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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.

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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.

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  • $\begingroup$ How do you know $r^n$, $\vert r \vert < 1$ is convergent without showing it is Cauchy? Cheers! $\endgroup$ – Robert Lewis Nov 15 '18 at 0:55
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    $\begingroup$ @RobertLewis Monotone convergence theorem. $\endgroup$ – Mohammad Riazi-Kermani Nov 15 '18 at 1:16
  • $\begingroup$ Not to put too fine a point on it, but reading en.wikipedia.org/wiki/Monotone_convergence_theorem it seems it is in effect using the Cauchy definition of convergence. $\endgroup$ – Robert Lewis Nov 15 '18 at 1:19
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    $\begingroup$ Yes, one can say so. $\endgroup$ – Mohammad Riazi-Kermani Nov 15 '18 at 1:29
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Note that if $n >m$ then $|r|^n = |r|^{n-m} |r|^m < |r|^m$.

Let $\epsilon>0$ and suppose $N$ is such that $(1+|r|)|r|^N < \epsilon$. Then if $m,n \ge N$ we have (assuming that $n\ge m$ without loss of generality) that $|r^n-r^m| = |r|^m||r^{n-m}-1| \le |r|^m (1+|r|) \le (1+|r|)|r|^N < \epsilon$.

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  • $\begingroup$ What! YOU again??? Cheers, old chap! And a +1 for good measure! $\endgroup$ – Robert Lewis Nov 15 '18 at 0:37
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    $\begingroup$ @RobertLewis: Still haven't recovered from the demise of the Med :-(. Only Moe's left from my old days... $\endgroup$ – copper.hat Nov 15 '18 at 0:38
  • $\begingroup$ Yeah, me neither. I'm stuck in a Starfucks in Goleta, wherever that is! $\endgroup$ – Robert Lewis Nov 15 '18 at 0:39
  • $\begingroup$ @RobertLewis: Unfortunately their presence seems to discourage local, more quirky cafes. $\endgroup$ – copper.hat Nov 15 '18 at 0:42
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    $\begingroup$ Well, I take what I can get these days! $\endgroup$ – Robert Lewis Nov 15 '18 at 0:43
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We have, with $m \ge n$,

$\vert r^n - r^m \vert = \vert r^n(1 - r^{m - n}) \vert = \vert r^n \vert \vert 1 - r^{m - n} \vert = \vert r \vert^n \vert 1 - r^{m - n} \vert; \tag 1$

also, again with $m \ge n$,

$\vert 1 - r^{m - n} \vert \le \vert 1 \vert + \vert r^{m - n} \vert = 1 + \vert r \vert^{m - n} \le 2; \tag 2$

we combine (1) and (2) and find

$\vert r^n - r^m \vert \le 2\vert r \vert^n; \tag 3$

now with $n$ sufficiently large we have

$2 \vert r \vert^n < \epsilon \tag 4$

for any $0 < \epsilon \in \Bbb R$; in fact, (4) is obtained when

$\ln 2 + n \ln \vert r \vert < \ln \epsilon, \tag 5$

or

$n \ln \vert r \vert < \ln \epsilon - \ln 2 = \ln \left ( \dfrac{\epsilon}{2} \right ), \tag 6$

or, since $\vert r \vert < 1$ implies $\ln \vert r \vert < 0$,

$n > (\ln \epsilon - \ln 2) / \ln \vert r \vert = \ln \left ( \dfrac{\epsilon}{2} \right ) / \ln \vert r \vert; \tag 6$

thus, given $\epsilon$, taking $N \in \Bbb N$ such that

$N > \ln \left ( \dfrac{\epsilon}{2} \right ) / \ln \vert r \vert; \tag 7$

we find that

$m \ge n > N \Longrightarrow \vert r^n - r^m \vert < \epsilon, \tag 8$

that is, that $r^n$ is Cauchy.

Actually, our OP hiroshin's approach isn't all that far off, since by taking $n$ large enough we will obtain $\vert r \vert^n < \epsilon$, a decisive contradiction.

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