Proof for the Cauchy sequence

Question

Q
Prove the following :
$[{}^\exists r\in(0,1)　s.t.　{}^\forall n\in \mathbb{N},|a_{n+2}-a_{n+1}|\leq r|a_{n+1}-a_{n}|]\Rightarrow $The sequence $\{a_n\}$ is a Cauchy sequence

I understand that $|a_{n+2}-a_{n+1}|\leq r^n|a_{2}-a_{1}|$,but I don't found the arbitrariness of the Cauchy sequence and how to choose a good epsilon.

Thanks,for help.

Answer 1

Here is an attempt:

Recall that for a sequence to be Cauchy, we need to find an $N$ such that $|a_n - a_m| < \varepsilon$ for all $n,m \geq N$.

Now, pick an arbitrary $n,m \in \mathbb{N}$, and let $n>m$ without loss of generality.

Examining $|a_n - a_m|$ we see that we can rewrite it as

$$|a_n - a_m| = |a_n -a_{n-1} + a_{n-1} - a_{n-2} + \cdots + a_{m+1} - a_m|$$

$$\leq ||a_n - a_{n-1}| + |a_{n-1} - a_{n-2}| + \cdots + |a_{m+1} - a_{m}||$$

$$\leq r^{n-1}|a_{2} - a_1| + \cdots + r^m|a_{2}-a_1|| \leq (n-m)r^m|a_{2}-a_1|$$

where the last line follows from repeated application of the assumption stated in the problem.

Since $r \in (0,1)$, we can pick an $N$ such that $\forall n,m \geq N$, $(n-m)r^m|a_2 - a_1| < \varepsilon$

And so we are done!

Edit: Fixed an inequality thanks to @infinity_hunter

Answer 2

Hint:

For $n > m >1$ $$ \mid a_n -a_m \mid= \mid a_n - a_{n-1} + a_{n-1} - \cdots + a_{m+1} -a_{m} \mid \le \mid a_n - a_{n-1} \mid + \cdots + |a_{m+1} - a_m| \le (r^{n-2} + \cdots + r^{m-1} +r^{m-2})(|a_{2}-a_{1}|)$$

Since $0 <r<1$, the sequence $\langle1+r + r^2 + \cdots + r^{n}\rangle$ is converges,hence it is Cauchy.

Can you finish it?