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

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)

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.

• What is $\ell$ in the definition of $\mathscr S$? More precisely, should the index $\ell-1$ in this definition read $\infty$?
– Did
Sep 18, 2016 at 9:39
• @Did It was a typo, I edited. Sep 18, 2016 at 9:40
• OK. Did you try to estimate the contribution of the slice of integers $n$ such that $$1\leqslant\log^{(k)}n<e$$ for each $k$?
– Did
Sep 18, 2016 at 9:44
• @Did Ok, I'll edit to add my progress in that direction. Sep 18, 2016 at 10:03
• Starting from where you stopped, note that $e^{(k+1)}\gg e^{(k)}$ when $k\to\infty$ hence the series at the end of your post converges/diverges if and only if $\sum x_k$ does, where $$x_k=e^{(k+1)}\left(\prod_{\ell=0}^{k} e^{(\ell)}\right)^{-1}$$ Now, $$\frac{x_{k}}{x_{k-1}}=\frac{e^{(k+1)}}{(e^{(k)})^2}$$ hence it suffices to note that $e^{(k+1)}\gg (e^{(k)})^2$ when $k\to\infty$ to conclude that the series diverges.
– Did
Sep 18, 2016 at 11:54

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 $\log^*(x)$.

• Can we hope to generalize this, with some sort of argument of closed-ness of the set of divergent series (so that the limit has to be divergent too) ? Sep 18, 2016 at 19:37
• In this setting we have some positive divergent series which descend pointwise to another divergent series. There are similar cases where positive divergent series descend pointwise to a convergent series. For example, start with a positive convergent series $a_1+a_2+\dots$, and consider divergent series form $a_1+a_2+a_3+a_4+a_4+a_4+a_4+\dots$. There are results like the monotone convergence theorem or the bounded convergence theorem about when you can interchange limits and summations, but if anything they assert that convergent series are closed in some restricted sense. Sep 18, 2016 at 22:17
• I was thinking to something like : for every $k$, $\ \sum_{n=1}^\infty a_k(n)$ diverges, $a_k(n)> a_k(n+1) >0$, $\frac{a_{k+1}(n)}{a_k(n)} \le 1$ and is decreasing in $n$, and $\sum_{n=1}^\infty \frac{a_{k+1}(n)^2}{a_k(n)}$ converges, then $\sum_{n=1}^\infty a_\infty(n)$ diverges Sep 18, 2016 at 22:29