# How do you find the sum of the infinite geometric series 64,-16,4,-1....?

Jan 8, 2016

$\frac{256}{5}$

#### Explanation:

Assuming that the series is convergent, let it be $S$. Therefore

$S = 64 - 16 + 4 - 1 + \ldots$

$= {4}^{3} - {4}^{2} + {4}^{1} - {4}^{0} + \ldots$

Then, dividing $S$ by 4 gives

$\frac{S}{4} = 16 - 4 + 1 - \frac{1}{4} + \ldots$

$= {4}^{2} - {4}^{1} + {4}^{0} - {4}^{- 1} + \ldots$

Now comes the magic. Observe that

$S = {4}^{3} - \left({4}^{2} - {4}^{1} + {4}^{0} - \ldots\right)$

$= {4}^{3} - \left(\frac{S}{4}\right)$

$\frac{5 S}{4} = 64$

$S = \frac{256}{5}$