MATH 201 Other Solution for Math 201 Exercises, Spring 2018

INSTRUCTOR’S SOLUTIONS MANUAL MULTIVARIABLE WILLIAM ARDIS Collin County Community College THOMAS’ CALCULUS TWELFTH EDITION BASED ON THE ORIGINAL WORK BY George B. Thomas, Jr. Massachusetts Institute of Technology AS REVISED BY Maurice D. Weir Naval Postgraduate School Joel Hass University of California, Davis The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Addison-Wesley from electronic files supplied by the author. Copyright © 2010, 2005, 2001 Pearson Education, Inc. Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-321-60072-1 ISBN-10: 0-321-60072-X 1 2 3 4 5 6 BB 14 13 12 11 10 PREFACE TO THE INSTRUCTOR This Instructor's Solutions Manual contains the solutions to every exercise in the 12th Edition of THOMAS' CALCULUS by Maurice Weir and Joel Hass, including the Computer Algebra System (CAS) exercises. The corresponding Student's Solutions Manual omits the solutions to the even-numbered exercises as well as the solutions to the CAS exercises (because the CAS command templates would give them all away). In addition to including the solutions to all of the new exercises in this edition of Thomas, we have carefully revised or rewritten every solution which appeared in previous solutions manuals to ensure that each solution ì conforms exactly to the methods, procedures and steps presented in the text ì is mathematically correct ì includes all of the steps necessary so a typical calculus student can follow the logical argument and algebra ì includes a graph or figure whenever called for by the exercise, or if needed to help with the explanation ì is formatted in an appropriate style to aid in its understanding Every CAS exercise is solved in both the MAPLE and MATHEMATICA computer algebra systems. A template showing an example of the CAS commands needed to execute the solution is provided for each exercise type. Similar exercises within the text grouping require a change only in the input function or other numerical input parameters associated with the problem (such as the interval endpoints or the number of iterations). For more information about other resources available with Thomas' Calculus, visit http://pearsonhighered.com. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. TABLE OF CONTENTS 10 Infinite Sequences and Series 569 10.1 Sequences 569 10.2 Infinite Series 577 10.3 The Integral Test 583 10.4 Comparison Tests 590 10.5 The Ratio and Root Tests 597 10.6 Alternating Series, Absolute and Conditional Convergence 602 10.7 Power Series 608 10.8 Taylor and Maclaurin Series 617 10.9 Convergence of Taylor Series 621 10.10 The Binomial Series and Applications of Taylor Series 627 Practice Exercises 634 Additional and Advanced Exercises 642 11 Parametric Equations and Polar Coordinates 647 11.1 Parametrizations of Plane Curves 647 11.2 Calculus with Parametric Curves 654 11.3 Polar Coordinates 662 11.4 Graphing in Polar Coordinates 667 11.5 Areas and Lengths in Polar Coordinates 674 11.6 Conic Sections 679 11.7 Conics in Polar Coordinates 689 Practice Exercises 699 Additional and Advanced Exercises 709 12 Vectors and the Geometry of Space 715 12.1 Three-Dimensional Coordinate Systems 715 12.2 Vectors 718 12.3 The Dot Product 723 12.4 The Cross Product 728 12.5 Lines and Planes in Space 734 