I'm facing this problem,
Let $\{ x_n \}_{ n\in \mathbb{N} }$ be a sequence of real numbers and $0 < r< 1$ such that $|x_{n+1} - x_n | < r^n$ for all $n \in \mathbb{N}$. Prove that $\{ x_n \}_{ n\in \mathbb{N}}$ is a Cauchy sequence.
My work based on other similar problems is:
Let's consider $|x_m-x_n|$ where $m$ y $n$ do not need to be consecutive numbers, assume $m\leq n$. Then $k=n-m\geq 0$,
$\\ \begin{aligned} |x_m-x_n| &=|x_m-x_{m+k}| \\ & =|x_m-x_{m+1}+x_{m+1}-x_{m+2}+...+x_{m+k-1}-x_{m+k}| \\ & \leq|x_{m}-x_{m+1}|+|x_{m+1}-x_{m+2}|+...+|x_{m+k-1}-x_{m+k}| \\ & < r^{m}+r^{m+1}+...+r^{m+k-1} \\ & = r^{m}(1+r^1+...+r^{k-1}) \\ & = (r^{m})\sum_{i=0}^{k-1}r^{i} \\ & \leq r^m(1/(1-r)). \\ \end{aligned}$
$\\ $ If $m>N_{\epsilon}$, then we have $r^{m}<r^{N_{\epsilon}}$ because $0<r<1$ (I think I need a strong argument to guarantee this). Then $r^m(1/(1-r))<r^{N_{\epsilon}}(1/(1-r))$
$\\ $ Can I guarantee that always exists $N_{\epsilon}\in \mathbb{N}$, such that $r^{N_{\epsilon}}(1/1-r)<\epsilon$ ? If so, then I get what I want
$$\\ |x_m-x_n|<r^m(1/(1-r))<r^{N_{\epsilon}}(1/(1-r))<\epsilon$$ $\ $ Whenever $m,n>N_{\epsilon}$
My greatest doubt. If $m>N_{\epsilon}$, then we have $r^{m}<r^{N_{\epsilon}}$ (Whenever $0<r<1$). Will this always be true? I'm thinking of mathematical induction to prove this.
Generally if you didn't express doubts about some steps this proof would have been quite widely accepted.
Your first concern is if you can assume that if $m>N_\epsilon$ then $r^m<r^{N_\epsilon}$ or a proof of that. Now we have that $N_\epsilon = m + d$ for some $d>0$ so $r^m-r^{N_epsilon} = r^m - r^{m+d} = r^m(1-r^d)$. So what you need to know is that $0<r^d<1$ which is quite straight forward to prove with induction.
Your second concern is if you can guarantee that $r^{N_\epsilon}(1/(1-r))$ can be made arbitrarily small by selecting $N_\epsilon$. First of all $r$ is just a constant so we should focus on $r^{N_\epsilon}$. It's a well known fact and you could use the estimate that if $h>0$ then $(1+h)^n \ge 1+nh$ (which can be shown unsing induction) to see that
$$r^{N_\epsilon} \le {1\over (1+ (1/r-1))^{N_\epsilon}} \le {1\over1 + N_\epsilon(1/r-1)} \le {1\over N_\epsilon}{1\over1/r-1}$$
Which can be made arbitrarily small (while we know that it's also positive).