Taylor Expansion $\log(\frac{1+z}{1-z})$

Question

Taylor Expansion of $\log(\frac{1+z}{1-z})$ around $z=0$

But $\log(1+z)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{z^n}{n}$

How can I continue from here?

Answer 1

You can use the fact that $\log(\frac{1+z}{1-z}) = \log(1+z) - \log(1-z)$, then do the Taylor expansion of each of these terms separately and then take the difference.