Convergence of $\sum\limits_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)} $ for nonnegative $\alpha$ and $\beta$

I think you should use the integral test. In order to establish the convergence/divergence of $$ \int_2^{\infty} \frac{dx}{x^a\log^b{x}} $$ I suggest you to use the substitution $x=e^t$ which transforms the integral into $$ \int_{\log 2}^{\infty} \frac{dt}{e^{(a-1)t}t^b} $$ which is easier. I think you should be able to conclude by yourself, now.

You can apply Cauchy's condensation test. Moving to $$ \sum_{n = 2}^{\infty} \frac{2^n}{2^{n\alpha} n^{\beta}} = \sum_{n = 2}^{\infty} \left( 2^{1-\alpha} \right)^n \frac{1}{n^{\beta}}, $$ the series surrenders to the ratio test when $\alpha \neq 1$. When $\alpha=1$, this is a well-known series.