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

Apr 14, 2017

$\frac{6}{5}$

Explanation:

First, take the $2$ out of the sigma so it's easier to deal with:

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

This is in the form:

$\sum {\left(u\right)}^{n}$ where $| u | < 1$. That means, that $\sum {u}^{n} = \frac{1}{1 - u}$

Our $u$, in this case, is $- \frac{2}{3}$. Simply plug into the formula.

$2 \left(\frac{1}{1 - \left(- \frac{2}{3}\right)}\right) = 2 \left(\frac{1}{\frac{5}{3}}\right) = 2 \left(\frac{3}{5}\right) = \frac{6}{5}$