# Prove the geometric sequence $(r^n)$ is Cauchy if $|r|<1$

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 $$0, so $$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.

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.

• How do you know $r^n$, $\vert r \vert < 1$ is convergent without showing it is Cauchy? Cheers! – Robert Lewis Nov 15 '18 at 0:55
• @RobertLewis Monotone convergence theorem. – Mohammad Riazi-Kermani Nov 15 '18 at 1:16
• 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. – Robert Lewis Nov 15 '18 at 1:19
• Yes, one can say so. – Mohammad Riazi-Kermani Nov 15 '18 at 1:29

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 $$|r^n-r^m| = |r|^m||r^{n-m}-1| \le |r|^m (1+|r|) \le (1+|r|)|r|^N < \epsilon$$.

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

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.