12.6 Cylinders and Quadric Surfaces 741 Practice Exercises 746 Additional Exercises 754 13 Vector-Valued Functions and Motion in Space 759 13.1 Curves in Space and Their Tangents 759 13.2 Integrals of Vector Functions; Projectile Motion 764 13.3 Arc Length in Space 770 13.4 Curvature and Normal Vectors of a Curve 773 13.5 Tangential and Normal Components of Acceleration 778 13.6 Velocity and Acceleration in Polar Coordinates 784 Practice Exercises 785 Additional Exercises 791 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 14 Partial Derivatives 795 14.1 Functions of Several Variables 795 14.2 Limits and Continuity in Higher Dimensions 804 14.3 Partial Derivatives 810 14.4 The Chain Rule 816 14.5 Directional Derivatives and Gradient Vectors 824 14.6 Tangent Planes and Differentials 829 14.7 Extreme Values and Saddle Points 836 14.8 Lagrange Multipliers 849 14.9 Taylor's Formula for Two Variables 857 14.10 Partial Derivatives with Constrained Variables 859 Practice Exercises 862 Additional Exercises 876 15 Multiple Integrals 881 15.1 Double and Iterated Integrals over Rectangles 881 15.2 Double Integrals over General Regions 882 15.3 Area by Double Integration 896 15.4 Double Integrals in Polar Form 900 15.5 Triple Integrals in Rectangular Coordinates 904 15.6 Moments and Centers of Mass 909 15.7 Triple Integrals in Cylindrical and Spherical Coordinates 914 15.8 Substitutions in Multiple Integrals 922 Practice Exercises 927 Additional Exercises 933 16 Integration in Vector Fields 939 16.1 Line Integrals 939 16.2 Vector Fields and Line Integrals; Work, Circulation, and Flux 944 16.3 Path Independence, Potential Functions, and Conservative Fields 952 16.4 Green's Theorem in the Plane 957 16.5 Surfaces and Area 963 16.6 Surface Integrals 972 16.7 Stokes's Theorem 980 16.8 The Divergence Theorem and a Unified Theory 984 Practice Exercises 989 Additional Exercises 997 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 SEQUENCES 1. a 0, a , a , a " # $ % � "� " � � # œ œ œ œ � œ œ � œ œ � 1 1 2 1 3 2 1 4 3 1 4 3 9 4 16 # # # # 2. a 1, a , a , a " # $ % " " # œ œ œ œ œ œ œ œ 1 1 1 1 1 1! ! 2 3! 6 4! 24 3. a 1, a , a , a " # $ % � �" � � # � � � � " " " œ œ œ œ � œ œ œ œ � ( 1) ( ) ( 1) ( 1) 1 4 1 3 6 1 5 8 1 7 # $ % & 4. a 2 ( 1) 1, a 2 ( 1) 3, a 2 ( 1) 1, a 2 ( 1) 3 " # $ % " # $ % œ � � œ œ � � œ œ � � œ œ � � œ 5. a , a , a , a " # $ % # # # # # " " " " # œ œ œ œ œ œ œ œ 2 2 2 2 2 2 # $ & # $ % % 6. a , a , a , a " # $ % � " " � � � " # # œ œ œ œ œ œ œ œ 2 2 1 3 2 1 7 2 15 2 4 2 8 16 2 # $ % # $ % 7. a 1, a 1 , a , a , a , a , " # $ % & ' " " " " # # # # # # œ œ � œ œ � œ œ � œ œ � œ œ 3 3 7 7 15 15 31 63 4 4 8 8 16 32 # $ % a , a , a , a ( ) * "! œ œ œ œ 127 255 511 1023 64 128 256 512 8. a 1, a , a , a , a , a , a , a , " # $ % & ' ( ) " " " " " " " # # # # œ œ œ œ œ œ œ œ œ œ œ ˆ ‰ ˆ ‰ ˆ ‰ " " " # # 3 6 4 4 5 1 0 7 0 5040 40,320 6 4 a , a * "! " " œ œ 362,880 3,628,800 9. a 2, a 1, a , a , a , " # $ % & � � # # # # " " " � � � � œ œ œ œ œ � œ œ � œ œ ( 1) (2) ( 1) (1) 2 4 8 ( 1) ( 1) # $ % " # & " ˆ ‰ ˆ ‰ 4 a , a , a , a , a ' ( ) * "! " " " " " # # œ œ � œ � œ œ 16 3 64 1 8 256 10. a 2, a 1, a , a , a , a , " # $ % & ' � � # # � � " " œ � œ œ � œ œ � œ œ � œ œ � œ � 1 ( 2) 2 ( 1) 3 3 4 5 5 3 2 2 3 4 † † † † ˆ ‰ ˆ ‰ 2 3 " # a , a , a , a ( ) * "! " " œ � œ � œ � œ � 2 2 7 4 9 5 11. a 1, a 1, a 1 1 2, a 2 1 3, a 3 2 5, a 8, a 13, a 21, a 34, a 55 " # $ % & ' ( ) * "! œ œ œ � œ œ � œ œ � œ œ œ œ œ œ 12. a 2, a 1, a , a , a 1, a 2, a 2, a 1, a , a " # $ % & ' ( ) * "! " " " " # � # # # � � œ œ � œ � œ œ œ œ � œ � œ œ � œ � œ ˆ ‰ ˆ ‰ ˆ ‰ " " # # " # 1 13. a ( 1) , n 1, 2, 14. a ( 1) , n 1, 2, n n n 1 n œ � œ á œ � œ á � 15. a ( 1) n , n 1, 2, 16. a , n 1, 2, n n n 1 ( ) n œ � œ á œ œ á � # �" n 1 � # 17. a , n 1, 2, 18. a , n 1, 2, n n 2 2n 5 3 n 2 n n 1 œ œ á œ œ á n � 1 a b a b � � � 19. a n 1, n 1, 2, 20. a n 4 , n 1, 2, n n œ � œ á œ � œ á # 21. a 4n 3, n 1, 2, 22. a 4n 2 , n 1, 2, n n œ � œ á œ � œ á 23. a , n 1, 2, 24. a , n 1, 2, n n 3n 2 n n! 5 œ œ á œ œ á � 3 n 1 � Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 570 Chapter 10 Infinite Sequences and Series 25. a , n 1, 2, 26. a , n 1, 2, n n 1 ( 1) n ( 1) n œ œ á œ œ Ú Û œ á � � # # # � � � n 1 n � " " # # ˆ ‰ 27. lim 2 (0.1) 2 converges (Theorem 5, #4) n Ä _ � œ Ê n 28. lim lim 1 1 converges n n Ä _ Ä _ n ( ) ( 1) n n � �" � n n œ � œ Ê 29. lim lim lim 1 converges n n n Ä _ Ä _ Ä _ " � � � # # � � 2n 2 1 n 2 2 œ œ œ � Ê ˆ ‰ ˆ ‰ " " n n 30. lim lim diverges n n Ä _ Ä _ 2n 1 3 n 2 n 3 � " � � � È È Š ‹ Š ‹ œ œ �_ Ê " " È È n n 31. lim lim 5 converges n n Ä _ Ä _ " � � � � 5n n 8n 5 1 % % $ " % œ œ � Ê Š ‹ ˆ ‰ n 8 n 32. lim lim lim 0 converges n n n Ä _ Ä _ Ä _ n 3 n 3 n 5n 6 (n 3)(n 2) n � � " � � � � � # # œ œ œ Ê 33. lim lim lim (n 1) diverges n n n Ä _ Ä _ Ä _ n 2n 1 n 1 n 1 (n 1)(n 1) # � � � � � � œ œ � œ _ Ê 34 lim lim diverges n n Ä _ Ä _ " � � � � n 70 4n n 4 $ # œ œ _ Ê Š ‹ Š ‹ " # # n 70 n 35. lim 1 ( 1) does not exist diverges 36. lim ( 1) 1 does not exist diverges n n Ä _ Ä _ a b ˆ ‰ � � Ê � � Ê n n " n 37. lim 1 lim 1 converges n n Ä _ Ä _ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ n n n n n � " " " " " " # # # # � œ � � œ Ê 38. lim 2 3 6 converges 39. lim 0 converges n n Ä _ Ä _ ˆ ‰ ˆ ‰ � � œ Ê œ Ê " " # # # � �" n n n 1 ( ) n 1 � 40. lim lim 0 converges n n Ä _ Ä _ ˆ � ‰ œ œ Ê " # # �" n ( )n n 41. lim lim lim 2 converges n n n Ä _ Ä _ Ä _ É É Ê Š ‹ È 2n 2n n 1 n 1 � � œ œ œ Ê 2 1 � " n 42. lim lim diverges n n Ä _ Ä _ " " (0.9) 9 0 n œ œ _ Ê ˆ ‰ n 43. lim sin sin lim sin 1 converges n n Ä _ Ä _ ˆ ‰ ˆ ‰ Š ‹ 1 1 1 # # # " " � œ � œ œ Ê n n 44. lim n cos (n ) lim (n )( 1) does not exist diverges n n Ä _ Ä _ 1 1 1 œ � Ê n 45. lim 0 because converges by the Sandwich Theorem for sequences n Ä _ sin n sin n n n n n œ � Ÿ Ÿ Ê " " 46. lim 0 because 0 converges by the Sandwich Theorem for sequences n Ä _ sin n sin n # # # # # " n n n œ Ÿ Ÿ Ê 47. lim lim 0 converges (using l'Hopital's rule) ^ n n Ä _ Ä _ n ln 2 # # " n n œ œ Ê Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.1 Sequences 571 48. lim lim lim lim diverges (using l'Hopital's rule) ^ n n n n Ä _ Ä _ Ä _ Ä _ 3 3 ln 3 n 3n 6n 6 3 (ln 3) 3 (ln 3) n n n n $ # # $ œ œ œ œ _ Ê 49. lim lim lim lim 0 converges n n n n Ä _ Ä _ Ä _ Ä _ ln (n ) n 2 n n 1 1 � " � � È È Š ‹ Š ‹ œ œ œ œ Ê ˆ ‰ Š ‹ " � " # " n 1 n 2 n n È È 50. lim lim 1 converges n n Ä _ Ä _ ln n ln 2n œ œ Ê ˆ ‰ ˆ ‰ " n 2 2n 51. lim 8 1 converges (Theorem 5, #3) n Ä _ 1 n Î œ Ê 52. lim (0.03) 1 converges (Theorem 5, #3) n Ä _ 1 n Î œ Ê 53. lim 1 e converges (Theorem 5, #5) n Ä _ ˆ ‰ � œ Ê 7 n n ( 54. lim 1 lim 1 e converges (Theorem 5, #5) n n Ä _ Ä _ ˆ ‰ ’ “ � œ � œ Ê " �" �" n n ( ) n n 55. lim 10n lim 10 n 1 1 1 converges (Theorem 5, #3 and #2) n n Ä _ Ä _ È n œ œ œ Ê 1 n 1 n Î Î † † 56. lim n lim n 1 1 converges (Theorem 5, #2) n n Ä _ Ä _ È ˆ ‰ È n n # # # œ œ œ Ê 57. lim 1 converges (Theorem 5, #3 and #2) n Ä _ ˆ ‰ 3 n 1 1 n lim 3 lim n Î " œ œ œ Ê n n Ä_ Ä_ 1 n 1 n Î Î 58. lim (n 4) lim x 1 converges; (let x n 4, then use Theorem 5, #2) n x Ä _ Ä _ � œ œ Ê œ � 1 n 4 1 x ÎÐ � Ñ Î 59. lim diverges (Theorem 5, #2) n Ä _ ln n n lim n lim ln n 1 1 n 1 n Î Î œ œ œ _ Ê n n Ä_ Ä_ _ 60. lim ln n ln (n 1) lim ln ln lim ln 1 0 converges n n n Ä _ Ä _ Ä _ c d ˆ ‰ Š ‹ � � œ œ œ œ Ê n n n 1 n 1 � � 61. lim 4 n lim 4 n 4 1 4 converges (Theorem 5, #2) n n Ä _ Ä _ È È n n n œ œ œ Ê † 62. lim 3 lim 3 lim 3 3 9 1 9 converges (Theorem 5, #3) n n n Ä _ Ä _ Ä _ È n 2n 1 2 1 n 1 n � � Î # Î œ œ œ œ Ê a b † † 63. lim lim lim 0 and 0 lim 0 converges n n n n Ä _ Ä _ Ä _ Ä _ n! n! n! n n n n n n n n n 2 3 (n 1)(n) n n n œ Ÿ œ Ê œ Ê " â � â " † † † † † ˆ ‰ 64. lim 0 converges (Theorem 5, #6) n Ä _ ( 4) n! � n œ Ê 65. lim lim diverges (Theorem 5, #6) n n Ä _ Ä _ n! 106n (10 )n n! œ œ _ Ê " Š ‹ ' 66. lim lim diverges (Theorem 5, #6) n n Ä _ Ä _ n! 2 3 n n 6n n! œ œ _ Ê " ˆ ‰ 67. lim lim exp ln lim exp e converges n n n Ä _ Ä _ Ä _ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ " " " � ÎÐ Ñ �" n ln n n ln n 1 ln n ln 1 ln n œ œ œ Ê Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 572 Chapter 10 Infinite Sequences and Series 68. lim ln 1 ln lim 1 ln e 1 converges (Theorem 5, #5) n n Ä _ Ä _ ˆ ‰ ˆ ‰ Š ‹ � œ � œ œ Ê " " n n n n 69. lim lim exp n ln lim exp n n n Ä _ Ä _ Ä _ ˆ ‰ ˆ ‰ ˆ ‰ Š ‹ 3n 3n 3n 1 3n 1 n ln (3n 1) ln (3n 1) � " � " � � � � � œ œ " n lim exp lim exp exp e converges œ œ œ œ Ê n n Ä _ Ä _ � � Š ‹ ˆ ‰ 3 3 3n 1 3n 1 n � � " # � Š� ‹ 6n 6 (3n 1)(3n 1) 9 # � � #Î$ 70. lim lim exp n ln lim exp lim exp n n n n Ä _ Ä _ Ä _ Ä _ ˆ ‰ ˆ ‰ ˆ ‰ Š ‹ � � n n n 1 n 1 n ln n ln (n 1) � � � � œ œ œ ˆ ‰ Š ‹ " " " � " # n n n 1 n � � lim exp e converges œ � œ Ê n Ä _ Š ‹ n n(n 1) # � �" 71. lim lim x x lim exp ln x lim exp n n n n Ä _ Ä _ Ä _ Ä _ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ Š ‹ x 2n 1 n 1 n n 1 n 1 n 1 n ln (2n 1) n � # � # � Î " " " Î � � œ œ œ x lim exp xe x, x 0 converges œ œ œ � Ê n Ä _ ˆ ‰ � � ! 2 2n 1 72. lim 1 lim exp n ln 1 lim exp lim exp n n n n Ä _ Ä _ Ä _ Ä _ ˆ ‰ ˆ ‰ ˆ ‰ � � – — � œ � œ œ " " n n n # # ln 1 1 Š ‹ ‚ ˆ ‰ Š ‹ Š ‹ Š ‹ � � � " # " $ # " " # n n 2 n n n lim exp e 1 converges œ œ œ Ê n Ä _ ˆ ‰ � � ! 2n n 1 # 73. lim lim 0 converges (Theorem 5, #6) n n Ä _ Ä _ 3 6 36 2 n! n! n n n n† � † œ œ Ê 74. lim lim lim 0 converg n n n Ä _ Ä _ Ä _ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ 10 12 10 120 11 11 11 121 n n n n 9 11 12 9 12 11 108 10 12 11 10 11 12 110 n n n n n n n � � � œ œ œ Ê 1 es (Theorem 5, #4) 75. lim tanh n lim lim lim lim 1 converges n n n n n Ä _ Ä _ Ä _ Ä _ Ä _ œ œ œ œ " œ Ê e e e 2e e e e 1 2e n n 2n 2n n n 2n 2n � � " � � � � 76. lim sinh (ln n) lim lim diverges n n n Ä _ Ä _ Ä _ œ œ œ _ Ê e e 2 n ln n ln n n � � # � " ˆ ‰ 77. lim lim lim lim converges n n n n Ä _ Ä _ Ä _ Ä _ n sin 2n 1 # " ˆ ‰ n � # " œ œ œ œ Ê sin cos cos ˆ ‰ ˆ ‰ Š ‹ Š ‹ ˆ ‰ ˆ ‰ Š ‹ ˆ ‰ " " " # # $ " " # n n 2 2 2 n n n n n n 2 n � � � � � �# � 78. lim n 1 cos lim lim lim