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