Proving Cauchy when given a sequence

Question

Let $\left\{x_n\right\}$ be a sequence and $0 < a < 1$. Suppose that for all $n \ge 3$ we have

$$ \left\lvert x_n - x_{n-1}\right\rvert \le a\left\lvert x_{n-1} - x_{n-2} \right\rvert. $$

Prove that $\left\{x_n\right\}$ is Cauchy.

I don't even know where to start here. To prove a sequence is cauchy I have to somehow reach the conclusion of $\left\lvert x_m - x_{k}\right\rvert < \varepsilon$ right? How do I even do that with this inequality. I'm completely lost.

The triangle inequality is very useful here. – Harambe Mar 23 '17 at 05:03 — Harambe, Mar 23 '17 at 05:03
Consider $|x_n-x_{n-1}|$ as a geometric sequence – Mengchun Zhang Mar 23 '17 at 05:06 — Mengchun Zhang, Mar 23 '17 at 05:06

Mark Viola · Accepted Answer · 2017-03-23T05:25:57.723

2

HINT:

$$|x_m-x_k|\le |x_m-x_{m-1}|+|x_{m-1}-x_{m-2}|+\cdots|x_{k+1}-x_k|$$

and

$$|x_n-x_{n-1}|\le a|x_{n-1}-x_{n-2}|\le a^2|x_{n-2}-x_{n-3}|\le\cdots \le a^{n-1}|x_1-x_0|$$

edited Mar 23 '17 at 05:25

answered Mar 23 '17 at 05:06

Mark Viola

179,405

Proving Cauchy when given a sequence

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