# How do you find the sum of the infinite geometric series 1/3 + 4/9 + 16/27 + 64/81?

Dec 14, 2015

The series diverges, as

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

#### Explanation:

An infinite geometric series is a series of the form

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

where ${a}_{0}$ is the initial term and $r$ is the ratio between terms.

To find $r$, then, we can divide any term after the first by the term prior, as
$\frac{{a}_{0} {r}^{k}}{{a}_{0} {r}^{k - 1}} = r$

Doing so for the given series, we find that

$r = \frac{\frac{4}{9}}{\frac{1}{3}} = \frac{4}{3}$

As $| r | \ge 1$, that means the terms do not tend towards $0$ as $n \to \infty$, meaning the series diverges.