This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 < ....
View
Full
Document
This note was uploaded on 01/11/2012 for the course MATH 100 taught by Professor Loveys during the Fall '11 term at McGill.
 Fall '11
 Loveys
 Math

Click to edit the document details