This question has been asked here (limits of two equivalent sequences), but I want to check if my proof is right. If $x_n \sim y_n$ and $y_n$ converges, prove that $x_n$ converges.
Let $1/M$ be a bound for $|y_n|$. Given $\varepsilon > 0$, we can find $N_1 \in \mathbb N$ such that for all $n > N_1$,
$$\left|\frac{x_n}{y_n} – 1 \right| = \left|\frac{1}{y_n}\right| |x_n – y_n| < M|x_n – y_n| < \frac{M \varepsilon}{2},$$
implying $|x_n – y_n| < \frac{\varepsilon}{2}$. We can also choose $N_2 \in \mathbb N$ so that for all $n > N_2$, $|y_n – L| < \frac{\varepsilon}{2}$, where $L$ is the limit. Now choose $N = \max\{N_1,N_2\}$. For any $n > N$, we have
$$|x_n – L| \leq |x_n – y_n| + |y_n – L| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon,$$
so $x_n \to L$.
Best Answer
There is a problem: you divide by $y_n$, and it is not guaranteed that $y_n\ne0$ for all $n$ large enough if $L=0$. So it is better to use this definition of equivalent sequences, which requires no division:
With this definition, it is perfectly clear that $(x_n)$ and $(y_n)$ both converge to the same limit or both diverge.