# How do you find the sum of the infinite geometric series given Sigma (3/8)(3/4)^(n-1) from n=1 to oo?

Feb 6, 2017

${S}_{\infty} = \frac{3}{2}$

#### Explanation:

For a GP $\text{ } a + a r + a {r}^{2} + a {r}^{3} + \ldots .$

$\text{providing } | r | < 1$

$\text{ the series will have a sum to infinity}$
$\text{ and is calculated by the formula}$

${S}_{\infty} = \frac{a}{1 - r}$

in this case

${\sum}_{0}^{\infty} \left(\frac{3}{8}\right) {\left(\frac{3}{4}\right)}^{n - 1} = \left(\frac{3}{8}\right) {\sum}_{0}^{\infty} {\left(\frac{3}{4}\right)}^{n - 1}$

if we write out the series for $n = 1 , 2 , 3 , 4 , \ldots .$

${S}_{\infty} = \left(\frac{3}{8}\right) \textcolor{b l u e}{\left\{1 + \frac{3}{4} + {\left(\frac{3}{4}\right)}^{2} + {\left(\frac{3}{4}\right)}^{3} + \ldots \ldots .\right\}}$

The GP is coloured blue.

$a = 1 , r = \frac{3}{4}$

$\because | \frac{3}{4} | < 1 \text{ the sum to infinity exists}$

${S}_{\infty} = \left(\frac{3}{8}\right) \left(\frac{1}{1 - \left(\frac{3}{4}\right)}\right)$

${S}_{\infty} = \left(\frac{3}{8}\right) \left(\frac{1}{\frac{1}{4}}\right)$

${S}_{\infty} = \left(\frac{3}{8}\right) \times \left(\frac{4}{1}\right) = \frac{12}{8} = \frac{3}{2}$