# What is the sum of the infinite geometric series 3-1+1/3...?

$3 {\sum}_{i = 0}^{\infty} {\left(- \frac{1}{3}\right)}^{i} = \frac{9}{4}$
The sum of sequence $\left\{3 , - 1 , \frac{1}{3} , - \frac{1}{9} , \frac{1}{27} , \ldots\right\} = 3 \left\{1 , - \frac{1}{3} , \frac{1}{9} , - \frac{1}{27} , \ldots\right\}$ or more compactly
$S = 3 {\sum}_{i = 0}^{\infty} {\left(- \frac{1}{3}\right)}^{i} = 3 \left(\frac{1}{1 + \frac{1}{3}}\right) = \frac{9}{4}$
$\frac{1 - {x}^{n + 1}}{1 - x} = 1 + x + {x}^{2} + {x}^{3} + \ldots + {x}^{n}$. Here $x = - \frac{1}{3}$ and $\left\mid x \right\mid < 1$ so ${\lim}_{n \to \infty} = \frac{1}{1 - x}$