# What is the sum of an infinite geometric series with a first term of 6 and a common ratio of 1/2?

Feb 9, 2017

$12$

#### Explanation:

The general term of a geometric series can be represented by the formula:

${a}_{n} = a {r}^{n - 1}$

where $a$ is the initial term and $r$ the common ratio.

We find:

$\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1} = {\sum}_{n = 1}^{N} a {r}^{n - 1} - r {\sum}_{n = 1}^{N} a {r}^{n - 1}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1}} = {\sum}_{n = 1}^{N} a {r}^{n - 1} - {\sum}_{n = 2}^{N + 1} a {r}^{n - 1}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1}} = a + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N} a {r}^{n - 1}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N} a {r}^{n - 1}}}} - a {r}^{N}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1}} = a \left(1 - {r}^{N}\right)$

Dividing both ends by $\left(1 - r\right)$ we find:

${\sum}_{n = 1}^{N} a {r}^{n - 1} = \frac{a \left(1 - {r}^{N}\right)}{1 - r}$

If $\left\mid r \right\mid < 1$ then ${\lim}_{N \to \infty} {r}^{N} = 0$ and we find:

${\sum}_{n = 1}^{\infty} a {r}^{n - 1} = {\lim}_{N \to \infty} \frac{a \left(1 - {r}^{N}\right)}{1 - r} = \frac{a}{1 - r}$

In the given example, $a = 6$ and $r = \frac{1}{2}$. So $\left\mid r \right\mid < 1$ and:

${\sum}_{n = 1}^{\infty} 6 \cdot {\left(\frac{1}{2}\right)}^{n - 1} = \frac{6}{1 - \frac{1}{2}} = 12$