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A floor series

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

  1. \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}.
  2. \displaystyle x = \sum_{n \geq 1 , 2 \nmid \left \lfloor 2^n x \right \rfloor} \frac{1}{2^n}.
  3. \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

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