$\left|x_{n+1} - x_n\right| < C\left|x_n - x_{n-1}\right|$ prove this is Cauchy [closed]
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.
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 $\;0<C<1\;$