Proving Cauchy Given Given Sequence Terms Arbitrarily Close

Question

I need to show that $\{x_{n}\}$ is Cauchy given that there exists $0<C<1$ s.t. $|x_{n+1}-x_{n}|\leq C|x_{n}-x_{n-1}|$. Intuitively, that statement clearly implies $\{x_{n}\}$ is Cauchy, since it implies the sequence terms become arbitrarily close. But how to make it precise?

Couldn't it also be said from the given information that the sequence is either monotone increasing or monotone decreasing and bounded? Then it would converge, which means it is Cauchy.

Thanks for any assistance!

Answer 1

Expanding robjohn's hint, if $|x_{n+1}-x_{n}|\leq C|x_{n}-x_{n-1}| $, then $|x_{n+2}-x_{n+1}| \leq C|x_{n+1}-x_{n}| \leq C(C|x_{n}-x_{n-1}|) \leq C^2|x_{n}-x_{n-1}| $.

Continuing this, what can you show about $|x_{n+3}-x_{n+2}| $? $|x_{n+4}-x_{n+3}| $? $|x_{n+k}-x_{n+k-1}| $?

Why is it essential that $0 < C < 1$?

Answer 2

Hint: The triangle inequality gives $$ |x_n-x_m|\le\sum_{k=m}^{n-1}|x_{k+1}-x_k| $$ Can you bound $|x_{k+1}-x_k|$ in terms of $|x_1-x_0|$ (think inductively)?

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Since everything is in absolute values, there is no way to tell if the sequence is increasing or decreasing.