# How do you find the sum of the infinite geometric series 1/8-1/16+1/32-1/64+...?

Apr 28, 2018

${S}_{\infty} = \frac{1}{12}$

#### Explanation:

#1/8-1/16+1/32-1/64--(1)

a GP is in the form

$a + a r + a {r}^{2} + . . a {r}^{n - 1} + . .$

for $\left(1\right)$

we have $a = \frac{1}{8}$

to find the common ratio $r$ we divide adjacent terms

$r = \frac{- \frac{1}{16}}{\frac{1}{8}} = - \frac{1}{2}$

for a GP to have a sum to infinity

$| r | < 1$

$| r | = | - \frac{1}{2} | < 1$

$\therefore \exists {S}_{\infty}$

${S}_{\infty} = \frac{a}{1 - r}$

${S}_{\infty} = \frac{\frac{1}{8}}{1 - - \frac{1}{2}}$

${S}_{\infty} = \frac{\frac{1}{8}}{\left(\frac{3}{2}\right)} = \frac{1}{8} \div \frac{3}{2}$

${S}_{\infty} = \frac{1}{12}$