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!
 A: 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 $r$.
A: 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.
