Chapter 11 - Infinite Series - 11.2 Summing and Infinite Series - Exercises - Page 547: 27

$$\frac{59049}{3328}$$

Given $$ \sum_{n=-4}^{n=\infty} \left(\frac{-4}{9}\right)^{n}$$ Since the series is a geometric series with $|r|=\left|\frac{-4}{9}\right|<1$ , then the series converges and has the sum \begin{align*} S&= \frac{a_1}{1-r}\\ &=\frac{ \left(\frac{-4}{9}\right)^{-4} }{1-\frac{-4}{9}}\\ &=\frac{59049}{3328} \end{align*}