We are in ZFC & CH. Given family $Y=\{y\_\alpha\}\_{\alpha<\omega\_1}$ of infinitesimal $\omega$-sequences (i.e. $\lim\_{n\to\infty}y\_{\alpha n}=0$) of rational numbers with the property: $\forall\alpha<\beta: \lim\_{n\to\infty}\frac{y\_{\beta n}}{y\_{\alpha n}}=0$. Can we prove that for any infinitesimal sequence $y$ there exists $y\_\alpha$ such that $\lim\_{n\to\infty}\frac{y\_{\alpha n}}{y\_{n}}=0$?



Existence of $\alpha$-sequence of infinitesimal sequences with $\alpha>\omega_1$
The answer is no. First, let be construct a family of $\omega\_1$ functions $f\_\alpha:\omega\to\omega$ with the following three properties: each $f\_\alpha$ tends to infinity, $f\_\alpha\ll f\_\beta$ whenever $\alpha<\beta$ (where $\ll$ is *eventual domination*: $f\_\alpha(n)<f\_\beta(n)$ for large enough $n$) and $f\_\alpha(n)\ll n-k$ for all $k>0$.


We start with, for example, $f\_0(n)=\lfloor\sqrt{n}\rfloor$. For any $\alpha$, we let $f\_{\alpha+1}(n)=f\_\alpha(n)+1$. For $\alpha$ a countable limit, pick a sequence $\alpha\_k,k\in\omega$ such that $\alpha\_k<\alpha$ and $\alpha\_k\to\alpha$ (a so-called *fundamental sequence* for $\alpha$). For any $k$, pick $n\_k$ large enough so that $f\_{\alpha\_k}(n)<n-k$ for $n>n\_k$. Define
$$f\_{\alpha}(n)=f\_{\alpha\_k}(n)+1 \text{ for the largest $k$ such that } n\geq k,n>n\_k$$
Clearly, for any $k$ we have $f\_\alpha\gg f\_{\alpha\_k}$. Since for any $\beta<\alpha$ there is $k$ with $\alpha\_k>\beta$, it follows $f\_\alpha\gg f\_{\alpha\_k}\gg f\_\beta$, so $f\_\alpha\gg f\_\beta$. It follows $f\_\alpha$ tends to infinity. Moreover, for any $k$ and any $n\geq k,n>n\_k$ we easily see $f\_\alpha<n-k+1$, so $f\_\alpha$ has the desired property. By transfinite recursion we get $f\_\alpha$ for all $\alpha$.


Now define $y\_{\alpha n}=1/f\_\alpha(n)!$. Then $y\_{\alpha n}\to 0$ (which is what, I think, it means for the sequence to be infinitesimal) and $y\_{\beta n}/y\_{\alpha n}=f\_\alpha(n)!/f\_\beta(n)!\leq 1/f\_\beta(n)\to 0$ for $\beta>\alpha$, since eventually $f\_\beta(n)\geq f\_\alpha(n)+1$. Finally, since for each $\alpha$ we eventually have $f\_\alpha(n)<n$, we have $y\_{\alpha n}>1/n!$, so letting $y\_n=1/n!$ we see $y\_{\alpha n}/y\_n\geq 1$ and doesn't tend to zero.


