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.
 A: Sum should be infinite, because for finite sum you should get:
$$\sum_{k=0}^{n}(\frac{1}{3})^k=1+\frac{1}{3}+\dots+(\frac{1}{3})^n=\frac{1-(\frac{1}{3})^{n+1}}{1-\frac{1}{3}},$$ where we used formula $1+q+\dots+q^n=\frac{1-q^{n+1}}{1-q}$ for $q=\frac{1}{3}$.
Therefore,
$$\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=\frac{1}{3}\sum_{k=0}^\infty \left(\frac{1}{3}\right)^{k}=\frac{1}{3}\lim_{k\to\infty}\sum_{i=0}^k (\frac{1}{3})^i=\frac{1}{3}\lim_{k\to\infty}\frac{1-(\frac{1}{3})^{k+1}}{1-\frac{1}{3}}=\frac{1}{3}\frac{1}{\frac{2}{3}}=\frac{1}{2},$$
since
$$\lim_{k\to\infty} (\frac{1}{3})^{k+1}=0.$$
A: 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 $S = \dfrac{1}{3} + \dfrac{1}{3}S \Rightarrow \dfrac{2}{3} S = \dfrac{1}{3} \Rightarrow S = \dfrac{1/3}{2/3}$.
