# How do you find the sum of the infinite geometric series given -8-4-2-...?

##### 1 Answer
May 19, 2018

$- 8 - 4 - 2 - \ldots = - 16$

#### Explanation:

Note that the given series is a geometric series with initial term $- 8$ and common ratio:

$\frac{- 4}{- 8} = \frac{- 2}{- 4} = \frac{1}{2}$

We find:

$- 8 - 4 - 2 - \ldots = \left(2 - 1\right) \left(- 8 - 4 - 2 - \ldots\right)$

$\textcolor{w h i t e}{- 8 - 4 - 2 - \ldots} = \left(- 16 - 8 - 4 - \ldots\right) - \left(- 8 - 4 - 2 - \ldots\right)$

$\textcolor{w h i t e}{- 8 - 4 - 2 - \ldots} = - 16$

...assuming the sum converges.

In the case of a general infinite series, we should usually look at partial sums and make sure that they converge to a limit.

The general term of a geometric series can be written:

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

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

Then we find:

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

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

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

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

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

Then (assuming $r \ne 1$) dividing both ends by $\left(1 - r\right)$ we find:

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

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

${\sum}_{n = 1}^{\infty} {a}_{r} = {\lim}_{N \to \infty} {\sum}_{n = 1}^{N} {a}_{r}$

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

$\textcolor{w h i t e}{{\sum}_{n = 1}^{\infty} {a}_{r}} = \frac{a}{1 - r}$

So a geometric series has a convergent sum if and only if its common ratio is of absolute value less than $1$.

In our example with $a = - 8$ and $r = \frac{1}{2}$ we find:

${\sum}_{n = 1}^{\infty} \left(- 8\right) {\left(\frac{1}{2}\right)}^{n - 1} = \frac{- 8}{1 - \frac{1}{2}} = - 16$