How do you find the sum of the infinite geometric series 1/2+1/4+1/8+1/16..?

Dec 16, 2015

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

Explanation:

Let the series be represented by ${\sum}_{n = 1}^{\infty} \left({x}_{n}\right)$
The common ratio of this infinite geometric series is
$r = {x}_{n + 1} / {x}_{n} = \frac{1}{4} / \frac{1}{2} = \frac{1}{2}$

Since $| r | = \frac{1}{2} < 1$, it implies that this series converges to the value $\frac{a}{1 - r}$, where $a$ is the first term in the series.

$\therefore {\sum}_{n = 1}^{\infty} \left(\frac{1}{2}\right) \cdot {\left(\frac{1}{2}\right)}^{n - 1} = \frac{a}{1 - r} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1$.