For all $n$ in $\mathbb N^*$, let $f(n) := n*\ln(n)*\ln(\ln(n))*...*\ln^{(k_n)}(n)$, with $\ln^{(k)}$ being the logarithm iterated $k$ times, and $k_n$ being the largest natural integer $k$ such that $\ln^{(k)}(n)≥1$.
Study the nature of the series $\sum 1/f(n)$.
One can show that when $k_n$ is a constant, the series diverges (by comparison with integral) but here it is not the case. I think the series also diverges. How to prove it ?
For any $k\geq 0$, we have $k_n=k$ for $$ a_k:=\begin{matrix} &\underbrace{e^{e^{{}^{.\,^{.\,^{.\,^e}}}}}} & \\ & k\mbox{ copies of }e \end{matrix} \leq n < a_{k+1}:=\begin{matrix} &\underbrace{e^{e^{{}^{.\,^{.\,^{.\,^e}}}}}} & \\ & k+1\mbox{ copies of }e \end{matrix} $$
The sum of $1/f(n)$ on this range of $n$, is at least $$ \int_{a_k}^{a_{k+1}} \frac1{x \ln(x) \ln(\ln(x))\cdots \ln^{(k)}(x) } dx=\ln^{(k+1)}(x) \Bigg\vert_{a_k}^{a_{k+1}} = 1. $$
Therefore, the sum of $1/f(n)$ over all $n$ diverges.
