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

Feb 1, 2016

Use the formula for the sum of an infinite geometric series to find:

$1 + \frac{1}{3} + \frac{1}{9} + \ldots = \frac{3}{2}$

#### Explanation:

The general term of a geometric series is given by the formula:

${a}_{n} = a {r}^{n - 1}$

where $a$ is the initial term and $r$ the common ratio.

The sum of an infinite geometric series is given by the formula:

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

when $\left\mid r \right\mid < 1$. Otherwise the series does not converge.

For a derivation of this formula see: socratic.org/questions/how-do-you-find-the-sum-of-the-infinite-geometric-series-1-5-25-125

In our current example $a = 1$ and $r = \frac{1}{3}$, so we find:

${\sum}_{n = 1}^{\infty} \left(1 \cdot {\left(\frac{1}{3}\right)}^{n - 1}\right) = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}$