# How do you find the sum of the infinite geometric series Sigma (1/10)^n from n=0 to oo?

Feb 14, 2017

${\sum}_{n = 0}^{\infty} {\left(\frac{1}{10}\right)}^{n} = \frac{10}{9}$

#### Explanation:

The sum of a geometric series is given by:

${\sum}_{n = 0}^{\infty} {x}^{n} = \frac{1}{1 - x}$ for $\left\mid x \right\mid < 1$

In our case:

$x = \frac{1}{10} < 1$

so:

${\sum}_{n = 0}^{\infty} {\left(\frac{1}{10}\right)}^{n} = \frac{1}{1 - \frac{1}{10}} = \frac{10}{9}$