Context
From the Riemann criterion, we know that
$$\sum \frac 1{n^s}$$
converges when $s>1$.
Continuing in the same path, we know thanks to Bertrand series that
$$\sum \frac 1{n\ln (n)^s}$$
converges when $s>1$.
And so on, i.e. if we note $\ln^{(k)}$ the $k$-th iteration of the logarithm:
$$\ln^{(k)}:=\underbrace{\ln\circ\ln\circ\cdots\circ\ln}_{k \text{ times}}$$
then for all $\ell$
$$\sum \left(\ln^{(\ell)}(n)^s\prod_{k=0}^{\ell-1} \ln^{(k)}(n)\right)^{-1}$$
converges when $s>1$.
The problem
So I was wondering what happens at the limit. Does it go infinity or does it converge ?
Let's consider the following series:
$$\mathscr S:=\sum_{n=1}^\infty \left(\prod_{\substack{k=0 \\ \ln^{(k)}(n)\geqslant 1}}^{\infty} \ln^{(k)}(n)\right)^{-1},$$
where the apparently infinite product is in fact finite for all $k$.
Basically, we take as much logarithm as we can so it doesn't get smaller than $1$.
Concretely
We have $\ln^{(0)}(1)=1\geqslant 1$ and $\ln^{(1)}(1)=0< 1$ so we stop there.
And $\ln^{(0)}(2)=2\geqslant 1$ and $\ln^{(1)}(2)\approx 0.69< 1$ so we stop there.
And $\ln^{(1)}(3)\approx 1.1\geqslant 1$ and $\ln^{(2)}(3)\approx 0.1< 1$ so we stop there.
$\vdots$
And $\ln^{(1)}(15)\approx 2.7\geqslant 1$ and $\ln^{(2)}(15)\approx 0.996< 1$ so we stop there.
And $\ln^{(2)}(16)\approx 1.02\geqslant 1$ and $\ln^{(3)}(16)\approx 0.02< 1$ so we stop there.
$\vdots$
So it starts like this:
$$\mathscr S=\frac 11+\frac 1{2}+\frac 1{3\ln (3)}\ldots+\frac 1{15\ln (15)}+\frac 1{16\ln (16)\ln\ln (16)}+\ldots.$$
The question
Does $\mathscr S$ converge ?
What could work
- We can prove Riemann criterion and Bertrand's one using the Cauchy condensation test:
$$\sum f(n)<\infty \iff \sum 2^nf(2^n)<\infty.$$
But I didn't get anywhere.
- We can also notice that since we want $n$ such that
$$1\leqslant \log^{(k)}(n)<e$$
we want $n$ sucht that
$$e^{(k)}\leqslant n < e^{(k+1)}.$$
So we can rewrite the series:
$$\mathscr S:=\sum_{k=0}^\infty \sum_{n=[e^{(k)}]+1}^{[e^{(k+1)}]}\left(\prod_{\ell=0}^{k} \ln^{(\ell)}(n)\right)^{-1}.$$
Therefore, if we take $n\in \{[e^{(k)}]+1, \ldots, [e^{(k+1)}]\}$ we have:
$$\sum_{n=[e^{(k)}]+1}^{[e^{(k+1)}]}\left(\prod_{\substack{\ell=0 \\ \ln^{(k)}(n)\geqslant 1}}^{k} \ln^{(\ell)}(n)\right)^{-1}\leqslant \frac{e^{(k+1)}-e^{(k)}}{\displaystyle\prod_{\ell=0}^{k} \ln^{(\ell)}(e^{(k+1)})}=\frac{e^{(k+1)}-e^{(k)}}{\displaystyle\prod_{\ell=0}^{k} e^{(\ell)}}.$$
Now we only need to understand whether or not
$$\sum_{k=0}^{\infty} (e^{(k+1)}-e^{(k)})\left(\prod_{\ell=0}^{k} e^{(\ell)}\right)^{-1}$$
converges.
Define recursively $$ f(x)= \begin{cases} 1 & x \leq 1 \\ x^{-1} f(\log x) & x > 1 \end{cases} $$
You're interested in the sum $\sum_{n=1}^\infty f(n).$ It's fairly easy to check that $f$ is a bounded nonincreasing function, so that sum converges if and only if the integral $\int_0^\infty f(x)\,dx$ converges.
Consider the increasing sequence given by $a_0 = 0$ and $a_{n+1} = e^{a_n}$. We have
$$I_n = \int_{a_n}^{a_{n+1}} f(x) \, dx = \int_{a_n}^{a_{n+1}} x^{-1} f(\log x)\,dx = \int_{\log a_n}^{\log a_{n+1}} f(u)\,du = \int_{a_{n-1}}^{a_{n}} f(u)\,du = I_{n-1}$$
and
$$I_0 = \int_0^1 f(x)\,dx = 1.$$
So $I_n = 1$ for all $n$, and therefore $$\int_0^{a_n} f(x)\,dx = I_0 + I_1 + \dots + I_{n-1} = n.$$
Hence, the integral we're interested in diverges. Specifically $$\int_0^T f(x)\,dx$$ grows roughly like the inverse of the tetration function $a_n$, i.e. roughly like $\log^*(T)$.
Indeed, you can get the result more directly by noting that $f$ is the derivative of the function given by
$$ g(x) = \begin{cases} x & x \leq 1 \\ 1+g(\log x) & x > 1, \end{cases} $$
which is a "smoothed" version of $\log^*(x)$.