## Calculus: Early Transcendentals (2nd Edition)

Converges to $4$
$a_{n+1} = \frac{1}{2}a_n + 2$; $a_0 = 3$ *Decimals are approximations $a_0 = 3$ $a_1 = 3.5$ $a_2 = 3.75$ $a_3 = 3.875$ $a_4 = 3.9375$ $a_5 = 3.96875$ $a_6 = 3.98438$ $a_7 = 3.99219$ $a_8 = 3.99609$ $a_9 = 3.99805$ $a_{10} = 3.99902$ The sequence seems to approach $4$.