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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.

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  • $\begingroup$ you don't choose $\varepsilon$. $\endgroup$ – Yorch Jul 22 at 3:23
  • $\begingroup$ You take a generic $\epsilon >0$ and then show there is some $N$ such that for all $n,m\geq N$, $|a_n-a_m|<\epsilon$ $\endgroup$ – Alan Jul 22 at 3:25
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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

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  • $\begingroup$ Taking $n=2, m=1$ in your final inequality gives $|a_2 - a_1| < r^2|a_2-a_1|$ which means that $r^2>1$ and so $r>1$ which is of course not possible $\endgroup$ – Infinity_hunter Jul 22 at 4:08
  • $\begingroup$ @Infinity_hunter You're correct, thanks. I believe It should be $|r^{n}|a_{2}-a_1| + r^{(n-1)}|a_{2} - a_1| + \cdot + r^m|a_{2}-a_1|| \leq (n-m)r^m|a_{2}-a_1|$, where we exclude your case n=2, m=1 (as the hypothesis doesn't apply to this case.) And this last expression still converges to zero, so the result still holds. $\endgroup$ – Jesse Irwin Jul 22 at 4:38
  • $\begingroup$ I think still there is a problem. It is not clear why you can choose $N$ such that for given $\varepsilon$ , $\forall n,m \geq N ,(n-m)r^m|a_2 - a_1| < \varepsilon$. If it is possible, for given $\varepsilon$, keeping $m$ fixed and by using Archimedian property we can find a $n$ such that $(n-m)r^m |a_2-a_1| > \varepsilon$ $\endgroup$ – Infinity_hunter Jul 22 at 5:52
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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?

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  • $\begingroup$ I think your calculation is true only if $n$ and $m$ are either both even or both odd, not when one is even and the other is odd. Isn't it? $\endgroup$ – Math-Learner Jul 22 at 6:51
  • $\begingroup$ @ Math-Learner Why do you think so? $\endgroup$ – Infinity_hunter Jul 22 at 6:57
  • $\begingroup$ Because when you apply triangle inequality you have paired all the $a_i$'s. Hence there must be an even number of $a_i$! $\endgroup$ – Math-Learner Jul 22 at 7:14
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    $\begingroup$ @Math-Learner Let me give you an example, $|a_5 -a_2| = |a_5 -a_4 +a_4 - a_3 +a_3 -a_2| \le |a_5-a_4| + |a_4-a_3| + |a_3-a_2|$. I am just adding and subtracting $a_i$ for $i = m+1, \dots ,n-2,n-1$ $\endgroup$ – Infinity_hunter Jul 22 at 7:27
  • $\begingroup$ Thanks for clearing my doubt $\endgroup$ – Math-Learner Jul 22 at 15:25

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