## Calculus with Applications (10th Edition)

$\frac{1+x}{1-x}=1+2x+2x^2+2x^3+...+x^n(x+1)+...$
The Taylor series for $\frac{1}{1-x}$ is $1+x+x^2+x^3+...+x^n+...$ So, the Taylor series for $\frac{x}{1-x}$ is $x+x.x+x.x^2+x.x^3+...+x.x^n...$ $x+x^2+x^3+x^4+...+x^{n+1}$ Hence, $\frac{1}{1-x}+\frac{x}{1-x}$ $=1+x+x^2+x^3+...+x^n+...+(x+x^2+x^3+x^4+...+x^{n+1})$ $=1+2x+2x^2+2x^3+...+x^n(x+1)+...$ We are given $\frac{1+x}{1-x}=\frac{1}{1-x}+\frac{x}{1-x}$ so $\frac{1+x}{1-x}=1+2x+2x^2+2x^3+...+x^n(x+1)+...$