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

##### 1 Answer
Nov 29, 2015

Apply the geometric sum formula to find that
${\sum}_{n = 0}^{\infty} 3 {\left(- \frac{1}{3}\right)}^{n} = \frac{9}{4}$

#### Explanation:

Given a geometric series with initial value $a$ and common ratio $r$ with $| r | < 1$ the sum is given by the formula

${\sum}_{n = 0}^{\infty} a {r}^{n} = \frac{a}{1 - r}$

(A short derivation of this formula is included in Can an infinite series have a sum? )

In the given series, the initial value is $a = 3$ and the common ratio, found by dividing any term by its preceding term, is $r = - \frac{1}{3}$.

As $| - \frac{1}{3} | < 1$ we can apply the formula to obtain

${\sum}_{n = 0}^{\infty} 3 {\left(- \frac{1}{3}\right)}^{n} = \frac{3}{1 - \left(- \frac{1}{3}\right)} = \frac{3}{\frac{4}{3}} = \frac{9}{4}$