Answer
\[\begin{align}
& \sum\limits_{k=0}^{\infty }{{{\left( -4x \right)}^{k}}} \\
& \left( -\frac{1}{4},\frac{1}{4} \right) \\
\end{align}\]
Work Step by Step
$\begin{align}
& f\left( -4x \right)=\frac{1}{1+4x} \\
& \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=-4x \\
& f\left( -4x \right)=\frac{1}{1-\left( -4x \right)} \\
& f\left( 3x \right)=\sum\limits_{k=0}^{\infty }{{{\left( -4x \right)}^{k}}} \\
& \text{The series converges for }\left| u \right|<1,\text{ then} \\
& \left| -4x \right|<1 \\
& \text{Solving the inequality} \\
& -1<-4x<1 \\
& \frac{1}{4}>x>-\frac{1}{4} \\
& or \\
& -\frac{1}{4}
