Convergence of $\sum _{n=3} ^\infty \frac 1 {(\ln \ln n)^{\ln \ln n}}$

Question

$$\sum _{n=3} ^\infty \frac 1 {(\ln \ln n)^{\ln \ln n}} .$$

I believe the series diverges. I am thinking of using the integral test to show this, but I am not sure if that is right.

Answer 1

This series is divergent.

The general term is decreasing then one may apply the Cauchy condensation test, thus the initial series behaves like $$ \sum_{n=3}^\infty\frac{2^p}{(\ln p+\ln(\ln 2))^{\ln p+\ln(\ln 2)}} $$ which is divergent since $$ \lim_{n \to +\infty}\frac{2^p}{(\ln p+\ln(\ln 2))^{\ln p+\ln(\ln 2)}}=+\infty\neq0. $$

Answer 2

Since $\log(\log(n))$ is monotonically increasing, we can write

$$\lim_{n\to \infty}\frac{\log(\log(n))^{\log(\log(n))}}{n}=\lim_{x\to \infty}\frac{\log(\log(x))^{\log(\log(x))}}{x}$$

Then, letting $x=e^{e^u}$ reveals

$$\begin{align} \lim_{n\to \infty}\frac{\log(\log(n))^{\log(\log(n))}}{n}&=\lim_{u\to \infty}\frac{u^u}{e^{e^u}}\\\\ &=\lim_{u\to \infty}e^{u\log(u)-e^{u}}\\\\ &=0 \end{align}$$

Thus, for sufficiently large $n$, $\frac{1}{\log(\log(n))^{\log(\log(n))}}>\frac1n$. Since the harmonic series diverges, then the series

$$\sum_{n=3}^\infty \frac{1}{\log(\log(n))^{\log(\log(n))}}\,\,\text{diverges also!}$$