Chapter 10: Infinite Sequences and Series - Section 10.8 - Taylor and Maclaurin Series - Exercises 10.8 - Page 618: 13

$\Sigma_{n=0}^\infty (-1)^n x^n$

Consider the Maclaurin Series for $\dfrac{1}{1-x}$ as follows: $\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n$ Plug $-x$ instead of $x$. we get $\dfrac{1}{1-(-x)}=\Sigma_{n=0}^\infty (-x)^n$ This implies that $\dfrac{1}{1+x}=\Sigma_{n=0}^\infty (-1)^n x^n$