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.

  • $\begingroup$ 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

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

  • $\begingroup$ 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


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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.