This is a geometric series with first term $a=\frac{1}{3}$ and common ratio $r=\frac{1}{2}$.
The sum of an infinite geometric series is given by the formula $\frac{a}{1-r}$, as long as $|r|<1$.
In this case, $|r|=\left|\frac{1}{2}\right|=\frac{1}{2}<1$, so the series converges.
Using the formula, we find that the sum of the series is $\frac{\frac{1}{3}}{1-\frac{1}{2}}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{1}{3}\cdot\frac{2}{1}=\boxed{\frac{2}{3}}$.
The answer is: \frac{2}{3}
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Evaluate the infinite geometric series: $$\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\dots$$

### Answer:
