Evaluate the infinite geometric series: ​

First term is 1/3.

Common ratio is (1/6)/(1/3) = 1/2.

Using the formula sum = a/(1 - r), sum = $\dfrac{\frac13}{1 - \frac12} = \dfrac23$

Observe that we can say

$\displaystyle\frac{1}{3} + \frac{1}{6} + \frac{1}{12} + \cdots = \frac{1}{3}\left(1 + \frac{1}{2} + \frac{1}{4}+\cdots\right) = \frac{1}{3}\sum^{\infty}_{n=0}\left(\frac{1}{2}\right)^n.$

Given this, our common ratio $r$ is $r = \frac{1}{2}$. Then, by the well-known infinite geometric series formula, we have

$\displaystyle \frac{1}{3} \sum^{\infty}_{n=0}\left(\frac{1}{2}\right)^n = \frac{1}{3} \cdot\frac{1}{1-\frac{1}{2}} = \frac{2}{3}.$