Convergence of series based on iteratively applying the logarithm function [duplicate]

Question

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.

Answer 1

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 $\log^*(x)$.