Answer
$|x| \displaystyle \lt \frac{1}{3}$
$f(x)=\displaystyle \frac{3x}{1-3x}, \displaystyle \quad|x| \lt \frac{1}{3}$
Work Step by Step
We aim for terms of Th.9.6 (Convergence of a Geometric Series).
A geometric series with ratio $r$ diverges when $|r| \geq 1$.
If $0 \lt |r| \lt 1$, then the series converges to the sum
$\displaystyle \sum_{n=0}^{\infty}ar^{n}=\frac{a}{1-r},\quad 0 \lt |r| \lt 1$.
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Adjusting for the subscript,
$\displaystyle \sum_{n=1}^{\infty}(3x)^{n}=(3x)^{1}+(3x)^{2}+(3x)^{3}+...$
$=3x[1+(3x)^{1}+(3x)^{2}+(3x)^{3}+...]$
$=(3x)\displaystyle \sum_{n=0}^{\infty}(3x)^{n}$
The series is geometric, $r=3x$, and
it converges for $|3x| \lt 1$ $\Rightarrow$ $|x| \displaystyle \lt \frac{1}{3}$
$f(x)=(3x)\displaystyle \sum_{n=0}^{\infty}(3x)^{n}=(3x)\frac{1}{1-3x}=$
$\displaystyle=\frac{3x}{1-3x}, \displaystyle \quad|x| \lt \frac{1}{3}$
