On convergence of problematic series. Determine if the following series is converges or not
$$\sum_{n=2}^{\infty}\frac{1}{\ln^{\ln(\ln n)} (n)}$$
 A: Note that $$\ln\left(\frac{1}{\ln^{\ln(\ln n)}(n)}\right)=\ln1-\ln(\ln^{\ln(\ln n)}(n))=-\ln(\ln n)\ln(\ln n)=-\left[\ln(\ln n)\right]^2$$
Therefore
$$\sum_{k=2}^\infty\frac{1}{\ln^{\ln(\ln n)}(n)}=\sum_{k=2}^\infty e^{\ln\left(\frac{1}{\ln^{\ln(\ln n)}(n)}\right)}=\sum_{k=2}^\infty e^{-\left[\ln(\ln n)\right]^2}$$
It's not too hard to show that $\left[\ln(\ln n)\right]^2<\ln n$ for sufficiently large $n$. Therefore, for sufficiently large $n$, we have $e^{-\left[\ln(\ln n)\right]^2}>e^{-\ln n}=\frac{1}{n}$, so the series diverges (by comparison with the harmonic series).
A: Not too hard seeing that the following limit is infinite:
$$\lim_{n\to\infty}n^1\times\frac{1}{\ln^{\ln(\ln n)}(n)}=+\infty$$ So according to a result of the Comparison Test, the series diverges.
A: Do it by comparison with $\sum_a^\infty \frac{1}{n}$. We want to show that ultimately, $(\ln n)^{\ln\ln n}\lt n$. Equivalently, we want to show that ultimately, $(\ln\ln n)^2\lt \ln n$. 
This can be done in various ways, like L'Hospital's Rule. 
Or else we can let $\ln\ln n=t$. So we want to show that for $t$ large, we have $t^2\lt e^t$. This is easy, for positive $t$, by the power series expansion, $e^t\gt \frac{t^3}{3!}$. 
