## Calculus: Early Transcendentals (2nd Edition)

Converges to $4$
$a_{n+1} = \frac{1}{2}\sqrt a_n + 3$; $a_0 = 1000$ *Decimals are approximations $a_0 = 1000$ $a_1 = 18.8114$ $a_2 = 5.1686$ $a_3 = 4.13673$ $a_4 = 4.01695$ $a_5 = 4.00212$ $a_6 = 4.00026$ $a_7 = 4.00003$ $a_8 = 4.0$ $a_9 = 4.0$ $a_{10} = 4.0$ The terms seem to approach closer and closer to 4 as the sequence continues.