# What is the sum of the infinite geometric series 9+6+4+...?

##### 1 Answer
May 16, 2016

$S = 27$

#### Explanation:

The series can be written as
$S = 9 + 6 + 4 + \frac{8}{3} + \frac{16}{9} + \frac{32}{27} + \ldots + 4 {\left(\frac{2}{3}\right)}^{n}$
$S = 15 + 4 {\sum}_{i = 0}^{i = \infty} {\left(\frac{2}{3}\right)}^{i}$
Now we will use a polynomial identity which says
$\frac{1 - {x}^{n + 1}}{1 - x} = 1 + x + {x}^{2} + {x}^{3} + \ldots + {x}^{n}$
So for $n$
${\sum}_{i = 0}^{i = n} {\left(\frac{2}{3}\right)}^{i} = \frac{1 - {\left(\frac{2}{3}\right)}^{n + 1}}{1 - \left(\frac{2}{3}\right)}$
but as $n \to \infty$, $\left(\frac{2}{3}\right) \to 0$, because $\left(\frac{2}{3}\right) < 1$
Finally we get
${\sum}_{i = 0}^{i = \infty} {\left(\frac{2}{3}\right)}^{i} = \frac{1}{1 - \left(\frac{2}{3}\right)} = 3$
Finally we obtain
$S = 15 + 4 \times 3 = 27$