Derivation of series of $\ln(x)$ for $x > 0$ How is the following expansion obtained?
$$
\ln(z) = 2 \left[\frac{z-1}{z+1} + \frac{1}{3} \left( \frac{z-1}{z+1} \right)^3  + \frac{1}{5} \left( \frac{z-1}{z+1} \right)^5  + \frac{1}{7} \left( \frac{z-1}{z+1} \right)^7 + \cdots \right]
$$
As far as I understand from http://www.math.com/tables/expansion/log.htm it is valid for $z > 0$?

Hint: For $x\in (-1,1), \ln (1+x) = x-x^2/2 + x^3/3 - \cdots .$ So for such $x,$
$$\ln \left (\frac{1+x}{1-x}\right ) = \dots\ ?$$
