Answer
$$\sum_{n=0}^{\infty}\frac{x^n}{3^{n+1}}$$
$|x|<3$
Work Step by Step
Given $$ f(x)=\frac{1}{3- x}$$
Since
$$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n},\ \ \ \ |x|<1\tag{1}$$
By using (1), we get:
\begin{align*}
\frac{1}{3- x}&=\frac{1}{3(1-(x/3))}\\
&=\frac{1/3}{(1-(x/3))}\\
&=\frac{1}{3} \sum_{n=0}^{\infty}( x/3)^{n}\\
&= \frac{1}{3} \sum_{n=0}^{\infty}\frac{x^n}{3^n}\\
&= \sum_{n=0}^{\infty}\frac{x^n}{3^{n+1}}\\
\end{align*}
Such that $ |x/3|<1$, or $|x|<3$
