## Precalculus (6th Edition) Blitzer

False. Given the geometric series $a_1=\frac{1}{2}, r=\frac{1}{2}, a_n=\frac{1}{512}$ we have $a_n=a_1r^{n-1}$ or $\frac{1}{512}=\frac{1}{2}(\frac{1}{2})^{n-1}=(\frac{1}{2})^n$ Thus, $n=9$ and the sum is $S_9=\frac{a_1(1-r^9)}{1-r}$ We can calculate, not just estimate, the sum, without precisely knowing the middle values.