$\{x_n\}$ be a bounded sequence of distinct real numbers , $|x_{n+1}-x_n|<|x_n-x_{n-1}|,\forall n\in \mathbb N$ , then is $\{x_n\}$ convergent? Let $\{x_n\}$ be a bounded sequence of distinct real numbers such that $|x_{n+1}-x_n|<|x_n-x_{n-1}|,\forall n\in \mathbb N$ , then is it true that $\{x_n\}$ converges ?
The motivation for this comes from the fixed point theorem that if $X$ is compact metric space and $f:X\to X$ is a function such that $d(f(x),f(y))<d(x,y),\forall x,y \in X , x\ne y$ then $f$ has a fixed point.
 A: Following @dxiv's comment: Consider the sequence defined recursively by putting $x_1=1,x_2=\frac{1}{2}$ and 
$$x_{n+1}=\begin{cases}x_n+\frac{1}{n+1} &\text{if $\left(x_{n-1}<x_n \text{ and } x_{n+1}+\frac{1}{n+1}\leq 1\right)$ or $\left( x_{n-1}>x_n \text{ and }x_n-\frac{1}{n+1}<0\right)$}\\
x_n-\frac{1}{n+1} &\text{otherwise}\end{cases}$$
The idea is that this sequence represents the walk of a person who make the $n$-th step of size $\frac{1}{n}$, and changing direction if there is not enough room for the next step.
For every $n\geq 1$, we have $|x_{n+1}-x_n|=\frac{1}{n+1}$, the sequence is bounded by $0$ and $1$, but it is not convergent since it has subsequences converging to $0$ and $1$ (namely, those points when the person change direction).
A: $$x_n=(-1)^n\left(1+\frac{1}{n}\right)$$
A: It does not converge : Consider a sequence $r_n$ s.t. it is monotone strictly decreasing and it converges to $\frac{1}{2}$ 
Then define $$ x_1=0,\ x_2=r_1,\ x_3=r_1-r_2,\ x_4=r_1-r_2+r_3,\ \cdots $$
That is $$ x_n=r_1-r_2+\cdots + (-1)^{n}r_{n-1} $$
Note that this sequence has two subsequences which converges to different limit
