MATH222 A2 solution

# MATH222 A2 solution - MATH 222 HOMEWORK 2 DUE SEPTEMBER 26,...

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Unformatted text preview: MATH 222 HOMEWORK 2 DUE SEPTEMBER 26, 2011 1. PROBLEMS Problem 1.1. Find the interval of convergence. (a.) X n =1 (3 n + 2) x n (b.) X n =0 (8 x ) n n ! (c.) X n =0 ( n + 2) 2 x n (d.) 100 X n =1 x n n Solutions Part a.) Set a n = (3 n + 2) x n . The series has the form n =1 a n . The ratio test tells us that the series converges if lim n a n +1 a n < 1 . Notice that lim n a n +1 a n = lim n (3 n +1 + 2) x n +1 (3 n + 2) x n = | x | lim n (3 + 2 3 n ) (1 + 2 3 n ) = | x | (3 + 0) (1 + 0) =3 | x | . The series certainly converges if 3 | x | < 1 , 1 2 MATH 222 HOMEWORK 2 DUE SEPTEMBER 26, 2011 which is equivalent to saying- 1 3 < x < 1 3 . Dont forget to check the convergence at the endpoints of the interval (- 1 / 3 , 1 / 3) . When x =- 1 / 3 , the series becomes X n =1 (3 n + 2) (- 1) n 3 n . The summand does not approach zero as n . Therefore the series diverges by the di- vergence test. A similar argument applies when x = 1 / 3 . The interval of convergence is- 1 / 3 < x < 1 / 3 . Part b.) Let a n = (8 x ) n n ! , then use the ratio test. We have lim n a n +1 a n = lim n (8 x ) n +1 ( n + 1)! n ! (8 x ) n = lim n (8 x ) n + 1 = 0 . The limit is always less that 1 . We conclude that the series converges for all x . The interval of convergence is- < x < ....
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## This note was uploaded on 01/11/2012 for the course MATH 100 taught by Professor Loveys during the Fall '11 term at McGill.

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MATH222 A2 solution - MATH 222 HOMEWORK 2 DUE SEPTEMBER 26,...

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