$$\frac{1}{x-1}=\frac{1}{x+1}+\frac{2}{x^2-1}=\frac{1}{x+1}+2(\frac{1}{x^2+1}+\frac{2}{x^4-1})=\frac{1}{x+1}+\frac{2}{x^2+1}+\frac{4}{x^4-1}=\frac{1}{x+1}+\frac{2}{x^2+1}+4(\frac{1}{x^4+1}+\frac{2}{x^8-1})=\frac{1}{x+1}+\frac{2}{x^2+1}+\frac{4}{x^4+1}+\frac{8}{x^8-1}=\frac{1}{x+1}+\frac{2}{x^2+1}+\frac{4}{x^4+1}+\frac{8}{x^8+1}....$$

$$\frac{1}{x-1}=\sum_{n=0}^\infty \frac{2^n}{x^{2^n}+1}$$

Hy I have only 23 hours to give a 50 rep bounty, who want it?
http://math.stackexchange.com/questions/384622/theory-of-structures-with-infinite-partial-orders-langle-h-sqsubset-i