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 $x_n \to L$.

  • $\begingroup$ What does it mean that they are equivalent... in this particular context? There are many equivalence relations. For example, you might require that all sequences are equivalent (this is a very loose relation) or you might require that sequences are equivalent only if they are equal (this is a very strict relation). $\endgroup$ – Squirtle Sep 5 '15 at 19:33
  • $\begingroup$ Without a clear definition, we might as well assume all sequences are equivalent. $a_n=n$ has no limit therefore $b_n=\frac{1}{n}$ should have no limit either since $(a_n)\sim (b_n)$. $\endgroup$ – Squirtle Sep 5 '15 at 19:52
  • $\begingroup$ The link gives the definition, as does the first chain of inequalities. $\endgroup$ – user217285 Sep 5 '15 at 20:52

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:

$(x_n)\sim (y_n)$ if there exists a sequence $(\varepsilon_n)$, converging to $0$ such that $x_n=y_n(1+\varepsilon_n)$.

With this definition, it is perfectly clear that $(x_n)$ and $(y_n)$ both converge to the same limit or both diverge.

  • $\begingroup$ Well, it is implicitly guaranteed that $y_n \neq 0$ because otherwise $x_n/y_n$ wouldn't be well-defined, right? $\endgroup$ – user217285 Sep 5 '15 at 20:52
  • $\begingroup$ @Nitin; Considering $\dfrac{x_n}{y_n}$ is not part of the hypotheses. $\endgroup$ – Bernard Sep 5 '15 at 20:57
  • $\begingroup$ The question starts with "If $x_n \sim y_n$" and the link gives the definition as $x_n/y_n \to 1$. $\endgroup$ – user217285 Sep 5 '15 at 21:01
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    $\begingroup$ Yes, but that is a simplified definition, that does not take into account the possibility that $y_n=0$ for an infinity of values of $n$. The true definition is the one I mentioned. $\endgroup$ – Bernard Sep 5 '15 at 21:04

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