When does this iterated logarithm series converge / diverge?

Question

Question: Let $b > 1$ be a positive real, let $\ell_b(n) = \max(1, \lfloor \log_b n \rfloor)$, and let $f_b(n) = n \ell_b(n) \ell_b^2(n) \dots $ (where we iterate $\ell_b$ until we hit $1$). For what values of $b$ does the series

$$\sum_{n=1}^{\infty} \frac{1}{f_b(n)}$$

converge or diverge?

This previous question asks about the case $b = 10$. There I gave an argument which shows that the series diverges for all $b > e$. I thought I had convinced myself awhile ago that this series ought to diverge for all $b > 1$ but surprisingly the argument does not show it; actually it suggests that the series converges for $b < e$ and it's unclear what happens if $b = e$.

My motivation is the same as Alan's in the linked question: this is an interesting test of the limits of common convergence tests. The integral test or Cauchy condensation shows that the analogue of this series where we only iterate $\ell_b(n)$ a fixed number of times diverges for any $b > 1$ but that argument doesn't work here. Actually Cauchy condensation produces nearly the same series again, which is basically the idea I use in my argument in the linked question.

Answer 1

Let the superlog base $b$ be

$slog(1,b) = 0$

,$slog(b,b) = 1$,

$slog(x,b) = slog(\ln_b(x),b)+1$

$slog(x) = slog(x,e)$

So basicly $slog(x,b)$ tell us how many times we can take the log base $b$ of $x$ to remain positive.

We can approximate the sum by the similar integral. This gives a close approximation and is certainly sufficient to decide convergeance.

by using the chain rule we get

$\int_1^{x} (t \ln(t) \ln(\ln(t)))^{-1} = \ln(\ln(\ln(x))$

as long as $slog(x) > 3$

This is similar to

$$\sum_{i=1}^x (t \ln(i) \ln(\ln(i)))^{-1} = \ln(\ln(\ln(x)) + O(e)$$

as long as $slog(x) > 3$

We Notice this can be generalized :

by using the chain rule we get

$\int_1^{x} (t \ln(t) \ln(\ln(t))) ... \ln(t)^{[k]})^{-1} = \ln^{k+1}(x)$

as long as $slog(x) > k+1$

this implies

$\int_1^{x} (t \ln(t) \ln(\ln(t))) ... \ln(t)^{[slog(x)]})^{-1} = \ln^{slog(x)}(x) = log^{*}(x)$

Where $log^{*}(x)$ means taking the log as many times as possible.

Hence $0<log^{*}(x)<e$ thus a bounded result.

And thus $\int_1^{x} (t \ln(t) \ln(\ln(t))) ... \ln(t)^{[slog(x)]})^{-1} = \ln^{slog(x)}(x) = log^{*}(x)$ does not diverge to infinity and neither does its related sum.

Now lets see what approximately happens when $1<b<e$. Define $c = (\ln(b))^{-1}$.

$$\ln_b(x) = \ln(x)/\ln(b) = c \ln(x)$$, $$\ln_b^{[2]}(x) = c \ln(\ln_b(x)) = c ( \ln(ln(x)) + \ln(c) )$$ $$\ln_b^{[3]}(x) = c \ln(c \ln(\ln_b(x))) = c ( \ln(c) + \ln(\ln(ln(x)) + \ln(c)) )$$ $$...$$

And we clearly have

$$ (c - o(1)) \ln^{v}(x) < \ln_b^{[v]}(x) < (c + o(1)) \ln^{[v]} + max(c,c \ln(c)) $$

And thus the asymptotic

$$\ln_b^{[v]}(x) = c \ln^{[v]}(x)$$

This implies that

$$\sum_{i=1}^x (t \ln_b(i) \ln_b(\ln_b(i)))^{-1} = (1 +/- o(1/2)) \sum_{i=1}^x (t c \ln(i) c \ln(\ln(i)))^{-1} = c^{-2}(\ln(\ln(\ln(x)) + O(e))$$

And in general

$$ \int_1^{x} (t \ln_b(t) \ln_b(\ln_b(t))) ... \ln_b(t)^{[slog(x)]})^{-1} = \int_1^{x} (t \ln_b(t) \ln_b(\ln_b(t))) ... \ln_b(t)^{[slog(x)]})^{-1} c^{slog(x,b)} = \ln_b^{slog(x)}(x) c^{slog(x,b)} = c^{- slog(x,b)} ln_b^{*}(x)$$

And it is clear that

$$ sup lim_{x \to +\infty} c^{- slog(x,b)} ln_b^{*}(x) < b$$

The same applies to the sum.

So it converges.

QED