I'm stuck with this summation problem How can I show that 


  $$\displaystyle\sum_{n=1}^\infty \left(-\frac{1}{2^n}\left(\bigg| {\rm sgn} \left( {\rm round}\ (n \cdot \frac{1}{3})-n \cdot \frac{1}{3}\right)\right)-1\bigg|\right) = \frac{1}{7}?$$
Hint:  Unpack it.  The result of the sgn function depends only on $n \pmod 3$.  Take three terms of the sum and evaluate it.  You will get a geometric sum.  You might notice  that $\frac 17=\frac 1{8(1-\frac 18)}$
Looking at $\bigg| {\rm sgn}\ ( {\rm round}\ (n \cdot \frac{1}{3})-n \cdot \frac{1}{3})\bigg|$, if n is divisible by 3 then the result is 0, otherwise it's 1. Then after subtracting 1 if n is divisible by 3 it's -1, otherwise it's 0.
Then the whole equation becomes an infinite sum of $\frac{1}{8} + \frac{1}{64} + ... = \frac{1}{7}$.