## Calculus: Early Transcendentals (2nd Edition)

Converges to $-4$
$a_n = \frac{1}{4}a_{n-1}-3$; $a_0 = 1$ *Decimals are approximations $a_0 = 1$ $a_1 = -2.75$ $a_2 = -3.6875$ $a_3 = -3.92188$ $a_4 = -3.98047$ $a_5 = -3.99512$ $a_6 = -3.99878$ $a_7 = -3.99969$ $a_8 = -3.99992$ $a_9 = -3.99998$ $a_{10} = -4.0$ The terms of the sequence seems to be approaching $-4$