Mark M. answered 01/01/24
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
1 / (1 - x) = 1 + x + x2 + x3 + ... for -1 < x < 1
d/dx[5ln(4 - x)] = -5/(4 - x) = -5 / [4(1 - x/4)] = (-5/4)[1 / (1 - (x/4))]
So, 5ln(4 - x) = (-5/4) ∫ [1 / (1 - (x/4)]dx = (-5/4) ∫ [1 + (x/4) + (x/4)2 + (x/4)3 + (x/4)4 + ...]dx, for -1 < x/4 < 1
= (-5/4)[x + x2/8 + x3/48 + x4/256 + ...] for -4 < x < 4
= (-5/4)x + (-5/32)x2 + (-5/192)x3 + (-5/1024)x4 + ... for -4 < x < 4
c0 = 0, c1 = -5/4, c2 = -5/32, c3 = -5/192, c4 = -5/1024
Interval of convergence = (-4,4). Radius of convergence = (4 - (-4)) / 2 = 4.