\[\ln\left(z+a\right)=\ln a+2\left(\left(\frac{z}{2a+z}\right)+\frac{1}{3}\left(%
\frac{z}{2a+z}\right)^{3}+\frac{1}{5}\left(\frac{z}{2a+z}\right)^{5}+\cdots%
\right),\]