Answer
\[\begin{align}
& -\sum\limits_{k=1}^{\infty }{\frac{{{\left( 3x \right)}^{k}}}{k}\text{ }} \\
& \left[ -\frac{1}{3},\frac{1}{3} \right) \\
\end{align}\]
Work Step by Step
\[\begin{align}
& f\left( 3x \right)=\ln \left( 1-3x \right) \\
& \text{Using the geometric series } \\
& f\left( u \right)=\ln \left( 1-u \right)=-\sum\limits_{k=1}^{\infty }{\frac{{{u}^{k}}}{k},\text{ for }-1\le u\le 1},\text{ then} \\
& \text{Let }u=3x \\
& \underbrace{f\left( 3x \right)}_{f\left( u \right)}=\underbrace{\ln \left( 1-3x \right)}_{\ln \left( 1-u \right)} \\
& f\left( 3x \right)=-\sum\limits_{k=1}^{\infty }{\frac{{{\left( 3x \right)}^{k}}}{k},\text{ }}\text{ for }-1\le 3x<1 \\
& \text{The series converges for }-1\le 3x<1,\text{ then} \\
& -1\le 3x<1 \\
& \text{Solving the inequality} \\
& -\frac{1}{3}\le x<\frac{1}{3} \\
& \text{The interval of convergence is:} \\
& \left[ -\frac{1}{3},\frac{1}{3} \right) \\
\end{align}\]
