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Suppose that for the sequence $\{x_n\}$ there exists $0 < C < 1$ such that $$\left|x_{n+1} - x_n\right| < C\cdot \left|x_n - x_{n-1}\right|.$$ Prove that ${x_n}$ is Cauchy.

TAPLON
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  • A useful formatting guide can be found here http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – lulu Apr 25 '16 at 19:39

3 Answers3

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Use condition several times

$$|x_{n+1}-x_n|\le C |x_n-x_{n-1}|\le C^2|x_{n-1}-x_{n-2}|\le...\le C^n|x_1-x_0|$$

and for $\;p>0\;$

$$|x_{n+p}-x_n|=|x_{n+p}-x_{n+p-1}+x_{n+p-1}-x_{n+p-2}+...+x_{n+1}-x_n|\le$$

$$\le\sum_{k=0}^{p-1}|x_{n+p-k}-x_{n+p-1-k}|\le\sum_{k=0}^{p-1}|x_1-x_0|C^{n+p}$$

and last sum is converging since it is geometric and $\;0<C<1\;$

  • how do you know that the condition implies that you can apply several times like that? – TAPLON Apr 25 '16 at 19:58
  • @JustinStevenson I'm unsure I understand, In first line I use condition every time. –  Apr 25 '16 at 20:15
  • I know this has been inactive for a while, but im going back and reviewing your response and in the last sum should your index be from k=0 or from n=0. Just wondering cause there is not k's in the formula. – TAPLON May 01 '16 at 19:29
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Sketch of proof:

  1. Prove that $\lvert x_{n+k}-x_{n+k-1}\rvert\le C^k\lvert x_n-x_{n-1}\rvert$

  2. Use the above to prove that $\lvert x_{n+m}-x_n\rvert\le \frac{1-C^{m}}{1-C}\lvert x_{n+1}-x_n\rvert\le\frac{1}{1-C}\lvert x_{n+1}-x_n\rvert$

  3. Use these inequalities (in particular, the first one with $n=1$) to prove the sequence is Cauchy.

  • How do you know that you are able to generalize the condition and still have it hold true. – TAPLON Apr 25 '16 at 19:59
  • @JustinStevenson Typically, properties on sequences written as $$\lvert x_{n+1}-x_n\rvert <C\lvert x_n-x_{n-1}\rvert$$ are (not so) implicitly meant to hold for all $n$. –  Apr 25 '16 at 20:03
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We see that $ x_{n+1}- x_n \le c|x_n-x_{n-1}| \le c^2|x_{n-1}-x_{n-2}| \le ... \le c^{n-1}|x_2-x_1|$ Let$ m\gt n$ Then$ |u_m-u_n|\le|u_m -u_{m-1}|+...+|u_{n+1}-u_n| \le|u_2-u_1|\{c^{m-3}+c^{m-4}+...+c^{n-2}\} \le|u_2-u_1|c^{n-2}.\frac{1-c^{m-n-1}}{1-c} \le\frac{c^{n-2}}{1-c}|u_2-u_1|$ Let$\epsilon >0$, Since$0 \lt c\lt1$,the sequence $\{c_{n-2}\}$ is a convergent sequence. Therefore there exists a natural number P such that $$\frac{c^{n-1}}{1-c}|u_2-u_1|\lt\epsilon$$ for all m,n$\gt P$ It follows that $|u_m-u_n|\lt\epsilon$ for all m,n $\gt P$ and this proves that,$\{x_n\}$ is a Cauchy sequence and it's convergent.

Nimu Basak
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