Depending on the parameter $\alpha$ study the convergence of the series $\sum_{n=2}^\infty\frac{\log(\log{n})}{\log^\alpha{n}}$ Depending on the parameter $\alpha$ study the convergence of the series $$\sum_{n=2}^\infty\frac{\log(\log{n})}{\log^\alpha{n}}$$
So for $\alpha \gt 0$ and $x \gt 1$ the function $f(x)=\frac{\log(\log{x})}{\log^\alpha{x}}$ is decreasing so I can use the integral test 
which is divergent for $\alpha =1$
Is my reasoning right? I don't know what to do next
 A: We have that eventually (notably for $n>e^e$)
$$\frac{\log(\log{n})}{\log^\alpha{n}}\ge \frac{1}{\log^\alpha{n}}$$
and for $\alpha>0$
$$\sum_{n=2}^\infty\frac{1}{\log^\alpha{n}}$$
diverges by limit comparison test with $\sum_{n=2}^\infty \frac1n$ indeed
$$\frac{\frac{1}{\log^\alpha{n}}}{\frac1n}=\frac{n}{\log^\alpha{n}}\to \infty$$
A: For all $x\gt0$, $\log(x)\lt x$. Therefore, for all $x\gt1$ and $\alpha\gt0$,
$$
\begin{align}
\log(x)^\alpha
&=\alpha^\alpha\log\!\left(x^{1/\alpha}\right)^\alpha\\
&\lt\alpha^\alpha x^{1/\alpha\cdot\alpha}\\[3pt]
&=\alpha^\alpha x
\end{align}
$$
Therefore, since $\log(\log(n))\ge1$ for $n\ge16$, it suffices to compare
$$
\sum_{n=2}^\infty\frac1{\log(n)^\alpha}\ge\frac1{\alpha^\alpha}\sum_{n=2}^\infty\frac1n
$$
A: If you know Bertrand's series, the answer is obvious by the comparison test:
A Bertrand's series is a series
$$\sum_{n=2}^\infty\frac1{n^\alpha\,\log^\beta n}\qquad (\alpha,\beta\in\bf R)$$
and it is known to  converge if and only if


*

*either $\alpha>1$,

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

