## Calculus: Early Transcendentals (2nd Edition)

$a_1 = \frac{3}{2}$ $a_2 = \frac{7}{4}$ $a_3 = \frac{15}{8}$ $a_4 = \frac{31}{16}$ Converges. The limit of the sequence is $2$.
$a_{n+1} = 1+\frac{a^n}{2}$; $a_0 = 2$ $a_1 = 1+\frac{a_0}{2} = \frac{3}{2}$ $a_2 = 1+\frac{a_1}{2} = \frac{7}{4}$ $a_3 = 1+\frac{a_2}{2} = \frac{15}{8}$ $a_4 = 1+\frac{a_3}{2} = \frac{31}{16}$ Converges and approaches $2$. Looking at the terms as $n$ increases shows that the terms are approaching the value of $2$.