## Thomas' Calculus 13th Edition

$x+\dfrac{x^2}{2}+\dfrac{5x^3}{6}+\dfrac{7x^4}{12}+...$
Consider the Taylor Series for $\ln (1+x)$ and $\dfrac{1}{1-x}$ as follows: $\ln (1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-...$; $\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n =1+x+x^2+....+x^n$ This implies that $\ln (1+x) \cdot (\dfrac{1}{1-x})=(x-\dfrac{1}{2}x^2+\dfrac{1}{3}x^3-..)\cdot (1+x+x^2+....+x^n)$ Hence, $\ln (1+x) \cdot (\dfrac{1}{1-x})=x+\dfrac{x^2}{2}+\dfrac{5x^3}{6}+\dfrac{7x^4}{12}+...$