How to determine convergence or divergence of this series depending on $\alpha$ and $\beta$? $$\sum_{n=3}^{\infty} n^{\alpha}\cdot \ln^{\beta}\left(\frac{n}{n-1}\right) \qquad(1)$$
When $\alpha=\beta=0$, $\ \ \ (1) = \sum_{n=3}^{\infty}1$ which is divergent.
I've tried using the ratio test. $$\lim_{n \to \infty} \frac{ a_{n+1}}{a_{n}} =\lim_{n \to \infty} {\left( 1+\frac{1}{n} \right)^\alpha}\left( \frac{\ln \left( 1+ \frac{1}{n}\right)}{\ln{\left( 1+\frac{1}{n-1} \right) }} \right)^\beta = 1.$$ for $\alpha,\beta \in \mathbb{R} \setminus \{0 \}$.
To calculate the limit of logarithms I have used L'Hôpital's rule. But this is useless because the ratio is $1$. Can you give some hint?
Hint. Note that as $n\to +\infty$, $$0<\ln\left(\frac{n}{n-1}\right)=\ln\left(1+\frac{1}{n-1}\right)\sim\frac{1}{n}.$$ Can you take it from here?