## Calculus (3rd Edition)

$$\frac{8}{5}$$
Given $$\frac{2^{3}}{7}+\frac{2^{4}}{7^{2}}+\frac{2^{5}}{7^{3}}+\frac{2^{6}}{7^{4}}+\cdots=\sum_{n=0}^{n=\infty} \frac{8}{7}\left(\frac{2}{7}\right)^{n}$$ Since the series is a geometric series with $|r|= \frac{2}{7}<1$, then the series converges and has the sum \begin{align*} S &=\frac{a_1}{1-r} \\ &=\frac{\frac{8}{7}}{1-\frac{2}{7}} \\ &=\frac{8}{5} \end{align*}