Is there a divergent monotone non-decreasing continuous positive real-function $f$ such that $$\lim\limits_{x\to +\infty} \frac{f(x)}{\log^{(k)}(x)} = 0$$ for all $k\geqslant 1$? (By $\log^{(k)}(x)$ I mean $\underbrace{\log\log\cdots\log}_{k \text{ times}}(x)$.)
I have a strange feeling that this is false, and, although, very unlikely to be false at the same time. A cool implication of a "no" would be that if you consider the poset $\langle \mathbb{P},\prec\rangle$ consisting of divergent monotone non-decreasing continuous positive real-functions defined in $[a,+\infty)$ for some $a\geqslant 0$ (note that $a$ is not fixed) with the order
$$f\prec g \iff f = o(g),$$ the set $\{\log^{(k)} : k\in\mathbb{N}\}$ would be a countable dense subset of $\mathbb{P}$ (in the sense of dense in a partial order). This is exactly what "feels" to be unlikely to me.
Any tips in prove, disprove or even in to find a dense subset of $\mathbb{P}$ are welcome.
In fact if $g_k:[a_k,\infty) \to (0,\infty)$ is any sequence of functions such that $\lim_{x\to \infty}g_k(x) = \infty$ for all $k,$ then there exists an increasing continuous $f: [0,\infty) \to (0,\infty)$ with $\lim_{x\to \infty}f(x) = \infty$ such that
$$\lim_{x\to \infty}\frac{f(x)}{g_k(x)} = 0$$
for all $k.$
Proof idea: Choose $0<b_1<b_2 < \cdots \to \infty $ such that
$$\min(g_1,\dots ,g_k)> 3^k \text {on} \,[b_k,\infty).$$
Define $f$ so that $2^k\le f \le 2^{k+1}$ on $[b_k, b_{k+1}].$