Past final exam question for an intro to Real Analysis course:
Let $C > 0$, $0<r<1$ and suppose that $\forall n\in \mathbb N, |x_{n+1} – x_n| \leq Cr^n$. Please help me prove that $(x_n)$ is a Cauchy sequence. (We can assume the $\lim\limits_{n\to\infty} r^n=0,$ for $0<r<1$)
So, I know that a Cauchy series must satisfy $|x_{n}-x_m| < \epsilon$ for any $\epsilon>0, \in \mathbb R$ and for all $n,m \gt H(\epsilon) \in \mathbb N$ Note that there can't be any conditions on n and m (I saw somewhere else someone required $m>n$ which you can only do if you're proving its not at Cauchy sequence, right?)
Another way of doing this is showing that it is contractive (and thus a Cauchy series) if there is a constant $a$ such that $|x_{n+1}-x_n| \leq a|x_n-x_{n-1}|$
Clearly I'm supposed to make use of $\lim\limits_{n\to\infty} r^n=0,$ for $0<r<1$… But I don't even know how where to start with this. As I am bumbling through this problem, a more thorough answer would be much appreciated. Thanks!
Best Answer
Let $\epsilon>0$ be given. Then we can find $N$ such that $r^N\frac C{\color{red}{1-r}}<\frac\epsilon2$ because $r^n\color{red}\to0$ as $n\to \infty$. Then for any $n>N$ we have $$\begin{align} |x_n-x_N|&\le|x_{N+1}-x_N|+\ldots+ |x_n-x_{n-1}|\\&\le Cr^N+Cr^{N+1}+\ldots +Cr^{n-1} \\&=Cr^N\cdot(1+r+\ldots + r^{n-N-1}) \\&\color{red}<Cr^N\sum_{k=0}^\infty r^k \color{red}= Cr^N\cdot\frac1{1-r}<\frac\epsilon2,\end{align}$$ hence for $n,m>N$ $$ |x_n-x_m|\le|x_n-x_N|+|x_m-x_N|<\frac\epsilon2+\frac\epsilon2=\epsilon.$$
Remark: I've marked all places in red where we made use of $0<r<1$.