A floor series

Let $x \in [0, 1)$ . Prove that

$\displaystyle x = \sum_{n=1}^{\infty} \frac{\left \lfloor 2^n x \right \rfloor - 2 \left \lfloor 2^{n-1} x \right \rfloor}{2^n}$ .
$\displaystyle x = \sum_{n \geq 1 , 2 \nmid \left \lfloor 2^n x \right \rfloor} \frac{1}{2^n}$ .
$\displaystyle \sum_{n=1}^{\infty} \frac{1 - \mathrm{sgn} \left ( \tan 2^n \right )}{2^{n+2}} = \frac{1}{\pi}$ where $\mathrm{sgn}$ denotes the sign function.

Solution

We have successively:
$\begin{align*} \sum_{n=1}^{\infty} \frac{\left \lfloor 2^n x \right \rfloor - 2 \left \lfloor 2^{n-1} x \right \rfloor}{2^n} &= \lim_{N \rightarrow +\infty} \sum_{n=1}^{N} \frac{\left \lfloor 2^n x \right \rfloor - 2 \left \lfloor 2^{n-1} x \right \rfloor}{2^n} \\ &=\lim_{N \rightarrow +\infty} \sum_{n=1}^{N} \left (\frac{\left \lfloor 2^n x \right \rfloor}{2^n} - \frac{ \left \lfloor 2^{n-1} x \right \rfloor}{2^{n-1}} \right ) \\ &=\lim_{N \rightarrow +\infty} \frac{\left \lfloor 2^N x \right \rfloor}{2^N} \end{align*}$

However,

$x-\frac{1}{2^N} = \frac{2^Nx-1}{2^N} < \frac{\lfloor 2^Nx \rfloor}{2^N} \leq \frac{2^Nx}{2^N}=x$

and the result follows.
Let $a_n=\lfloor 2^nx \rfloor$ . Then, $a_n \leq 2^n x < a_n+1$ and $a_{n-1} \leq 2^{n-1}x < a_{n-1}+1$ . It follows that $-1 < a_n-2a_{n-1} < 2$ . Hence,
$\lfloor 2^nx \rfloor -2\lfloor 2^{n-1}x \rfloor =a_n-2a_{n-1} \in \{0,1\}$

since $a_n$ is integer for all $n$ . The result follows from the previous question as well as the fact that $a_n-2a_{n-1}=1$ if-f $n$ is odd; $2 \nmid a_n$ .
It follows from the previous question since $1-\mathrm{sgn}(\tan 2^n)= \left\{\begin{matrix} 2 & , & 2 \nmid \left \lfloor \frac{2^{n+1}}{\pi} \right \rfloor \\ 0 & , & \text{otherwise} \end{matrix}\right.$ .