# How do you find the sum of the infinite geometric series Sigma 2(2/3)^n from n=0 to oo?

Apr 27, 2018

$\sum 2 {\left(\frac{2}{3}\right)}^{n} = 6$

#### Explanation:

$\sum 2 {\left(\frac{2}{3}\right)}^{n}$ for n=0 to infinity

=2$\left(\sum {\left(\frac{2}{3}\right)}^{n}\right)$

Sum of an infinite geometric series (where the ratio r is $- 1 \le r \le 1$) is given by $\frac{a}{1 - r}$ where r is the common ratio and ‘a’ is the first term.

Therefore,
$\sum {\left(\frac{2}{3}\right)}^{n} = \frac{1}{1 - \left(\frac{2}{3}\right)}$

$= \frac{1}{\frac{1}{3}} = 3$

Substituting this value in our original equation gives us -

2$\left(\sum {\left(\frac{2}{3}\right)}^{n}\right) = 2 \cdot 3 = 6$