sin 0 converges n n n n Ä _ Ä _ Ä _ Ä _ ˆ ‰ ˆ ‰ � œ œ œ œ Ê " " n n ˆ ‰ ˆ ‰ � ‘ ˆ ‰ Š ‹ Š ‹ " � cos sin " " " " # " # n n n n n 79. lim n sin lim lim lim cos cos 0 1 converges n n n n Ä _ Ä _ Ä _ Ä _ È Š ‹ Š ‹ 1 1 n n sin cos È È Š ‹ Š ‹Š ‹ œ œ œ œ œ Ê 1 1 1 n n 1 1 n 2n3 2 2n3 2 È È È � � Î Î 80. lim 3 5 lim exp ln 3 5 lim exp lim exp n n n n Ä _ Ä _ Ä _ Ä _ a b a b ’ “ ’ “ – — n n n n 1 n 1 n ln 3 5 n 1 � œ � œ œ Î Î � a b n n 3 ln 3 5 ln 5 n n 3 5 n � n � lim exp lim exp exp ln 5 5 œ œ œ œ n n Ä _ Ä _ ’ “ ’ “ a b Š ‹ ˆ ‰ ˆ ‰ ˆ ‰ 3n 5n 3 3 n n 5 5 n 3 5 n ln 3 ln 5 1 1 ln 3 ln 5 � � � � 81. lim tan n converges 82. lim tan n 0 0 converges n n Ä _ Ä _ �" �" # # " œ Ê œ œ Ê 1 1 Èn † Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.1 Sequences 573 83. lim lim 0 converges (Theorem 5, #4) n n Ä _ Ä _ ˆ ‰ Šˆ ‰ ‹ Š ‹ " " " " 3 3 n n 2 2 n � œ � œ Ê È È n 84. lim n n lim exp lim exp e 1 converges n n n Ä _ Ä _ Ä _ È ’ “ ˆ ‰ n # � � � ! � œ œ œ œ Ê ln n n n n n 2n 1 a b # # 85. lim lim lim lim 0 converges n n n n Ä _ Ä _ Ä _ Ä _ (ln n) 200 (ln n) 200 199 (ln n) n n n n 200! #!! "** "*) œ œ œ á œ œ Ê † 86. lim lim lim lim lim 0 converges n n n n n Ä _ Ä _ Ä _ Ä _ Ä _ (ln n) 10(ln n) 80(ln n) n n n n 3840 & % $ È È È È œ œ œ œ á œ œ Ê – — Š ‹ Š ‹ 5(ln n) n n % " #È 87. lim n n n lim n n n lim lim n n n n Ä _ Ä _ Ä _ Ä _ Š ‹ Š ‹ Š ‹ È È � � œ � � œ œ # # � � � � � � " � � n n n n n n n n n n 1 1 È È È É # # # " n converges œ Ê " # 88. lim lim lim n n n Ä _ Ä _ Ä _ " " � � � � � � � � � � � � � � � � � È È È È È È È È È È n 1 n n n 1 n n n 1 n n n 1 n n n 1 n n 1 n # # # # # # # # # # œ œ Š ‹ Š ‹ lim 2 converges œ œ � Ê n Ä _ É É ˆ ‰ 1 1 1 � � � � � " " # " n n n 89. lim dx lim lim 0 converges (Theorem 5, #1) n n n Ä _ Ä _ Ä _ " " " n x n n ln n ' 1 n œ œ œ Ê 90. lim dx lim lim 1 if p 1 converges n n n Ä _ Ä _ Ä _ ' 1 n n 1 " " " " " " � � � x 1 p x 1 p n p 1 p p 1 p 1 œ œ � œ � Ê ’ “ ˆ ‰ � � 91. Since a converges lim a L lim a lim L L 1 L 72 L L 72 0 n n n 1 72 72 1 a 1 L 2 Ê œ Ê œ Ê œ Ê � œ Ê � � œ n n n Ä _ Ä _ Ä _ � � � n a b L 9 or L 8; since a 0 for n 1 L 8 Ê œ � œ � Ê œ n 92. Since a converges lim a L lim a lim L L L 2 L 6 L L 6 0 n n n 1 a 6 a 2 L 2 L 6 2 Ê œ Ê œ Ê œ Ê � œ � Ê � � œ n n n Ä _ Ä _ Ä _ � � � � � n n a b L 3 or L 2; since a 0 for n 2 L 2 Ê œ � œ � Ê œ n 93. Since a converges lim a L lim a lim 8 2a L 8 2L L 2L 8 0 L 2 n n n 1 n 2 Ê œ Ê œ � Ê œ � Ê � � œ Ê œ � n n n Ä _ Ä _ Ä _ � È È or L 4; since a 0 for n 3 L 4 œ � Ê œ n 94. Since a converges lim a L lim a lim 8 2a L 8 2L L 2L 8 0 L 2 n n n 1 n 2 Ê œ Ê œ � Ê œ � Ê � � œ Ê œ � n n n Ä _ Ä _ Ä _ � È È or L 4; since a 0 for n 2 L 4 œ � Ê œ n 95. Since a converges lim a L lim a lim 5a L 5L L 5L 0 L 0 or L 5; since n n n 1 n 2 Ê œ Ê œ Ê œ Ê � œ Ê œ œ n n n Ä _ Ä _ Ä _ � È È a 0 for n 1 L 5 n � Ê œ 96. Since a converges lim a L lim a lim 12 a L 12 L L 25L 144 0 n n n 1 n 2 Ê œ Ê œ � Ê œ � Ê � � œ n n n Ä _ Ä _ Ä _ � ˆ ‰ È Š ‹ È L 9 or L 16; since 12 a 12 for n 1 L 9 Ê œ œ � � Ê œ È n 97. a 2 , n 1, a 2. Since a converges lim a L lim a lim 2 L 2 n 1 1 n n n 1 1 1 1 a a L � � œ � œ Ê œ Ê œ � Ê œ � n n n n n Ä _ Ä _ Ä _Š ‹ L 2L 1 0 L 1 2; since a 0 for n 1 L 1 2 Ê � � œ Ê œ „ � Ê œ � 2 n È È Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 574 Chapter 10 Infinite Sequences and Series 98. a 1 a , n 1, a 1. Since a converges lim a L lim a lim 1 a L 1 L n 1 n 1 n n n 1 n � � œ � œ Ê œ Ê œ � Ê œ � È È È È n n n Ä _ Ä _ Ä _ L L 1 0 L ; since a 0 for n 1 L Ê � � œ Ê œ � Ê œ 2 1 5 1 5 2 2 n „ � È È 99. 