This preview shows pages 1–2. 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 138 Calculus 2, Solutions to Assignment 10 1: Determine which of the following series converge. (a) X n =0 n 2 n 2 + 1 Solution: For n 1 we have 0 n 2 n 2 + 1 n 2 n 2 = 1 2 n 3 / 2 , and we know that X 1 2 n 3 / 2 converges ( since it is a constant multiple of the pseries with p = 3 2 ) , and so X n 2 n 2 + 1 converges by the C.T. (b) X n =2 1 n ln n Solution: Let f ( x ) = 1 x ln x so that a n = f ( n ). For x > 1, f ( x ) is positive, continuous and decreasing. Making the substitution u = ln x so that du = 1 x dx , we have Z 2 f ( x ) dx = Z 2 dx x ln x = Z ln 2 du u = h 2 u i ln 2 = . Since Z 2 f ( x ) dx diverges, the series X n =2 1 n ln n diverges too, by the I.T. (c) X n =1 n n 3 n n ! Solution: Let a n = n n 3 n n ! . Then a n +1 a n = ( n + 1) n +1 3 n +1 ( n + 1)! 3 n n ! n n = 1 3 n + 1 n n e 3 < 1, so X a n converges by the R.T. ( If you dont remember that lim n n + 1 n n = e , then use lH opitals Rule ) . 2: For each of the following series, determine whether it converges, and if so then determine whether it converges absolutely. (a) X n =1 ( 1) n n n + 2 Solution: Let a n = ( 1) n n n + 2 . Then  a n  = n n + 2 . Since the sequence  a n  decreases to 0, X a n converges by tha A.S.T. On the other hand, if we let b n = n n = 1 n , then we have  a n  b n 1 and X b n diverges ( it is a pseries with p =...
View Full
Document
 Spring '08
 MARSHMAN
 Calculus

Click to edit the document details