TITLE: Nature of $\sum 1/f(n)$ with $f(n) := n\ln(n)\ln(\ln(n))...*\ln^{(k_n)}(n)$ ; $k_n$ being the largest natural integer $k$ such that $\ln^{(k)}(n)≥1$
QUESTION [6 upvotes]: 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 ?

REPLY [2 votes]: 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.