# How do you find the sum of the infinite geometric series 1/3+1/9+1/27+1/81+...?

Dec 9, 2015

$S \infty = \frac{1}{2}$

#### Explanation:

Formula for sum of infinite geometric series is
${S}_{\infty} = {a}_{1} / \left(1 - r\right)$ ; $\text{ " " " } - 1 < r < 1$

We have a geometric series :$\frac{1}{3} + \frac{1}{9} + \frac{1}{81} + \ldots \ldots \ldots$

First we know ${a}_{1} = \frac{1}{3}$ (the first term)

Second: Identify $r$ , we know $r = {a}_{2} / {a}_{1}$ or r= a_n/a_(n-1

$r = \frac{\frac{1}{9}}{\frac{1}{3}}$ $\Leftrightarrow \frac{1}{9} \cdot \frac{3}{1} = \frac{1}{3}$

$r = \frac{1}{3}$

Substitute into the formula
$S \infty = \frac{\frac{1}{3}}{1 - \frac{1}{3}}$

$= \frac{\frac{1}{3}}{\frac{2}{3}}$

$= \left(\frac{1}{3}\right) \cdot \left(\frac{3}{2}\right)$

$S \infty = \frac{1}{2}$