Given a real number $x$, define the sequence $u$ by :
$$\forall n\in\mathbb{N},\quad u_n=\frac{\lfloor 10^nx\rfloor}{10^n}$$
It is well known that $\lim_{n\to\infty}u_n=x$ and that sequence $u$ is non-decreasing.
Now consider, instead of $u$, the sequence $v$ defined by :
$$\forall n\in\mathbb{N}^\star,\quad v_n=\frac{\lfloor nx\rfloor}n$$
This sequence converges to $x$ too. But in general, it is not monotonic. For example, if we choose $x=\frac32$, it is not difficult to show that the sequence $v$ is oscillating ...
My question : Given an arbitrary sequence $\epsilon$ whose terms are all in $\{-1,1\}$, does it exist $x\in\mathbb{R}$ such that $\epsilon_n\left(v_{n+1}-v_n\right)\geqslant0$ for all $n\geqslant1$ ?
If we allow $x\in\mathbb Z$, then $v_n=x$ for all $n$, so any $\epsilon$ can be obtained by any $x\in\mathbb Z$.
If we only allow $x\in\mathbb R\setminus\mathbb Z$, there are infinitely many $\epsilon$ with no corresponding $x$. An example is $\epsilon=(1,1,-1,-1,1,1,-1,-1,...)$.
To show this, let's first analyse $v_n$.
Define $d_n(x)=\lfloor(n+1)x\rfloor-\lfloor nx\rfloor$. Since $v_n(x)=\lfloor x\rfloor+v_n(x-\lfloor x\rfloor)$, it is enough to deal with $x\in(0,1)$, in which case $d_n(x)\in\{0,1\}$. Now we only have two cases:
$$d_n=0\ \Rightarrow\ v_{n+1}=\frac{\lfloor nx\rfloor}{n+1}<v_n\ \Rightarrow\ \epsilon_n=-1$$
$$d_n=1\ \Rightarrow\ v_{n+1}=\frac{\lfloor nx\rfloor+1}{n+1}>\frac{\lfloor nx\rfloor}n=v_n\ \Rightarrow\ \epsilon_n=1$$
where the second line is due to $\lfloor nx\rfloor<n$ and the fact $0<a<b\ \Rightarrow\ \frac{a+1}{b+1}>\frac ab$. By these two observations, we know $\epsilon_n=1$ whenever $\lfloor nx\rfloor$ increases in the next step, and $\epsilon_n=-1$ otherwise. This means if there exists $x\in(0,1)$ such that $\epsilon=(1,1,-1,-1,...)$, then $x$ must satisfy
$$0<x<1,\ 2x\geq1,\ 3x\geq2,\ 4x<3,\ 5x<3,...$$
Keep going in this pattern, it follows for all odd $n\in\mathbb N$ we have
$$2nx\geq n,\ \ (2n+1)x\geq n+1,\ \ (2n+2)x<n+2,\ \ (2n+3)x<n+2,\ \ ...$$
The first and last inequality imply that $\frac12\leq x<\frac{n+2}{2n+3}$. Sending $n\to\infty$, the RHS $\to1/2$ and therefore $x=1/2$. However, this is a contradiction since $x=1/2$ gives $\epsilon=(1,-1,1,-1,...)$.
With the same idea, you can construct infinitely many $\epsilon$ with no corresponding $x\in\mathbb R\setminus\mathbb Z$, another example is $\epsilon=(1,1,1,-1,-1,-1,...)$.