Why is $\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=\frac{1/3}{2/3}$ $\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=\frac{1/3}{2/3}$ what theorem or algebra leads to this equality?
EDIT: The sum should have been infinite.

Let $S = \sum_{k=1}^\infty \left(\frac{1}{3}\right)^k$. Then,
$$ S = \frac{1}{3} + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^3 + \ldots = \frac{1}{3} \left[ 1 + \underbrace{\frac{1}{3} + \left(\frac{1}{3}\right)^2  + \ldots}_{=S} \right] = 
\frac{1}{3} \left[ 1 + S \right]$$
So, we have $S = \dfrac{1}{3} + \dfrac{1}{3}S \Rightarrow \dfrac{2}{3} S = \dfrac{1}{3} \Rightarrow S = \dfrac{1/3}{2/3}$.
