# chep 9 calculas problem solve

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• 1.Instructors Resource Manual Section 9.1 513 CHAPTER 9 Infinite Series 9.1 Concepts Review 1. a sequence 2. lim n n a exists (finite sense) 3. bounded above 4. 1; 1 Problem Set 9.1 1. 1 2 3 4 5 1 2 3 4 5 , , , , 2 5 8 11 14 a a a a a= = = = = 1 1 1 lim lim ; 3 1 33 n n n n n = = converges 2. 1 2 3 4 5 5 8 11 14 17 , , , , 2 3 4 5 6 a a a a a= = = = = 2 1 33 2 lim lim 3; 1 1 n n n n n n ++ = = + + converges 3. 1 2 3 6 18 38 2, 2, , 3 9 17 a a a= = = = = 4 5 66 22 102 34 , 27 9 39 13 a a= = = = 2 2 22 2 3 1 44 2 lim lim 4; 1 3 1 n n n n n n n n ++ = = ++ converges 4. 1 2 3 4 5 14 29 50 77 5, , , , 3 5 7 9 a a a a a= = = = = 22 1 33 2 lim lim ; 2 1 2 n n n n nn n ++ = = diverges 5. 1 2 3 7 26 63 , , , 8 27 64 a a a= = = 4 5 124 215 , 125 216 a a= = 3 2 3 2 3 3 2 3 3 3 3 lim lim ( 1) 3 3 1n n n n n n n n n n n n + + + + = + + + + 2 2 3 3 3 3 3 1 1 lim 1 1 n n n n n n + + = = + + + 6. 1 2 3 5 14 29 , , , 3 5 7 a a a= = = 4 5 50 5 2 77 , 9 9 11 a a= = = 2 2 2 1 3 3 2 3 lim lim ; 2 1 22 n n n n n n + + = = + + converges 7. 1 2 3 4 1 2 1 3 4 2 , , , , 3 4 2 5 6 3 a a a a= = = = = = 5 5 7 a = 2 1 lim lim 1, 2 1n n n n n = = + + but since it alternates between positive and negative, the sequence diverges. 8. 1 2 3 4 5 2 3 4 5 1, , , , 3 5 7 9 a a a a a= = = = = 1 for odd cos( ) 1 for even n n n = 1 1 1 lim lim , 2 1 22 n n n n n = = but since cos(n ) alternates between 1 and 1, the sequence diverges. 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.