Chapter 11 - Infinite Series - 11.2 Summing and Infinite Series - Exercises - Page 548: 35

$$\frac{7}{15}$$

Given $$\frac{7}{8}-\frac{49}{64}+\frac{343}{512}-\frac{2401}{4096}+\cdots=\sum_{n=0}^{n=\infty} \frac{7}{8}\left(\frac{-7}{8}\right)^{n} $$ Since the series is a geometric series with $|r|= \frac{7}{8}<1$, then the series converges and has the sum \begin{align*} S &=\frac{a_1}{1-r} \\ &=\frac{\frac{7}{8}}{1-\frac{-7}{8}} \\ &=\frac{7}{15} \end{align*}