Answer
$$ \sum_{n=0}^{\infty}(-3) ^nx^ {n}$$
$|x|\lt 1/3$
Work Step by Step
Given $$ f(x)=\frac{1}{1+ 3x}$$ Since $$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n},\ \ \ \ |x|\lt 1\tag{1}$$ By using (1), we get
\begin{align*} \frac{1}{1+3 x}&=\frac{1}{1-(-3 x)}\\ &= \sum_{n=0}^{\infty}(-3 x)^{n}\\ &= \sum_{n=0}^{\infty}(-3) ^nx^ {n} \end{align*} Such that $ |-3x|\lt1$, or $|x|\lt 1/3$
