Chapter 11 - Infinite Series - 11.2 Summing and Infinite Series - Exercises - Page 547: 24

$$\frac{64}{25}$$

Given $$\frac{4^{3}}{5^{3}}+\frac{4^{4}}{5^{4}}+\frac{4^{5}}{5^{5}}+\cdots$$ Since the series is a geometric series with $|r|=\left|\frac{4}{5}\right|<1$ , then the series converges and has the sum \begin{align*} S&= \frac{a_1}{1-r}\\ &=\frac{\frac{4^{3}}{5^{3}}}{1-\frac{4}{5}}\\ &= \frac{4^{3}}{5^{2}}=\frac{64}{25} \end{align*}