Let $x_1=1$ and $x_{n+1}=\tfrac{1}{1+x_n}$. I want to proof that for every $n,m\in \mathbb{N}$ with $m\leq n$, the inequality $|x_{n}-x_{m}|\leq 2^{-m}$ is true.
This can be done by induction over $m$. Two facts I found out are
- $\tfrac{1}{2} \leq x_n \leq 1$
- For every $n,m\in \mathbb{N}$ the inequality $|x_{n+1}-x_{m+1}|\leq \tfrac{1}{2}|x_n-x_m|$ is true.
Now the induction: Begin: or $m=1$ we can see with 1. that it is true. The Hypothesis is $|x_{n}-x_{m+1}|\leq 2^{-(m+1)}$
With 2. we can see that $ |x_{n+1}-x_{m+1}| \leq \tfrac{1}{2}|x_n-x_m| \leq \tfrac{1}{2} 2^{-m} = 2^{-(m+1)}$
But how to get from $|x_{n}-x_{m+1}|$ to $|x_{n+1}-x_{m+1}|$?
Thanks in advance.
I guess I solved it by myself.
$|x_{n}-x_{m+1}| \leq |x_{n+1}-x_{m+1}|$ because $x_n \leq x_{n+1}$ for every $x_n \in (-1, \tfrac{-1+\sqrt{5}}{2}]$ and as $\lim(x_n)=\tfrac{-1+\sqrt{5}}{2}$ for $n \rightarrow ∞$ the statement is true.