studying the series $\sum_\limits{n=1}^\infty \frac{1}{n^ {\alpha}(\log n)^ {\beta}}$. I have to study the character of this series 
$$\sum_{n=1}^\infty \frac{1}{n^{\alpha}(\log  n)^{\beta}}.$$
$\alpha$ and $\beta$ are two parameters.
I'm considering the case $\alpha >1 $ and all real values for $\beta $ .
If $\beta \ge 0$, $\exists n_0 \in N $ such that $\forall n\ge n_0, \dfrac{1}{n^ {\alpha}(\log  n)^ {\beta}}<\dfrac{1}{n^ {\alpha}}$. 
For the comparison test $\dfrac{1}{n^{\alpha} (\log  n)^ {\beta}}$ converges.
I tried to analyze the case $-1<\beta<0,\beta=-1, \beta<-1$ but I'm not sure.
Anyway, for $\beta<-1$ the series is convergent for $\alpha > 1- \beta$, divergent for $\alpha < 1- \beta$.
For $-1<\beta<0$ the series is convergent for $\alpha > 2$, divergent for $\alpha <2$.
For $\beta=-1$ the series is convergent for $\alpha > 2$, divergent for $1<\alpha < 2$.
I've used the fact that $\log n<n$
 A: You can refer to  Cauchy condensation test
$$ 0 \ \leq\ \sum_{n=1}^{\infty} f(n)\ \leq\ \sum_{n=0}^{\infty} 2^{n}f(2^{n})\ \leq\ 2\sum_{n=1}^{\infty} f(n)$$
and study the equivalent condensed series
$$\sum_{n=1}^\infty \dfrac{2^n}{2^{n\alpha}(\log  2^n)^{\beta}}=\sum_{n=1}^\infty \dfrac{1}{2^{n(\alpha-1)}n^{\beta}\log^{\beta}2}$$
from which is clear that the series converges if and only if


*

*$\alpha>1$

*$\alpha=1$ and $\beta>1$

A: For $\beta<0$, so we are dealing with $\dfrac{(\log n)^{\gamma}}{n^{\alpha}}$ for $\gamma>0$.
Now use the fact that $(\log n)^{\gamma}\leq C_{\gamma,\xi}n^{\xi}$ where $\alpha-\xi>1$, $\xi>0$, where $C_{\gamma,\xi}>0$ is some constant, then $\dfrac{(\log n)^{\gamma}}{n^{\alpha}}\leq C_{\gamma,\xi}\dfrac{1}{n^{\alpha-\xi}}$ and $\displaystyle\sum\dfrac{1}{n^{\alpha-\xi}}<\infty$.
A: The results  for this series (which incidentally is known as a Bertrand's series) are much simpler: it converges if and only if


*

*$\alpha >1$ (by comparison with a Riemann series) 

*or $\alpha=1$ and $\beta>1$ (by the integral test).

