# How do you find the sum of the infinite geometric series 8 - 4 + 2 - 1 + 1/2 -...?

May 27, 2016

$8 - 4 + 2 - 1 + \frac{1}{2} - \ldots = \frac{16}{3}$

#### Explanation:

A geometric series is a series of the form ${\sum}_{k = 0}^{\infty} a {r}^{n}$ where $a$ is the first term in the sum and $r$ is the constant ratio between terms. The series will converge if and only if $| r | < 1$ (with the trivial exception being if $a = 0$.

Given a convergent geometric series, that is, a geometric series with $| r | < 1$, we have

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

(see the above link for a derivation of this formula)

In the given sum, the common ratio between terms is $- \frac{1}{2}$. As $| - \frac{1}{2} | = \frac{1}{2} < 1$, the series will converge and we can find the value it converges to with the above formula. Noting that the first term is $a = 8$, we have

$8 - 4 + 2 - 1 + \frac{1}{2} - \ldots = {\sum}_{n = 0}^{\infty} 8 {\left(- \frac{1}{2}\right)}^{n}$

$= \frac{8}{1 - \left(- \frac{1}{2}\right)}$

$= \frac{8}{\frac{3}{2}}$

$= \frac{16}{3}$