Depending on the parameter $\alpha$ study the convergence of the series $\sum_{n=2}^\infty\frac{\log(\log{n})}{\log^\alpha{n}}$

Question

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

Answer 1

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

$$\frac{\frac{1}{\log^\alpha{n}}}{\frac1n}=\frac{n}{\log^\alpha{n}}\to \infty$$

Answer 2

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 $$ \sum_{n=2}^\infty\frac1{\log(n)^\alpha}\ge\frac1{\alpha^\alpha}\sum_{n=2}^\infty\frac1n $$

Answer 3

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 $\beta>1$.