# How do you find the sum of the infinite geometric series 12+4+4/3+...?

Dec 9, 2015

$18$

#### Explanation:

An infinite geometric series of the form ${\sum}_{n = 1}^{\infty} a {r}^{n - 1}$ converges if and only if $| r | < 1$, where r is the common ration between terms and given by $r = {x}_{n + 1} / \left({x}_{n}\right)$.
In this case it converges to the value $\frac{a}{1 - r}$ where a is the first term in the corresponding sequence $\left({x}_{n}\right)$.

So in this case, $r = \frac{4}{12} = \frac{4}{3} / 4 = \frac{1}{3} < 1$ hence the series converges.

Its sum is hence ${\sum}_{n = 1}^{\infty} 12 \cdot {\left(\frac{1}{3}\right)}^{n - 1} = \frac{12}{1 - \frac{1}{3}} = 18$