Comparison of speed of convergence for two sequences Let ${y_{n}}$ and ${x_{n}}$ be two sequences of positive numbers that
both converge to s, and  $\frac{y_{n}-s}{x_{n}-s}\stackrel{n\rightarrow\infty}{\longrightarrow}0$.
Can we show that for every $ε>0$ there is an $m\in\mathbb{N}$ such that for $n\geq m, \left|y_{n}-s\right|\leq\varepsilon$ and $\left|x_{m}-s\right|>ε$ ?
 A: No. For example, define $x_1=1$ and define $x_{n+1}=\frac{x_n}{n^3}$. Let $y_n=\frac{x_n}{n}$ for all $n$. Then $s=\lim_n x_n=\lim_ny_n=0$ and $\lim_n y_n/x_n=\lim_n 1/n=0$. Note that $|x_n-s|=x_n$ and $|y_n-s|=y_n$ for all $n$. Also note that both sequences are strictly decreasing.
Fix $1<n\in \mathbb{N}$ and let $\varepsilon=x_n/n^2$. Note that $x_n>\varepsilon$, $y_n=x_n/n>\varepsilon$, and $y_{n+1}<x_{n+1}=x_n/n^3<\varepsilon$. So $|x_m-s|,|y_m-s|>\varepsilon$ for all $m\leqslant n$, and $|x_m-s|, |y_m-s|<\varepsilon$ for all $m>n$. So both sequences switch from greater than $\varepsilon$ to less than $\varepsilon$ at the same index.
This example uses the fact that $x_{n+1}/x_n<y_n/x_n$, which means $(x_n)_{n=1}^\infty$ goes to zero faster than a geometric sequence. Perhaps if you build in some condition (such as $\inf_n |\frac{x_{n+1}-s}{x_n-s}|>0$), one could hope for a positive answer.
Addendum: Here is the intuition: My counterexample above had $x_n$ and $y_n$ both switching from $>\epsilon$ to $<\epsilon$ at the same index. If $y_n$ is  much smaller than $x_n$ and both are above $\epsilon$, then $x_n$ must be much larger than $\epsilon$, while $x_{n+1}$ must be smaller than $\epsilon$. This means we must have a large step down from $x_n$ to $x_{n+1}$. More specifically, if $y_n/x_n<\vartheta$ and $x_{n+1}<\epsilon<y_n<x_n$, then $x_{n+1}/x_n<\vartheta$. Since $y_n/x_n\to 0$ in my example, this means that these steps down $x_{n+1}/x_n<\vartheta$ would occur for each $\vartheta>0$. If we require that $x_{n+1}/x_n$ is bounded away from zero, we can avoid this. When the sequence $(x_n)_{n=1}^\infty$ drops below $\epsilon$, it will go from not too far above $\epsilon$ to not too far below $\epsilon$. If we also require that $y_n/x_n$ is already small, we will force $y_n$ to be below $\epsilon$ while $x_n$ is still above it.
First note that since only $y_n-s$ and $x_n-s$ appear, we can define $u_n=y_n-s$ and $v_n=x_n-s$ and consider $u_n, v_n$. So $s$ doesn't matter. Moreover, everything we say here would hold if $(y_n)_{n=1}^\infty$ and $(x_n)_{n=1}^\infty$ were converging to different limits $s$ and $s'$. We also note that since $u_n/v_n\to 0$ if and only if $|u_n|/|v_n|\to 0$, we can assume $u_n,v_n\geqslant 0$.
So let us restate: if $(u_n)_{n=1}^\infty,(v_n)_{n=1}^\infty$ are positive sequences which converge to zero such that $u_n/v_n\to 0$, then for $\epsilon>0$, can we find $m\in\mathbb{N}$ such that $v_m>\epsilon$ and $u_n\leqslant \epsilon$ for all $n\geqslant m$.
First we note that we can't hope for this to be true for all $\epsilon>0$, because for large $\epsilon>0$, we won't have $v_m>\epsilon$ for any $m$. More specifically, if $(v_n)_{n=1}^\infty$ is a convergent sequence, it is bounded, so $M:=\sup_n v_n<\infty$. Then for any $\epsilon>M$, we will not be able to find any $m\in\mathbb{N}$ satisfying $v_m>\epsilon$. This has nothing to do with $(u_n)_{n=1}^\infty$. So the best we can hope for is some $\epsilon_0>0$ such that we can find the desired $m$ for all $\epsilon\in (0,\epsilon_0)$.
Assume that $(u_n)_{n=1}^\infty$, $(v_n)_{n=1}^\infty$ are positive sequences with $\lim_n u_n=\lim_n v_n=0$ and $\lim_n u_n/v_n=0$. Suppose also that $\vartheta>0$ is such that $\inf_n v_{n+1}/v_n\geqslant \vartheta$. Note that $\vartheta<1$, since otherwise the sequence $(v_n)_{n=1}^\infty$ would be non-decreasing, and would therefore not converge to zero.
Fix $n_0\in \mathbb{N}$ such that $u_n/v_n<\vartheta$ for all $n\geqslant n_0$.   Let $\epsilon_0=\sup_{n\geqslant n_0} v_n$. This supremum is actually a maximum since the sequence is positive and converges to zero. This choice of $\varepsilon_0$ yields that for any $\epsilon\in (0,\epsilon_0)$, $$N(\epsilon):=\{n\geqslant n_0:v_n\geqslant\epsilon\}$$ is non-empty. Since $\lim_n v_n=0$, each set $N(\epsilon)$ is finite.
Fix $\epsilon\in (0,\epsilon_0)$. Let $m=\max N(\epsilon)$. This $m$ exists, since $N(\epsilon)$ is finite and non-empty. Since $m\in N(\epsilon)$, $v_m\geqslant\epsilon$. Since $m$ is the maximum of $N(\epsilon)$, $v_{m+1}\notin N(\epsilon)$, and $v_{m+1}<\epsilon$. Since $v_{m+1}/v_m\geqslant \vartheta$, it follows that $$v_m\leqslant v_{m+1}/\vartheta<\epsilon/\vartheta.$$  Since $m\geqslant n_0$, $u_m/v_m<\vartheta$, so $u_m<v_m\vartheta<(\epsilon/\vartheta)*\vartheta=\epsilon$.  So $v_m\geqslant \epsilon$ and $u_m< \epsilon$.
Since $m=\max N(\epsilon)$, it follows that for all $n>m$, $n\notin N(\epsilon)$, and $v_n<\epsilon$. Since $\vartheta<1$ and $m\geqslant n_0$, it follows that $u_n\leqslant \vartheta v_n<\vartheta\epsilon<\epsilon$ for all $n>m$. So $v_m\geqslant \epsilon$ and $u_n<\epsilon$ for all $n\geqslant m$.
