Can we show that $\frac1{2^n}\lceil2^nx\rceil\downarrow x$ for $n\to\infty$?

Question

Let $x\in\mathbb R$ and $$\tilde x_n:=\frac1{2^n}\lfloor2^nx\rfloor\;\;\;\text{for }n\in\mathbb N\;.$$ It's easy to show that $\tilde x_n\uparrow x$ for $n\to\infty$. Now, let $$x_n:=\frac1{2^n}\lceil2^nx\rceil\;\;\;\text{for }n\in\mathbb N\;.$$ Can we analogously show that $x_n\downarrow x$ for $n\to\infty$? I've tried to use $$x-1<\lfloor x\rfloor\le x\le\lceil x\rceil<x+1\tag 1$$ together with the assumption $$2^nx=k+\delta\tag 2$$ for some $k\in\mathbb Z$ and $\delta\in[0,1)$. With this approach I was able to show $\tilde x_n\le\tilde x_{n+1}$, but I wasn't able to show $x_n\ge x_{n+1}$.

Answer 1

Well,

$$x_n - x_{n+1} = \frac{2\lceil 2^n x\rceil - \lceil 2^{n+1}x\rceil}{2^{n+1}} \geqslant 0$$

since $2\lceil y\rceil \geqslant \lceil 2y\rceil$ for all $y\in \mathbb{R}$.