Calculus Question. Help Me Please!!?

Question

Ok, I am doing an AP Calc BC Free response. The question is

Let f be the function defined by f(x) = sum of the terms to infiniti, with n =1 of ((x^n)(n^n))/((3^n)(n!))

a)Find the radius of convergence of this series
b)Use the first 3 terms of htis series to find an apporximation of f(-1)
c)Estimate the amount of error involved in the approxmiation in part b. Justify your answer

Thanks

Answer 1

Tricky. The key here is to realize that, up to factors that grow more slowly than any geometric series, n! ~ n^n/e^n. Therefore, [n=1, ∞]∑(x^n n^n)/(3^n n!) converges iff the series [n=1, ∞]∑(x^n n^n)/(3^n n^n/e^n) does, and [n=1, ∞]∑(x^n n^n)/(3^n n^n/e^n) = [n=1, ∞]∑(ex/3)^n. This series converges iff |ex/3| < 1, which is true iff |x|<3/e, so the radius of convergence is 3/e.

Part b is much easier, and amounts only to simple arithmetic:

f(-1) ≈ [n=1, 3]∑((-1)^n n^n)/(3^n n!) = -1/3 + 2/9 - 1/6 = -5/18

For part c, note that since [n=1, ∞]∑((-1)^n n^n)/(3^n n!) is an alternating series the absolute value of whose terms are monotone decreasing, we can expect the error after the first three terms to be less than or equal to the absolute value of the fourth term, which is (4/3)^4/4! ≈ 0.132.