Let $a_n$ and $b_n$ be strictly positive and decreasing sequences such that $\lim a_n = \lim b_n = 0$.
Is it true that one of the following has to hold: $a_n = \mathcal O (b_n)$ or $b_n = \mathcal O(a_n)$?
Looking for a counter example i managed to constructed the following sequence but I am not sure if it is useful for my purpose.
$c_n$ a decreasing positive function with limit zero. then
$a: \, c_2, c_2, c_4, c_4, c_6, c_6 \ldots$
$ b: \, c_1, c_3, c_3, c_5, c_5, c_7 \ldots$
Answer
With
$$a_n^{-1} = \begin{cases} 2\cdot (n+2)! &\text{if } n \text{ is odd} \\ (n+3)! &\text{if } n \text{ is even}\end{cases} \qquad\text{and}\qquad b_n^{-1} = \begin{cases} (n+3)! &\text{if } n \text{ is odd} \\ 2\cdot (n+2)! &\text{if } n \text{ is even} \end{cases}$$
we have two strictly decreasing sequences converging to $0$ with
$$\frac{a_n}{b_n} = \begin{cases} \dfrac{n+3}{2} &\text{if } n \text{ is odd} \\ \dfrac{2}{n+3} &\text{if } n \text{ is even}\end{cases}\,,$$
so neither $b_n = \mathcal{O}(a_n)$ nor $a_n = \mathcal{O}(b_n)$.
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