Show that sequence is Cauchy? $|x_{n+1} − x_n| < r^n$

Question

Let $0 ≤ r < 1$ be given and let $(x_n)$ be a sequence such that $|x_{n+1} − x_n| < r^n$ for any $n ∈ N$. Show that the sequence is Cauchy. My proof starts from this: We have Cauchy's definition $\forall \epsilon>0$ $|a_n-a_m|< \epsilon $. And I write like this $||x_{n+1} − x_n| - |x_{m+1}- x_m||< ?$. And here, the problem is that it must be less than what $r^n$ or $r^m$. Maybe I start wrong?

Answer 1

You should write correctly the definition of convergence in sense of Cauchy. It is: $\forall \epsilon > 0$ there exists $p \in \mathbb{N}$ such that for all $n, m \ge p, \hspace{0.2cm} |x_n - x_m| \le \epsilon$. This definition is the same as the following: $\underset{p \in \mathbb{N}}{\sup} |x_{n+p} - x_n| \to 0$ as $n \to + \infty$. That being said you have for all $p \ge 1$:

\begin{align*} |x_{n+p} - x_n| &\le \sum_{k = 0}^{p-1} |x_{n+k + 1}-x_{n+k}|\\ &\le \sum_{k = 0}^{p-1} r^{n+k}\\ &= r^n\frac{1 - r^p}{1 - r}\\ &\le \frac{r^n}{1 - r} \end{align*}

Then

$$\underset{p \in \mathbb{N}}{\sup} |x_{n+p} - x_n| \le \frac{r^n}{1 - r}$$

The right hand side of the inequality tends to $0$ as $n \to + \infty$. Which implies convergence in sense of Cauchy