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

Study the convergence of the following series: $$\sum_{n=2}^\infty \frac{1}{n^\alpha \cdot\ln^\beta(n)} \text{ where }\alpha,\beta \geq 0 $$

Applying d'Alembert criterion I have that $$ \lim_{n\to\infty} \frac{n^\alpha \ln^\beta(n)}{(n+1)^\alpha \ln^\beta (n+1)} = \lim_{n\to\infty} \left(\frac{n}{n+1}\right)^\alpha\left(\frac{\ln(n)}{\ln (n+1)}\right)^\beta = 1$$ so the nature of the series is inconclusive.

If $\alpha = \beta = 0$, then the series diverges, since $\sum_{n=2}^\infty 1 = \infty$.

Should I study the rest of the cases (i.e. if $\alpha = 0, \beta > 0$ the root test and the ratio test are also inconclusive). What is the best form to study the series?.

Thanks in advance

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.