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

Mar 5, 2016

Sum of the infinite geometric series $- 3 {\left(\frac{3}{4}\right)}^{n - 1}$ is $- 12$.

#### Explanation:

Sum of a geometric series $\left\{a , a r , a {r}^{2} , a {r}^{3} , \ldots . . , a {r}^{n - 1}\right\}$ where $a$ is first term and ratio is $r$ (note ${n}^{t h}$ term is $a {r}^{n - 1}$) is given by

$a \times \left(\frac{{r}^{n} - 1}{r - 1}\right)$ where $r > 1$ and $a \times \left(\frac{1 - {r}^{n}}{1 - r}\right)$ where $r < 1$.

In the above case, $a = - 3$ and ${n}^{t h}$ term is $- 3 {\left(\frac{3}{4}\right)}^{n - 1}$.

As $r < 1$ as $n \to \infty$, ${r}^{n} \to 0$.

Hence, sum of infinite series will be $\left(- 3\right) \times \frac{1}{1 - \frac{3}{4}}$ or

$- 3 \times \frac{1}{\frac{1}{4}}$ or $- 3 \times 4 = - 12$