I have the sum $$\sum\limits_{k=1}^n\frac{\lfloor2^k\omega\rfloor-2\lfloor2^{k-1}\omega\rfloor}{2^k}=\frac{\lfloor2\omega\rfloor}{2}-\lfloor\omega\rfloor+\frac{\lfloor4\omega\rfloor}{4}-\frac{\lfloor2\omega\rfloor}{2}+\frac{\lfloor8\omega\rfloor}{8}-\frac{\lfloor4\omega\rfloor}{4}+\dots+\frac{\lfloor2^{n-1}\omega\rfloor}{2^{n-1}}-\frac{\lfloor2^{n-2}\omega\rfloor}{2^{n-2}}+\frac{\lfloor2^n\omega\rfloor}{2^n}-\frac{\lfloor2^{n-1}\omega\rfloor}{2^{n-1}}=\frac{\lfloor2^n\omega\rfloor}{2^n}-\lfloor\omega\rfloor $$ Now I want to show that the equation $\omega-\sum\limits_{k=1}^n\frac{\lfloor2^k\omega\rfloor-2\lfloor2^{k-1}\omega\rfloor}{2^k}=\omega-\Big(\frac{\lfloor2^n\omega\rfloor}{2^n}-\lfloor\omega\rfloor\Big)$ is bounded above by $2^{-n}$ for $\omega\in[0,1]$.
I manage to show that it is smaller than $1$ but I need some help to show that it is smaller than $2^{-n}$.
Note, it is not true for $\omega=1$, so you want $\omega\in[0,1)$, in which case $\lfloor \omega \rfloor = 0$.
Now, more generally, prove that, for any $A>0$ if $\omega\in[0,1)$ then $0\leq \omega - \frac{\lfloor A\omega\rfloor}A < \frac{1}{A}$.
This follows since $0\leq \omega A -\lfloor \omega A\rfloor <1$.