Answer
$\Sigma_{k=0}^{\infty} (-1)^k (x)^k$
Work Step by Step
$\dfrac{1}{1-x}=\Sigma_{k=0}^{\infty} x^k$
We will replace $x$ with $-x$.
$\dfrac{1}{1-(-x)}=\Sigma_{k=0}^{\infty} (-x)^k$
$\dfrac{1}{1+x}=\Sigma_{k=0}^{\infty} (-1)^k (x)^k$