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