Chapter 11.10, Problem 50E

Use the formula ${\tanh }^{-1}x=\frac{1}{2}\ln \left(\frac{1+x}{1-x}\right)-1<x<1$ and the Maclaurin series for $\ln (1+x)$ to show that ${\tanh }^{-1}x={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{x^{2n+1}}{2n+1}}$