1, 1, 2, 4, 8, 16, 32, 1, 2 , 2 , 2 , 2 , 2 , 2 , x 1 and x 2 for n 2 á œ á Ê œ œ ! " # $ % & " n n 2 � 100. (a) 1 2(1) 1, 3 2(2) 1; let f(a b) (a 2b) 2(a b) a 4ab 4b 2a 4ab 2b # # # # # # # # # # � œ � � œ ß œ � � � œ � � � � � 2b a ; a 2b 1 f(a b) 2b a 1; a 2b 1 f(a b) 2b a 1 œ � � œ � Ê ß œ � œ � œ Ê ß œ � œ � # # # # # # # # # # (b) r 2 2 r 2 # � � � � � � „" " � � � # � � # n n � œ � œ œ œ Ê œ „ ˆ ‰ Ê Š ‹ a 2b a 4ab 4b 2a 4ab 2b a b (a b) (a b) y y a 2b # # # # # # # # # a b n n In the first and second fractions, y n. Let represent the (n 1)th fraction where 1 and b n 1 n � � a a b b for n a positive integer 3. Now the nth fraction is and a b 2b 2n 2 n y n. Thus, � � Ê a 2b a b � � n lim r 2. n Ä _ n œ È 101. (a) f(x) x 2; the sequence converges to 1.414213562 2 œ � ¸ # È (b) f(x) tan (x) 1; the sequence converges to 0.7853981635 œ � ¸ 1 4 (c) f(x) e ; the sequence 1, 0, 1, 2, 3, 4, 5, diverges œ � � � � � á x 102. (a) lim nf lim lim f (0), where x n x x Ä _ Ä ! Ä ! ˆ ‰ " " � � w n x x n f( x) f(0 x) f(0) œ œ œ œ ? ? � � ? ? ? ? ? (b) lim n tan f (0) 1, f(x) tan x n Ä _ �" w �" " " � ˆ ‰ n 1 0 œ œ œ œ # (c) lim n e 1 f (0) e 1, f(x) e 1 n Ä _ a b 1 n x Î � œ œ œ œ � w ! (d) lim n ln 1 f (0) 2, f(x) ln (1 2x) n Ä _ ˆ ‰ � œ œ œ œ � 2 2 n 1 2(0) w � 103. (a) If a 2n 1, then b 2n 2n 2n 2n, c 2n 2n œ � œ Ú Û œ Ú Û œ Ú � � Û œ � œ Ü Ý œ Ü � � Ý a 4n 4n 1 a # # # # # # # # � � " " # # # 2n 2n 1 and a b (2n 1) 2n 2n 4n 4n 1 4n 8n 4n œ � � � œ � � � œ � � � � � # # # # # # % $ # # a b 4n 8n 8n 4n 1 2n 2n 1 c . œ � � � � œ � � œ % $ # # # # a b (b) lim lim 1 or lim lim sin lim sin 1 a a a a 2 Ä _ Ä _ Ä _ Ä _ Ä Î Ú Û Ú Û Ü Ý Ü Ý a a a a # # # # # # # # œ œ œ œ œ 2n 2n 2n 2n 1 # # � � � ) ) ) 1 104. (a) lim (2n ) lim exp lim exp lim exp e 1; n n n n Ä _ Ä _ Ä _ Ä _ 1 1 2n ln 2n 2n n Î ! # # " a b Š ‹ œ œ œ œ œ ˆ ‰ ˆ ‰ � � 1 2 2n 1 1 n! 2n , Stirlings approximation n! (2n ) for large values of n ¸ Ê ¸ ¸ ˆ ‰ ˆ ‰ È È n n n e e e 1 2n n n 1 1 Î a b (b) n n! 40 15.76852702 14.71517765 50 19.48325423 18.393 È n n e 97206 60 23.19189561 22.07276647 105. (a) lim lim lim 0 n n n Ä _ Ä _ Ä _ ln n n cn cn c c 1 c œ œ œ ˆ ‰ " n � " (b) For all 0, there exists an N e such that n e ln n ln n ln % � œ � Ê � � Ê � �Ð ÑÎ �Ð ÑÎ " ln c ln c c ln c % % % % ˆ ‰ n 0 lim 0 n Ê � Ê � Ê � � Ê œ Ä _ c n n n " " " " % c c c % % ¸ ¸ 106. Let {a } and {b } be sequences both converging to L. Define {c } by c b and c a , where n n n 2n n 2n 1 n œ � œ n 1, 2, 3, . For all 0 there exists N such that when n N then a L and there exists N œ á � � � � % % " " # k k n such that when n N then b L . If n 1 2max{N N }, then c L , so {c } converges to L. � � � � � ß � � # " # k k k k n n n % % Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.1 Sequences 575 107. lim n lim exp ln n lim exp e 1 n n n Ä _ Ä _ Ä _ 1 n n n Î ! " " œ œ œ œ ˆ ‰ ˆ ‰ 108. lim x lim exp ln x e 1, because x remains fixed while n gets large n n Ä _ Ä _ 1 n n Î ! " œ œ œ ˆ ‰ 109. Assume the hypotheses of the theorem and let be a positive number. For all there exists a N such that % % " when n N then a L a L L a , and there exists a N such that when � � � Ê � � � � Ê � � " # k k n n n % % % % n N then c L c L c L . If n max{N N }, then � � � Ê � � � � Ê � � � ß # " # k k n n n % % % % L a b c L b L lim b L. � � Ÿ Ÿ � � Ê � � Ê œ % % % n n n n n k k n Ä _ 110. Let . We have f continuous at L there exists so that x L f(x) f(L) . Also, a L there % $ $ % � ! Ê � � Ê � � Ä Ê k k k k n exists N so that for n N a L . Thus for n N, f(a ) f(L) f(a ) f(L). � � � � � � Ê Ä k k k k n n n $ % 111. a a 3n 3n 4n 4 3n 6n n 2 n 1 n 3(n 1) 1 (n 1) 1 n 1 n n 1 3n 1 3n 4 3n 1 � � � � � � � # � � � � # # Ê � Ê � Ê � � � � � � � 4 2; the steps are reversible so the sequence is nondecreasing; 3 3n 1 3n 3 Ê � � Ê � � � 3n n 1 � " � 1 3; the steps are reversible so the sequence is bounded above by 3 Ê � 112. a a n 1 n (2(n 1) 3)! (2n 3)! (2n 5)! (2n 3)! (2n 5)! (n 2)! ((n 1) 1)! (n 1)! (n 2)! (n 1)! (2n 3)! (n 1)! � � � � � � � � � � � � � � � Ê � Ê � Ê � (2n 5)(2n 4) n 2; the steps are reversible so the sequence is nondecreasing; the sequence is not Ê � � � � bounded since (2n 3)(2n 2) (n 2) can become as large as we please (2n 3)! (n 1)! � � œ � � â � 113. a a 2 3 n 1 which is true for n 5; the steps are n 1 n 2 3 2 3 2 3 (n 1)! n! 2 3 n! (n 1)! � � � Ÿ Ê Ÿ Ê Ÿ Ê Ÿ � n 1 n 1 n n n 1 n 1 n n � � � � † reversible so the sequence is decreasing after a , but it is not nondecreasing for all its terms; a 6, a 18, & " # œ œ a 36, a 54, a 64.8 the sequence is bounded from above by 64.8 $ % & œ œ œ œ Ê 324 5 114. a a 2 2 ; the steps are n 1 n 2 2 2 2 2 n 1 n n n 1 n(n 1) � � # # � # # � # " " " " " Ê � � � � Ê � � Ê � n 1 n n 1 n n 1 � � � reversible so the sequence is nondecreasing; 2 2 the sequence is bounded from above � � Ÿ Ê 2 n " #n 115. a 1 converges because 0 by Example 1; also it is a nondecreasing sequence bounded above by 1 n œ � Ä " " n n 116. a n diverges because n and 0 by Example 1, so the sequence is unbounded n œ � Ä _ Ä " " n n 117. a 1 and 0 ; since 0 (by Example 1) 0, the sequence converges; also it is n 2 1 2 n n œ œ � � � Ä Ê Ä n n n n n � " " " " " # # # a nondecreasing sequence bounded above by 1 118. a ; the sequence converges to by Theorem 5, #4 n 2 1 2 3 3 3 n œ œ � ! n n n � " ˆ ‰ 119. a ( 1) 1 diverges because a 0 for n odd, while for n even a 2 1 converges to 2; it n n n n n 1 n n œ � � œ œ � a b ˆ ‰ ˆ ‰ � " diverges by definition of divergence 120. x max {cos 1 cos 2 cos 3 cos n} and x max {cos 1 cos 2 cos 3 cos (n 1)} x with x 1 n n 1 n n œ ß ß ß á ß œ ß ß ß á ß � Ÿ � so the sequence is nondecreasing and bounded above by 1 the sequence converges. Ê 121. a a n 1 2n 2n n 2n 2n n 1 n n n 1 1 2n n 2(n 1) n 1 Í Í � � � � � Í � � � " � � � # # È È È È È È È È È È and 2 ; thus the sequence is nonincreasing and bounded below by 2 it converges 1 2n n � È È Ê È È Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 576 Chapter 10 Infinite Sequences and Series 122. a a n 2n 1 n 2n 1 0 and 1; thus the sequence is n n 1 n 1 n 1 n n 1 n (n 1) Í Í � � � Í � � � � � " � # # nonincreasing and bounded below by 1 it converges Ê 123. 