Evaluate the infinite geometric series: $$\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\dots$$
I know that the sum of an infinite geometric series is $\frac{a}{1-r}$, where $a$ is the first term and $r$ is the common ratio. In this series, $a=\frac{1}{3}$ and $r=\frac{1}{2}$, because each term is half of the previous term. So, the sum is $\frac{\frac{1}{3}}{1-\frac{1}{2}}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}$. 