# How do you find the sum of the infinite geometric series 4(1/3)^(n - 1)?

${\sum}_{n = 1}^{\infty} 4 {\left(\frac{1}{3}\right)}^{n - 1} = \frac{3}{2}$
Any geometric series of the form ${\sum}_{n = 1}^{\infty} a {r}^{n - 1}$ converges to the value $\frac{a}{1 - r}$ provided $| r | < 1$.
In this case, $a = 4 \mathmr{and} r = \frac{1}{3} < 1$.
Hence the sum of the series converges to $\frac{4}{1 - \frac{1}{3}} = \frac{3}{2}$.