$\{x_n\}$ 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? [duplicate]

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.

marked as duplicate by user223391, Claude Leibovici, Did real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 26 '16 at 7:50

• Why the downvotes? – Darío G Oct 26 '16 at 4:44
• It should be straightforward to adapt this answer for the strict inequality. – dxiv Oct 26 '16 at 5:07
• Voting to reopen. This is not a duplicate $-$ the condition $|x_{n+1}-x_n|<|x_n-x_{n-1}|$ is not at all the same as $|x_n-x_{n+1}|\rightarrow 0$. – TonyK Oct 31 '16 at 12:08

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).

$$x_n=(-1)^n\left(1+\frac{1}{n}\right)$$

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