## Calculus: Early Transcendentals (2nd Edition)

$a_1 = \frac{2}{3}$ $a_2 =\frac{2}{3}$ $a_3 = \frac{2}{3}$ $a_4 = \frac{2}{3}$ Converges. The limit of the sequence approaches $\frac{2}{3}$.
$a_{n+1} = 1-\frac{a_n}{2}$; $a_0 = \frac{2}{3}$ $a_1 = 1- \frac{a_0}{2} = \frac{2}{3}$ $a_2 = 1- \frac{a_1}{2} =\frac{2}{3}$ $a_3 = 1- \frac{a_2}{2} = \frac{2}{3}$ $a_4 = 1- \frac{a_3}{2} = \frac{2}{3}$ Converges and approaches $\frac{2}{3}$. The terms of the sequence are constant.