# How do you find the sum of the infinite geometric series Sigma 4(1/4)^n from n=0 to oo?

Mar 20, 2017

${\Sigma}_{0}^{\infty} 4 {\left(\frac{1}{4}\right)}^{n} = 5 \frac{1}{3}$

#### Explanation:

The first term of the series ${\Sigma}_{0}^{\infty} 4 {\left(\frac{1}{4}\right)}^{n}$ is

$4 {\left(\frac{1}{4}\right)}^{0} = 4 \times 1 = 4$ and as each successive term $\frac{1}{4}$ times its immediately preceding term common ratio is $\frac{1}{4}$.

and let the sum of the series ${S}_{\infty}$ be $4 + 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \ldots \ldots \ldots . .$

As ${S}_{\infty} = 4 + 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \ldots \ldots \ldots . .$, we have

$4 {S}_{\infty} = 16 + 4 + 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \ldots \ldots \ldots . .$

Now subtracting first from second we get

$3 {S}_{\infty} = 16$ and ${S}_{\infty} = \frac{16}{3} = 5 \frac{1}{3}$

Note - In a geometric series with $a$ as first term and common ratio as $r$, ${S}_{\infty} = \frac{a}{1 - r}$, hence sum of series is $\frac{4}{1 - \frac{1}{4}} = \frac{4}{\frac{3}{4}} = 4 \times \frac{4}{3} = \frac{16}{3} = 5 \frac{1}{3}$.