# Proving a sequence is Cauchy given some qualities about the sequence

I've got a sequence $x_n$ such that I've proved $b\leq x_n \leq c$, and $|x_{n+1}-x_{n}|\leq \frac{4}{9}|x_n-x_{n-1}|$

However I'm not very familiar with Cauchy sequences, so I don't know how to exactly prove it's Cauchy. I've gotten so far:

Let $m,n>0$. Then $m=n+a$. So \begin{align*} |x_m-x_n| &= |x_{n+a}-x_n|\\ &= |(x_{n+a}-x_{n+a-1})+(x_{n+a-1}-x_{n+a-2})+\ ...\ +(x_{n+1}-x_n)|\\ &\leq a|x_n-x_{n-1}| \end{align*}

I know I'm close, but I'm not sure where to go from here.

EDIT: \begin{align*} |x_m-x_n| &= |x_{n+a}-x_n|\\ &= |(x_{n+a}-x_{n+a-1})+(x_{n+a-1}-x_{n+a-2})+\ ...\ +(x_{n+1}-x_n)|\\ &\leq \Sigma_{i=1}^{i=a}(\frac{4}{9})^i * |x_{n}-x_{n-1}|\\ &\leq (c-b)\frac{4}{9}\cdot \frac{1-(4/9)^a}{5/9}\\ &\leq \frac{4}{5}(c-b)(1-(\frac{4}{9})^a) \end{align*}

But now I'm not sure how to say it's Cauchy.

EDIT 2:

$$\frac{4}{5}(c-b)(1-(\frac{4}{9})^a) \leq \frac{4}{5}(c-b)$$

So let $n_0 = \epsilon * \frac{5}{4(c-b)}$. Then for any $\epsilon > 0$, $|x_m-x_n|<\epsilon$ whenever $m,n > \frac{5}{2}\epsilon$.

• Hint: geometric series of ratio $4/9$. – Julien Oct 22 '13 at 1:44
• Ah! I Think I get it. I'll get a sum from 1 to a of $\frac{4}{9}|x_n-x_{n-1}|$ which simplifies to (4/9) * ((4/9)^a -1) / ((4/9) - 1) |x_n - x_(n-1)| correct? – druckermanly Oct 22 '13 at 1:53
• I can't read your comment, but the idea is that $|x_{k}-x_{k-1}|\leq (4/9)^{k-1}|x_1-x_0|$ by induction. And then you can use that after you second equality to get a sharper estimate. – Julien Oct 22 '13 at 1:58
• Haha, sorry. Forgot to use tex for a bit. I just finished up the problem (I think) so do you think you could do a quick look-over to see if I've got any flaws in my logic / reasoning? – druckermanly Oct 22 '13 at 2:25
• It does not work, because your upper bound does not tend to $0$ when $a$ tends to $\infty$. You need to go all the way down to $|x_1-x_0|$ like I mentioned in my previous comment. And you will not need these $b, c$. – Julien Oct 22 '13 at 2:29

What you have is that $x_{n}$ is a contractive sequence, such that $|x_{n+1} - x_{n} | \le C |x_{n} - x_{n-1}|$ for all $n \ge N$, $0 < C < 1$. You have $C = \frac{4}{9}$. Theorem : A contractive sequence is a Cauchy sequence, so it is enough to use the theorem.
But if you also want to prove it (you should try), in general it is the same step as you have so far, then use the hint from @julien with geometric series of ratio $C$.
• I edited my post to show how far I've gotten-- what do I let $n_0$ equal so that for all $n>n_0, m>n_0$ I've got the full $\epsilon$ proof? – druckermanly Oct 22 '13 at 2:08
• Wait: $$\frac{4}{5}(c-b)(1-(\frac{4}{9})^a) \leq \frac{4}{5}(c-b)$$ So let $n_0 = \epsilon * \frac{5}{4(c-b)}$ Correct? – druckermanly Oct 22 '13 at 2:14
• for all $n$, $|x_{n+2}-x_{n+1}| \le C |x_{n+1} - x_{n}| \le C^2 |x_{n} - x_{n-1}| \le .... \le C^n |x_2 - x_1|$. Now for $m > n$, $$|x_{m} - x_{n}| \le |x_{m} - x_{m-1}| + |x_{m-1} - x_{m-2}| + ... + |x_{n+1} - x_n| \le .....$$ , Evaluate the right side limit as $n$ $\rightarrow$ $\infty$. – Arief Anbiya Oct 22 '13 at 2:29
• @user2899162 What my answer meant is that use the argument above, but your $\epsilon$ argument also suffice.. – Arief Anbiya Oct 22 '13 at 2:38