Chapter 10: Infinite Sequences and Series - Section 10.9 - Convergence of Taylor Series - Exercises 10.9 - Page 625: 28

$\Sigma_{n=0}^\infty \dfrac{2x^{(2n+1)}}{(2n+1)}$

Here, $ \ln (1+x)-\ln (1-x)=(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-...)-(-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}-...)$ This implies that $\ln (1+x)-\ln (1-x)=2x+\dfrac{2}{3}x^3+\dfrac{2}{5}x^5+...= \Sigma_{n=0}^\infty \dfrac{2x^{(2n+1)}}{(2n+1)}$