Suppose, towards a contradiction, that $(r^n)$ is not Cauchy. Then $\exists \epsilon >0$ such that for every $n\in \mathbb{N}$, $\exists m > n$ such that $|r^m – r^n| \geq \epsilon$. Then $|r^n| > |r^n(r^{m-n} – 1)| = |r^m – r^n| \geq \epsilon$.
This is as far as I got assuming only that the sequence is bounded by -1 and 1, which I'm not even sure if I'm allowed to assume.
The question is at the end of a chapter that goes over Cauchy sequences, and how they allow us to define products of reals by showing that for $x,y\in \mathbb{R}$, $\lim_{n\rightarrow \infty} (x_ny_n)$ converges, where $(x_n), (y_n)\subset \mathbb{Q}$ converge to $x,y$ respectively.
Any help would be appreciated.
Edit: Since $|r|< 1$, $|r|= (1 – k)$ for some $0<k<1$, so $r^{n+1} = r^n(1-k) < r^n$, and the sequence is strictly decreasing. Since $(|r^n|)$ is bounded below by 0, the sequence must converge, so the sequence is Cauchy.
Best Answer
You know that every convergent sequence is Cauchy.
You also know that the geometric sequence with $|r|<1$ is convergent.
Therefore the geometric sequence with $|r|<1$ is Cauchy.