If $x_n$ and $y_n$ are Cauchy and $y_n \neq 0$ for all $n \in \mathbb{N}$, is ${\frac{x_n}{y_n}}$ also Cauchy?

My current proof: As $x_n$ and $y_n$ are Cauchy, they are both convergent. Then, as $y_n \neq 0$ for all $n \in \mathbb{N}$, then $\displaystyle{\frac{x_n}{y_n}}$ is also convergent. Thus, as $\displaystyle \frac{x_n}{y_n}$ is convergent, $\displaystyle\frac{x_n}{y_n}$ is Cauchy. Therefore, the statement is true.

Does my reasoning make sense? Are there any steps that I am missing, or are there any significant mistakes?

I would appreciate any insight. Thank you for your help!

• You have also to assume then $y_n \not\to 0$. Otherwise you are wrong. For $x_n = 1 \to 1$, $y_n =\frac 1n \to 0$, but $\frac{x_n}{y_n} = n$ does not converge. – martini Nov 26 '12 at 8:23
• Thank you! I knew that I was missing something essential in my proof. – Jess Nov 26 '12 at 8:26
• I should point out that Cauchy sequences are not necessarily convergent, though since your sequences are real-valued, they are. I apologize if you already were aware of this; I just know that sometimes this is a point of confusion when students first generalize analysis to arbitrary metric spaces. – Christopher A. Wong Nov 26 '12 at 8:31
• A real sequence is Cauchy iff it is convergent. Thus your question reduces to: if x_n and y_n are convergent, is x_n / y_n convergent? This is true by the algebra of limits. One way of seeing this is that if y_n converges to B then 1/y_n converges to 1/B and if x_n converges to A then x_n/y_n = [x_n times (1/y_n)] converges to A/B by AOL – Adam Rubinson Nov 26 '12 at 9:16
• @Adam: you're missing an important case. $y_n$ can't converge to $0$, or else the fraction converges to $A/0$ (i.e., does not converge) – nomen Dec 25 '14 at 3:26

Suppose $\{x_n\}$, $\{y_n\}$ are Cauchy sequences where $y_n \neq 0$ for all $n \in \mathbb N$ and $y_n \not\to 0$.

Let $\varepsilon > 0$.

There exists $N_1$ such that if $n, m > N_1$ then $\displaystyle \vert x_n - x_m \vert < \frac{C^2 \varepsilon}{2M_y}$.

There exists $N_2$ such that if $n, m > N_2$ then $\displaystyle \vert y_n - y_m \vert < \frac{C^2 \varepsilon}{2M_x}$.

Since $\{x_n\}$, $\{y_n\}$ are Cauchy, they are bounded.

So there exist $M_x, M_y$ such that $\vert x_n \vert \leq M_x$ for all $n \in \mathbb N$ and $\vert y_n \vert \leq M_y$ for all $n \in \mathbb N$.

Since $y_n \not \to 0$ there exists $C > 0$ such that for all $n \in \mathbb N$ we have $\vert y_n \vert \geq C$. This implies that $\displaystyle \frac{1}{\vert y_ny_m \vert} \leq \frac{1}{C^2}$ for all $n \in \mathbb N$.

If $n, m > N := \max\{N_1, N_2\}$, then \begin{align*} \bigg \vert \frac{x_n}{y_n} - \frac{x_m}{y_m} \bigg \vert &= \bigg \vert \frac{x_ny_m - x_my_n}{y_ny_m} \bigg \vert \\ &= \bigg \vert \frac{x_ny_m - x_ny_n + x_ny_n - x_my_n}{y_ny_m} \bigg \vert \\ &\leq \frac{\vert x_ny_m - x_ny_n\vert + \vert x_ny_n - x_my_n\vert }{\vert y_ny_m\vert } \\ &=\frac{\vert x_n \vert \vert y_m - y_n\vert + \vert y_n \vert \vert x_n - x_m\vert }{\vert y_ny_m\vert } \\ &< \frac{M_x \cdot \frac{C^2 \varepsilon}{2M_x} + M_y \frac{C^2 \varepsilon}{2M_y}}{C^2} \\ &= \varepsilon. \end{align*}

Thus $\{x_n/y_n\}$ is a Cauchy sequence by definition.

Without the $y_n \not\to 0$ hypothesis, this will not work. We need to bound the denominator away from zero. Martini's counterexample shows this brilliantly: $$x_n = 1 \to 1, \qquad y_n = \frac{1}{n} \to 0, \qquad \frac{x_n}{y_n} = n \to \infty$$ and hence it's not Cauchy because it's not bounded, and we know that every Cauchy sequence is bounded.

Suppose $x_n = 1/n^2$ and $y_n = 1/n^4$.

Then $x_n$ and $y_n$ are Cauchy but $x_n/y_n =n^2$ which is not convergent and therefore not Cauchy.

So you also need to assume that $|y_n| > r$ for some positive real $r$.