The problem asks us to find the sum of the series $\sum_{n=1}^{\infty} \frac{x^n}{3^n}$.

1. Problem Description

The problem asks us to find the sum of the series $\sum_{n=1}^{\infty} \frac{x^n}{3^n}$.

2. Solution Steps

The given series can be rewritten as:

$\sum_{n=1}^{\infty} \frac{x^n}{3^n} = \sum_{n=1}^{\infty} (\frac{x}{3})^n$.

This is a geometric series with first term $a = \frac{x}{3}$ and common ratio $r = \frac{x}{3}$.

A geometric series $\sum_{n=1}^{\infty} ar^{n-1}$ converges to $\frac{a}{1-r}$ if $|r| < 1$. In our case, we have $\sum_{n=1}^{\infty} ar^{n}$.

Then, $\sum_{n=1}^{\infty} (\frac{x}{3})^n = \sum_{n=1}^{\infty} \frac{x}{3} (\frac{x}{3})^{n-1}$. So, $a = \frac{x}{3}$ and $r = \frac{x}{3}$.

The series converges if $|r| = |\frac{x}{3}| < 1$, which means $|x| < 3$ or $-3 < x < 3$.

The sum of the geometric series is given by:

$S = \frac{a}{1-r} = \frac{\frac{x}{3}}{1-\frac{x}{3}} = \frac{\frac{x}{3}}{\frac{3-x}{3}} = \frac{x}{3-x}$.

The sum of a geometric series $\sum_{n=0}^{\infty} r^n$ is $\frac{1}{1-r}$ for $|r|<1$.

Here we have $\sum_{n=1}^{\infty} (\frac{x}{3})^n = \sum_{n=0}^{\infty} (\frac{x}{3})^n - (\frac{x}{3})^0 = \frac{1}{1-\frac{x}{3}} - 1 = \frac{1}{\frac{3-x}{3}} - 1 = \frac{3}{3-x} - 1 = \frac{3-(3-x)}{3-x} = \frac{x}{3-x}$.

3. Final Answer

$\frac{x}{3-x}$