Answer
$\frac{1}{1+x}=\sum_{n=0}^\infty (-1)^nx^n$
Interval of convergence: $(-1, 1)$
Work Step by Step
We know the following power series.
$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$ where $|x|<1$
Replacing $x$ by $-x$,
$\frac{1}{1+x}=\sum_{n=0}^\infty (-x)^n$ where $|-x|<1$
$\frac{1}{1+x}=\sum_{n=0}^\infty (-1)^nx^n$ where $|x|<1$
The power series representing $f(x)=\frac{1}{1+x}$ is $\sum_{n=0}^\infty (-1)^nx^n$ with the interval of convergence is $(-1,1)$.
