#### Answer

$$\frac{{256}}{3}$$

#### Work Step by Step

$$\eqalign{
& 64 + 16 + 4 + 1 + \cdots \cr
& {\text{Sum of a Geometric Series }}\left( {{\text{see page 635}}} \right) \cr
& {\text{The infinite geometric series}} \cr
& a + ar + a{r^2} + a{r^3} + \cdots \cr
& {\text{converges}}{\text{, if }}r{\text{ is in }}\left( { - 1,1} \right),{\text{ to the sum }}\frac{a}{{1 - r}}.{\text{ And diverges if }}r{\text{ is not in }}\left( { - 1,1} \right) \cr
& {\text{then this is a geometric series}}{\text{, with }}a = {a_1} = 64 \cr
& r = \frac{{16}}{{64}} = \frac{1}{4} \cr
& {\text{Since }}r{\text{ is in the interval }}\left( { - 1,1} \right),{\text{ the series converges and has a sum }}\frac{a}{{1 - r}} \cr
& \frac{a}{{1 - r}} = \frac{{64}}{{1 - 1/4}} \cr
& {\text{simplifying}} \cr
& \frac{a}{{1 - r}} = \frac{{256}}{3} \cr} $$