$\left|x_{n+1} - x_n\right| < C\left|x_n - x_{n-1}\right|$ prove this is Cauchy

Question

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.

Answer 1

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 $\;0<C<1\;$

Answer 2

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.