Answer
$\dfrac{1}{1+x}=\Sigma_{n=0}^\infty (-1)^n x^n$
Work Step by Step
Since, we know that the Maclaurin Series for $\frac{1}{1-x}$ is defined as:
$\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n$
We need to replace in the above series $x$ with $-x$.
Now, $\dfrac{1}{1-(-x)}=\Sigma_{n=0}^\infty (-x)^n$
Thus, $\dfrac{1}{1+x}=\Sigma_{n=0}^\infty (-1)^n x^n$
