How to pick $N$ for Cauchy sequence?

Question

I have to prove the sequence $x^n$ is a Cauchy sequence when $|x| \lt 1$.

In the book example that I saw when they proved that the sequence $\frac1n$ was Cauchy they seemed to just pick $N$ as $\frac2\varepsilon$ as their first step without explanation other than the fact that if you did the followup work...it worked (basically the same process seen in example 1 on this page).

Do I need to pick some $N$ before proceeding, and how do I know what that $N$ is? If not, then how do I approach the problem? I know that I have to use the definition: for all $\varepsilon \gt 0$ there exists some $N$ such that if $n, m \ge N$ then $|x_n - x_m| \lt \varepsilon$.

Answer 1

Fix $\varepsilon>0$ and $x \in (-1, 1)$.

You want to find $N$ such that for all $n, m\geq N$ we have \begin{align} |x^n-x^m|<\epsilon. \end{align} WLOG, assume $n>m$, then this means \begin{align} |x^m||x^{n-m}-1|<|x|^m(|x|^{n-m}+1)<2|x|^m<2|x|^N. \end{align} In particular, if you want $|x_n-x_m|$ to be less than $\varepsilon$ then a sufficient condition would be to have $2|x|^N<\varepsilon$, which is equivalent to having $N<\frac{\log \frac{\varepsilon}{2}}{\log|x|}$.

Answer 2

Let $x $ such that $|x|<1$.

we have $\forall n,p\geq 0$

$$|x^{n+p}-x^n|=|x|^n|x^p-1|\leq 2|x^n|.$$

on the other hand $$\lim_{n\to+\infty} x^n=0$$

thus, given $\epsilon>0$, $$\exists N\geq 0 \;\; : \; n>N\implies |x^n|<\frac {\epsilon}{2} $$

or $$\forall p\geq 0 \;\;\forall n>N \;\;|x^{n+p}-x^n|<\epsilon $$

which means that $(x^n) $ is Cauchy.