Proving that $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence.

Solution 1:

For every $\epsilon>0$, take a natural number $N$ such that $r^N <(1-r)\epsilon$, for example by taking $N=\lfloor\frac{\ln (1-r)\epsilon}{\ln r}\rfloor+1$. Then, for all $m,n\geq N$, assume $m<n$, we have \begin{align} |x_n - x_m|&=|(x_n - x_{n-1}) + (x_{n-1} - x_{n-2}) + ... + (x_{m+1} - x_m)|\\ &\leq |(x_n - x_{n-1})| + |(x_{n-1} - x_{n-2})| + ... + |(x_{m+1} - x_m)|\\ &< r^{n-1}+\dots+r^m\\ &=r^m(1+r+r^2+\dots+r^{n-m-1})\\ &<\frac{r^m}{1-r}\\ &\leq\frac{r^N}{1-r}\\ &<\epsilon \end{align}

Solution 2:

You are definitely on the right track. Here's what you want to do: suppose without loss of generality that $m>n$ then

$$\begin{align}|x_m-x_n| &= |x_m-x_{m-1}+x_{m-1}-\cdots-x_n| \\ &\le |x_m-x_{m-1}|+\cdots+|x_{n+1}-x_n| \\ &= r^m+r^{m-1}+\cdots+r^n \\ &= r^n(1+r+\cdots+r^{m-n}) \\ &= r^n\frac{1-r^{m-n+1}}{1-r}.\end{align}$$

If you can make this less than $\varepsilon$, you'll be done.

Solution 3:

This is exactly that: $$ x_{n+p} - x_n = \sum_{k=1}^p x_{n_k} - x_{n_{k-1}} $$ $$ |x_{n+p} - x_n| \le \sum_{k=1}^p |x_{n+k} - x_{{n+k-1}}| \le \sum_{k=1}^p r^{n+k-1} \\ \sum_{k=1}^\infty r^{n+k-1} = \frac{r^{n}}{1-r} \to_{n\to\infty} 0 $$ So eventually, $$\sup_{p\in\mathbb N} |x_{n+p} - x_n| \to_{n\to\infty} 0 $$