## Calculus (3rd Edition)

$$\frac{4}{3}$$
Given $$\frac{4}{9}+\frac{8}{27}+\frac{16}{81}+\frac{32}{243}+\cdots$$ This is a geometric series with $|r|= \frac{8}{27}\frac{9}{4}=\frac{2}{3}<1$, so the series converges and has the sum \begin{align*} S&=\frac{a}{1-r}\\ &=\frac{4/9}{1-\frac{2}{3}}\\ &= \frac{4}{3} \end{align*}