Cauchy Sequence Proof using Power series

Question

Let {${x_n}$} be a sequence such that there exists a $0<C<1$, such that: $|x_{n+1}-x_n|\leq$$C|x_n-x_{n-1}|$. Prove that {$x_n$} is a Cauchy Sequence.

I'm given a hint that $1+C+C^2...+C^n=\frac{1-C^{n+1}}{1-C}$, which I've used to construct:

$|x_m-x_n|\leq |x_n-x_{n-1}|\frac{1-C^{k+1}}{1-C}$

where $k=m-n$,but I'm not sure how to use the definition of a Cauchy sequence to proceed. Any help is greatly appreciated.

Answer 1

You're almost there. Just observe that $\frac{1-C^{k+1}}{1-C} \leq \frac{1}{1-C}$ and $|x_n - x_{n-1}| \leq C^{n-2} \cdot |x_2 - x_1|$.

Are you able to conclude now?

Answer 2

By induction, we may obtain $$|x_{n+1}-x_n|\leq C^{n-1}|x_2-x_1|,$$where $n=1,2,\cdots.$Thus, for $p=1,2,\cdots$, we have \begin{align*}|x_{n+p}-x_n|&=|\sum_{k=n}^{n+p-1}(x_{k+1}-x_k)|\\&\leq \sum_{k=n}^{n+p-1}|x_{k+1}-x_k|\\&\leq |x_{2}-x_1|\cdot\sum_{k=n}^{n+p-1}C^{k-1}\\&\leq |x_{2}-x_1|\cdot \sum_{k=n}^{\infty}C^{k-1}\\&=|x_2-x_1|\cdot \frac{C^{n-1}}{1-C}\end{align*} Here, notice that $0<C<1$ and $\dfrac{|x_2-x_1|}{1-C}$ is a positive constant. Thus, for any $\varepsilon>0$ and all $p=1,2,\cdots$, we may always choose a $N>0$ such that $$|x_{n+p}-x_n|\leq|x_2-x_1|\cdot \frac{C^{n-1}}{1-C}<\varepsilon$$ when $n>N,$ which is just what Cauchy's convergence text requires.