int_2^{\infty}\frac{1}{x\log^p x}\mathrm{d}x <\infty, $$ where the last $<$ is motivated by the fact that a primitive is $\frac{1}{\log^{p-1} x}$ (and $p-1>0$).