I need to find out for what $\alpha, \beta$ the following sum converges: $$\sum_{n=2}^\infty n^\alpha (\log n)^\beta$$
I thought I'd do that with the help of the integral criterion, that is to say I considered (after I had substituted $x = e^u$): $$\int_2^\infty e^{u(\alpha+1)} \cdot u^\beta \mathrm du$$
Then, I realised that for $\alpha + 1 > 0$, the exponential function will always dominate the monomial, so I only considered $\alpha + 1 \leq 0$. First, I let $\alpha + 1=0$, then I would be left with: $$\int_2^\infty u^\beta \mathrm du$$
I solved the integral and substituted back and ended up with: $$\lim_{b \to \infty} \frac{(\log b)^{\beta+1}}{\beta+1}$$
I figured this would only converge for $\beta+1 \leq 0$ and thought: "Ok, now I have some cases, where the sum converges: $\alpha = -1 \land \beta \leq -1$". To be sure, I checked the integral with the help of wolframalpha, and it told me the integral would not converge.
What did I do wrong? Is there an easier way to find $\alpha$ and $\beta$ so that the sum converges?