Answer
\[\begin{align}
& \sum\limits_{k=0}^{\infty }{{{\left( 3x \right)}^{k}}} \\
& \left( -\frac{1}{3},\frac{1}{3} \right) \\
\end{align}\]
Work Step by Step
$\begin{align}
& f\left( 3x \right)=\frac{1}{1-3x} \\
& \text{Using the geometric series }f\left( u \right)=\frac{1}{1-u}=\sum\limits_{k=0}^{\infty }{{{u}^{k}},\text{ for }\left| u \right|<1},\text{ so} \\
& \text{Let }u=3x \\
& f\left( 3x \right)=\sum\limits_{k=0}^{\infty }{{{\left( 3x \right)}^{k}}} \\
& \text{The series converges for }\left| u \right|<1,\text{ then} \\
& \left| 3x \right|<1 \\
& \text{Solving the inequality} \\
& -1<3x<1 \\
& -\frac{1}{3}
