# How do you find the sum of 2/1+4/3+8/9+16/27+...+2^(n+1)/3^n+...?

May 26, 2017

${\sum}_{n = 0}^{\infty} {2}^{n + 1} / {3}^{n} = 6$

#### Explanation:

${\sum}_{n = 0}^{\infty} {2}^{n + 1} / {3}^{n} = {\sum}_{n = 0}^{\infty} \frac{2 \times {2}^{n}}{3} ^ n$

${\sum}_{n = 0}^{\infty} {2}^{n + 1} / {3}^{n} = 2 {\sum}_{n = 0}^{\infty} {\left(\frac{2}{3}\right)}^{n}$

This is the sum of a geometric series of ratio: $q = \frac{2}{3}$, and we know that for $q < 1$

${\sum}_{n = 0}^{\infty} {q}^{n} = \frac{1}{1 - q}$

so:

${\sum}_{n = 0}^{\infty} {2}^{n + 1} / {3}^{n} = 2 \left(\frac{1}{1 - \frac{2}{3}}\right) = 6$