# Proof for the Cauchy sequence

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.

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

• 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 – Infinity_hunter Jul 22 at 4:08
• @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. – Jesse Irwin Jul 22 at 4:38
• 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 $(n-m)r^m |a_2-a_1| > \varepsilon$ – Infinity_hunter Jul 22 at 5:52

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

Can you finish it?

• 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? – Math-Learner Jul 22 at 6:51
• @ Math-Learner Why do you think so? – Infinity_hunter Jul 22 at 6:57
• Because when you apply triangle inequality you have paired all the $a_i$'s. Hence there must be an even number of $a_i$! – Math-Learner Jul 22 at 7:14
• @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 $i = m+1, \dots ,n-2,n-1$ – Infinity_hunter Jul 22 at 7:27
• Thanks for clearing my doubt – Math-Learner Jul 22 at 15:25