If $(x_n)$ is a Cauchy sequence that does not tend to zero, prove that $\left(\frac{x_n}{\|x_n\|}\right)$ is Cauchy Let $X$ be a normed linear space. Prove that if $(x_n)$ is a Cauchy sequence of nonzero elements of $X$ that does not converge to zero, then $\left(\frac{x_n}{\|x_n\|}\right)$ is also a Cauchy sequence.
 A: Since $x_n$ does not converge to $0$, there is $\epsilon > 0$ such that $\|x_n\| > \epsilon$ for infinitely many $n$.  From the fact $x_n$ is Cauchy, take $N$ so $\|x_n - x_m\| < \epsilon/2$ for $m,n > N$, and conclude $\|x_n\| > \epsilon/2$ for all $n > N$.
Then for $m, n > N$,
$$\eqalign{\left\| \frac{x_n}{\|x_n\|} - \frac{x_m}{\|x_m\|}\right\| &= \left\|\frac{x_n - x_m}{\|x_n\|} + x_m \frac{\|x_m\|-\|x_n\|}{\|x_n\| \|x_m\|}\right\|\cr
&\le \frac{\|x_n - x_m\|}{\|x_n\|} + \frac{\left|\|x_n\| - \|x_m\|\right|}{\|x_n\|}\cr
&\le \frac{4}{\epsilon} \|x_n - x_m\|}$$
and this $\to 0$ as $m, n \to \infty$.
A: By the reverse triangle inequality, the real sequence $(\|x_n\|)$ is also Cauchy, hence converges. Moreover, its limit is nonzero by hypothesis. Hence $\left(\frac1{\|x_n\|}\right)$ is convergent.
Now by the usual method, the product of the two Cauchy sequences $\left(\frac1{\|x_n\|}\right)$ (of reals) and $(x_n)$ (of vectors) is Cauchy.
Note that we could as well avoid to use the completeness of $\mathbb R$, proving directly that $\left(\frac1{\|x_n\|}\right)$ is Cauchy.
