Let $\{x_n\}$ be a sequence such that there exists a $0 < C < 1$ such that $$|x_{n+1} - x_n| \le C |x_n - x_{n-1}|.$$ Prove that $\{x_n\}$ is Cauchy. Hint: You can freely use the formula (for $C \not = 0$) $$1+C+C^2+...+C^n = \frac {1-C^{n+1}}{1-C}.$$
My attempt : $\frac {|x_{n+1} -x _n|}{|x_n - x_{n-1}|}<C$. Then, $\frac {|x_{n+1} -x _n|}{|x_n - x_{n-1}|}<1$. So, I thought that it goes to zero as $n \to \infty$ because the difference between $x_n$ and $x_{n+1}$ is getting smaller as $n \to \infty.$ But, I don't know how to use the hint in the question, and how to write the answer properly. Could you give some hint?
Two ideas are needed. First the difference $x_n-x_{n-1}$ tends to $0$. This is an immediate consequence of the inequality given in question. Just apply it recursively to get $$|x_n-x_{n-1}|\leq C^{n-2}|x_2-x_1|$$ and you are done.
The next idea deals with the hint given in question. Let $p$ be a positive integer then $$|x_{n+p} - x_n|\leq \sum_{i=1}^{p}|x_{i+n}-x_{i-1+n}|\\\leq (1+C+C^2+\dots+C^{p-1})|x_{n+1}-x_n|<\frac{|x_{n+1}-x_n|}{1-C}$$ Now it should be obvious that $\{x_n\} $ is Cauchy as RHS tends to $0$.