Let $\{x_n\}_{n=1}^\infty \subseteq (0,\infty)$ and $\{y_n\}_{n=1}^\infty \subseteq (0,\infty)$ be some positive (possibly unbounded) sequences. I am trying to "cleanly" and "tightly" characterize the necessary and sufficient conditions under which $x_{n}>y_{n}$ holds for arbitrarily large $n$. Is the following statement true?
$$x_n>y_n \text{ holds for arbitrarily large } n \Leftrightarrow \limsup_{n\to\infty} \frac{x_n}{y_n} >1.$$
My proof is found below. Please let me know if it is correct. If it is incorrect, please let me know what statement should be to the right of "$\Leftrightarrow$" in the above statement. Thank you very much$\ $
My proof: First I prove the "$\Rightarrow$" part, via proof by contrapositive. Suppose that $\limsup\limits_{n\to\infty}\frac{x_n}{y_n} = \lim\limits_{n'\to\infty}\sup\limits_{n\geq n'}\frac{x_n}{y_n} \leq 1$. This implies that $\exists n'\in\mathbb{N}$ s.t. $\sup_{n\geq n'} \frac{x_n}{y_n} \leq 1$. It then follows that $x_{n}/y_{n}\leq 1$$\forall n\geq n'$, which is equivalent to saying that $x_n \leq y_n$ holds $\forall n\geq n'$. Therefore $x_n>y_n$ fails to hold for arbitrarily large $n$.
Now I prove the "$\Leftarrow$" part. Suppose that $\limsup\limits_{n\to\infty}\frac{x_n}{y_n} = \lim\limits_{n'\to\infty}\sup\limits_{n\geq n'}\frac{x_n}{y_n} = \sigma$ for some $\sigma>1$. I now proceed via proof by contradiction. Suppose that $\exists m\in\mathbb{N}$ s.t. $\forall n'\geq m$, $\frac{x_{n'}}{y_{n'}}\leq 1$ holds. (This is the negation of "$x_n>y_n$ holds for arbitrarily large $n$.") It then follows that $\sup_{n\geq n'}\frac{x_{n'}}{y_{n'}}\leq 1$ holds $\forall n'\geq m$, since the supremum is the smallest upper bound. This yields a contradiction to the supposition that $\lim\limits_{n'\to\infty}\sup\limits_{n\geq n'}\frac{x_n}{y_n} = \sigma$. Therefore, such an $m$ cannot exist, so $x_n>y_n$ must hold for arbitrarily large $n$. $\ \blacksquare$