# Let $\{x_n\}$ be a Cauchy sequence of real numbers, prove that a new sequence $\{y_n\}$, with $y_n$=$x_n^\frac{1}{3}$, is also a Cauchy sequence.

Let $\{x_n\}$ be a Cauchy sequence of real numbers, prove that a new sequence $\{y_n\}$, with $y_n$=$x_n^\frac{1}{3}$, is also a Cauchy sequence.

What I am thinking so far is following:

In order to prove $\{y_n\}$ is a Cauchy sequence, we need to find an $N \in \mathbb{N}$ such that for any $m, n \ge N$, $$|y_n-y_m| = |x_n^\frac{1}{3}-x_m^\frac{1}{3}| = \frac {\vert{x_n-x_m}\vert} {\vert{x_n^\frac{2}{3}+x_n^\frac{1}{3}x_m^\frac{1}{3}+x_m^\frac{2}{3}}\vert} < \epsilon.$$ How should I proceed then?

• A lazy but rigourous way to kill this problem: use the compleness of $\Bbb R$ to establish the equivalence between Cauchy sequences and convergent sequences. – GNUSupporter 8964民主女神 地下教會 Dec 19 '17 at 23:43
• @GNUSupporter it is not a lazy way, and I don't think it kills the problem. – Gribouillis Dec 19 '17 at 23:44
• @Gribouillis My point is to state (but not prove) these facts. Since the cubic root is continous ($x\mapsto x^3$ is strictly $\uparrow$ and continous on $\Bbb R$, so has a continous inverse.) $y_n=f(x_n)$ where $f$ is the cubic root – GNUSupporter 8964民主女神 地下教會 Dec 19 '17 at 23:48

An idea: if $\;\lim\limits_{n\to\infty}x_n=0\;$ , then clearly also $\;x^{1/3}\to0\;$ and thus $\;\left\{x^{1/3}\right\}\;$ is Cauchy and there's nothing more to prove.

Otherwise: if $\;\lim\limits_{n\to\infty}x_n=x\neq0\;$, then there exists $\;\delta>0\;$ and $\;N\in\Bbb N\;$ s.t. $\;n>N\implies |x_n|\ge\delta\;$ , and thus from your calculation and for $\;n,m>N\;$:

$$\left|x_n^{1/3}-x_m^{1/3}\right|=\frac{|x_n-x_m|}{\left|x_n^{2/3}+x_n^{1/3}x_m^{1/3}+x_m^{2/3}\right|}\le\frac{|x_n-x_m|}{3\delta}$$

and now use that $\;\{x_n\}\;$ is Cauchy...

• Once you say that $\lim x_n = x$, the only thing you need is $\lim x_n^{1/3} = x^{1/3}$ – Gribouillis Dec 19 '17 at 23:46
• @Gribouillis True, yet I wanted to use as similar a condition to the usual definition of Cauchy Sequences as possible. Observe that the first part is usually covered even before arithmetic of limits... – DonAntonio Dec 19 '17 at 23:51
• See however an alternate method below! – Gribouillis Dec 20 '17 at 0:28

Lemma: For $x, y\in{\mathbb R}$, one has $$|x^{1/3} - y^{1/3}|\le 3 |x - y|^{1/3}$$ Proof

Suppose that $|x|\le |x-y|$. It implies $|y|\le |x| + |y-x|\le 2|x - y|$. Hence $|x^{1/3} - y^{1/3}|\le |x|^{1/3}+ |y|^{1/3}\le (1 + 2^{1/3})|x-y|^{1/3}\le 3 |x-y|^{1/3}$. The same proof holds if one supposes $|y|\le |x-y|$

Suppose now that $|x-y|< \min(|x|, |y|)$. It follows that $x$ and $y$ have the same sign. We may suppose that they are positive. The mean value theorem gives for a $z\in (x, y)$

$$|x^{1/3} - y^{1/3}|\le \frac{1}{3} z^{-2/3}|x - y|\le \frac{1}{3} |x - y|^{-2/3}|x - y|\le \frac{1}{3}|x - y|^{1/3}$$ $\square$

The result is now obvious because for $\epsilon>0$, one can obtain $|x_n^{1/3} - x_m^{1/3}|\le \epsilon$ by imposing $|x_n - x_m|\le\epsilon^3/27$

• Very nice... +1 – DonAntonio Dec 20 '17 at 1:02

Hint:

Cauchy sequences are bounded.

Quick and dirty approach:

1. Since $\mathbb R$ is complete, $\{x_n\}$ converges to some $x\in\mathbb R$.
2. The function $z\mapsto z^{1/3}$ in continuous on $\mathbb R$, which implies that $x_n^{1/3}\to x^{1/3}$ as $n\to\infty$.
3. Therefore, the sequence $\{y_n\}$ is convergent and hence Cauchy.