Chapter 11 - Infinite Series - 11.2 Summing and Infinite Series - Exercises - Page 547: 33

$$4$$

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