I'm trying to show whether the below series converges or diverges, and I have very little clue on how to do it. I know about the Comparison Test, but I can't think of a sequence $b_n > a_n$ to perform a comparison with.
Esentially, my question is more how should I approach this problem, and is there a less "guess-based" approach for finding a $b_n$? $$\sum_{n=2}^\infty a_n =\sum_{n=2}^\infty\frac{1}{\log(\log(n))\log(n)n}$$
Use the integral test. The series you presented is convergence if and only if
$$\int_2^\infty \frac{1}{\log\log(x)\log(x) x} dx <\infty$$
Now substitute $x=e^x$ (and remember to multiply by the derivative of $e^x$ whihc is $e^x$) then we have
$$=\int_{\log(2)}^\infty \frac{1}{\log(x)xe^x}\cdot e^x dx = \int_{\log(2)}^\infty \frac{1}{\log(x)x} dx$$
substitue $x=e^x$ once again you will get
$$\int_{\log\log(2)}^\infty \frac{1}{x}dx$$ which diverge.
Note that this trick would work even if you increase the number of $\log$. For instance it would also mean that the series diverge for $a_n = \frac{1}{\log\log\log(n)\log\log(n)\log(n)n}$ and so on. For each component of $\log$ you will need to substitue $x=e^x$ one more time.