# How do you find the sum of the infinite geometric series 2/3 - 4/9 + 8/27 - 16/81 +...?

May 23, 2018

$\frac{2}{5}$

#### Explanation:

all geometric sequences have a starting term $a$ and common ratio $r$.

the common ratio $r$ is the number that one term in the sequence is multiplied by to find the next, hence the name 'ratio'.
it is the same for all terms in the sequence, hence the name 'common'.

the common ratio here would be the number that $\frac{2}{3}$ is multiplied by to make $- \frac{4}{9}$, as well as the number that $- \frac{4}{9}$ is multiplied by to make $\frac{8}{27}$.

$\frac{2}{3} \cdot \frac{2}{3} = \frac{2 \cdot 2}{3 \cdot 3}$, which is $\frac{4}{9}$.

$\frac{2}{3} \cdot - \frac{2}{3} = - \frac{2 \cdot 2}{3 \cdot 3}$, which is $- \frac{4}{9}$.

the number that $\frac{2}{3}$ is multiplied by to make $- \frac{4}{9}$ is $- \frac{2}{3}$.
therefore, the common ratio $r$ is $- \frac{2}{3}$.

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the infinite sum of a geometric series can only be found in certain cases: when $r$ is between $- 1$ and $1$.

when $r$ is between these values, the series, when being added, will converge - become closer on either side - to a certain number.
otherwise, the sums will diverge - become further away from each other - meaning that there is no certain value that the series can be said to approach.

here, $r$ is indeed in between $- 1$ and $1$. this means that the infinite series can be calculated as a number that the series converges to.

this calculation is carried out using the formula ${S}_{\infty} = \frac{a}{1 - r}$.

$a$, the starting term in the sequence, is $\frac{2}{3}$ here.
$r$, the common ratio, is $- \frac{2}{3}$ here.

here, $1 - r$ is $1 - \left(- \frac{2}{3}\right)$, which is $\frac{5}{3}$.
here, $\frac{a}{1 - r}$ is $\frac{2}{3} \div \frac{5}{3}$.

$\frac{2}{3} \div \frac{5}{3}$ is the same as $\frac{2}{3} \cdot \frac{3}{5}$.

cross-cancelling the $3$ out gives $\frac{2}{1} \cdot \frac{1}{5}$, which is $\frac{2}{5}$.

hence, the sum to infinity $\left({S}_{\infty}\right)$ is $\frac{2}{5}$.