# How do you find the sum of the infinite geometric series 5+5/6+5/6^2+5/6^3+...?

Feb 29, 2016

${\sum}_{n = 1}^{\infty} 5 {\left(\frac{1}{6}\right)}^{n - 1} = \frac{5}{1 - \frac{1}{6}} = 6$

#### Explanation:

The common ratio of this geometric series is $r = \frac{\frac{5}{6}}{5} = \frac{1}{6}$.

Since $| r | = \frac{1}{6} < 1$, it implies that the series converges to $\frac{a}{1 - r}$ where $a = 5$ is the first term.

This infinite geometric series, as well as its sum, may hence be written in the form

${\sum}_{n = 1}^{\infty} 5 {\left(\frac{1}{6}\right)}^{n - 1} = \frac{5}{1 - \frac{1}{6}} = 6$.