Chapter 11 - Section 11.2 - Series - 11.2 Exercises - Page 715: 3

$\Sigma^{\infty}_{n=1}a_{n}=2$

$s_{n} = 2-3(0 \times 8)^{n}$ $\lim\limits_{n \to \infty}s_{n} = \lim\limits_{n \to \infty}(2-3(0.8)^{n})$ $=\lim\limits_{n \to \infty}(2)-3 \times \lim\limits_{n \to \infty} (0.8)^{n}$ $=2-3(0)$ $=2$ $\Sigma^{\infty}_{n=1}a_{n}=2$