The Bertrand series, all cases

$\alpha>1$. Choose $\alpha'\in(1,\alpha)$. Whatever the sign of $\beta$, $(\ln n)^{-\beta}\le n^{\alpha-\alpha'}$ for large $n$ (a power of $\ln n$ is negligible against a positive power of $n$). So $u_n\le\frac1{n^{\alpha'}}$, and the series converges.

$\alpha<1$. Choose $\alpha'\in(\alpha,1)$. Whatever the sign of $\beta$, $(\ln n)^\beta\le n^{\alpha'-\alpha}$ for large $n$. So $u_n\ge\frac1{n^{\alpha'}}$, and the series diverges.

$\alpha=1$. For $\beta>0$ the function $\frac1{x(\ln x)^\beta}$ is positive and decreasing. With $u=\ln x$,
$$\int_2^\infty\frac{dx}{x(\ln x)^\beta}=\int_{\ln2}^\infty\frac{du}{u^\beta},$$
finite iff $\beta>1$. Fix an integer $n\geq 2$. On $[n:n+1]$, we have
$$\frac1{(n+1)^\alpha(\ln (n+1))^\beta} \leq f \leq \frac1{n^\alpha(\ln n)^\beta}.$$
Integrating on $[N:N+1]$ gives
$$\frac1{(n+1)^\alpha(\ln (n+1))^\beta} \leq \int_n^{n+1} f(x) dx \leq \frac1{n^\alpha(\ln n)^\beta}.$$
Taking the sum from $n=2$ to yields
$$\sum_{n\geq 2} \frac1{(n+1)^\alpha(\ln (n+1))^\beta} \leq \int_2^{=\infty} f(x) dx \leq \sum_{n=2}^{+\infty} \frac1{n^\alpha(\ln n)^\beta}.$$
Thus the serie $\sum_{n\geq 2} \frac1{(n+1)^\alpha(\ln (n+1))^\beta}$ and the integral $\int_2^{+\infty} f(x) dx$ have the same nature.

For $\beta\le0$, $\frac1{n(\ln n)^\beta}\ge\frac1n$ for $n\ge3$: divergence.

Conclusion. The series converges iff $\alpha>1$, or $\alpha=1$ and $\beta>1$.