4 so a a 4 4 1 and 4 3 3 3 3 3 3 3 4 4 4 4 4 4 4 n n n n n 1 n n 1 n 1 n n � � � �" � œ � Í � � Í Í ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ 4 4; thus the sequence is nonincreasing and bounded below by 4 it converges � Ê ˆ ‰ 3 4 n 124. a 1, a 2 3, a 2(2 3) 3 2 2 3, a 2 2 2 3 3 2 2 1 3, " # $ % # # $ $ œ œ � œ � � œ � � " œ � � " � œ � � a b a b a b a b 2 2 † † a 2 2 2 1 3 3 2 2 1 3, , a 2 2 1 3 2 3 2 3 & $ $ % % �" �" �" � œ � � � œ � � á œ � � œ � � c d a b a b a b n n n n † n 1 2 (1 3) 3 2 3; a a 2 3 2 3 2 2 1 2 œ � � œ � � Í � � � � Í � � Í Ÿ n 1 n n n 1 n n 1 n n 1 � � � � so the sequence is nonincreasing but not bounded below and therefore diverges 125. Let 0 M 1 and let N be an integer greater than . Then n N n n nM M � � � Ê � Ê � � M M 1 M 1 M � � n M nM n M(n 1) M. Ê � � Ê � � Ê � n n 1 � 126. Since M is a least upper bound and M is an upper bound, M M . Since M is a least upper bound and M " # " # # " Ÿ is an upper bound, M M . We conclude that M M so the least upper bound is unique. # " " # Ÿ œ 127. The sequence a 1 is the sequence , , , , . This sequence is bounded above by , n ( ) 3 3 3 œ � á �" # # # # # # " " n but it clearly does not converge, by definition of convergence. 128. Let L be the limit of the convergent sequence {a }. Then by definition of convergence, for there n % # corresponds an N such that for all m and n, m N a L and n N a L . Now � Ê � � � Ê � � k k k k m n % % # # a a a L L a a L L a whenever m N and n N. k k k k k k k k m n m n m n � œ � � � Ÿ � � � � � œ � � % % # # % 129. Given an 0, by definition of convergence there corresponds an N such that for all n N, % � � L a and L a . Now L L L a a L L a a L 2 . k k k k k k k k k k k k " # # " # " # " � � � � � œ � � � Ÿ � � � � � œ n n n n n n % % % % % L L 2 says that the difference between two fixed values is smaller than any positive number 2 . k k # " � � % % The only nonnegative number smaller than every positive number is 0, so L L 0 or L L . k k " # " # � œ œ 130. Let k(n) and i(n) be two order-preserving functions whose domains are the set of positive integers and whose ranges are a subset of the positive integers. Consider the two subsequences a and a , where a L , k n i n k n Ð Ñ Ð Ñ Ð Ñ " Ä a L and L L . Thus a a L L 0. So there does not exist N such that for all m, n N i n k n i n Ð Ñ Ð Ñ Ð Ñ # " # " # Ä Á � Ä � � � ¸ ¸ k k a a . So by Exercise 128, the sequence a is not convergent and hence diverges. Ê � � Ö × k k m n n % 131. a L given an 0 there corresponds an N such that 2k N a L . Similarly, 2k 2k Ä Í � � Ê � � % % " " c d k k a L 2k 1 N a L . Let N max{N N }. Then n N a L whether 2k 1 2k 1 n � # � " # Ä Í � � Ê � � œ ß � Ê � � c d k k k k % % n is even or odd, and hence a L. n Ä 132. Assume a 0. This implies that given an 0 there corresponds an N such that n N a 0 n n Ä � � Ê � � % % k k a a a 0 a 0. On the other hand, assume a 0. This implies that Ê � Ê � Ê � � Ê Ä Ä k k k k k k k k k k k k k k n n n n n % % % given an 0 there corresponds an N such that for n N, a 0 a a % % % % � � � � Ê � Ê � k k k k k k k k k k n n n a 0 a 0. Ê � � Ê Ä k k n n % 133. (a) f(x) x a f (x) 2x x x x œ � Ê œ Ê œ � Ê œ œ œ # w � � � � � � # # � n 1 n n 1 x a 2x x a x a x 2x 2x x # # # # n n n n n n n n n xa a b ˆ ‰ (b) x 2, x 1.75, x 1.732142857, x 1.73205081, x 1.732050808; we are finding the positive " # $ % & œ œ œ œ œ number where x 3 0; that is, where x 3, x 0, or where x 3 . # # � œ œ � œ È Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.2 Infinite Series 577 134. x 1, x 1 cos (1) 1.540302306, x 1.540302306 cos (1 cos (1)) 1.570791601, " # $ œ œ � œ œ � � œ x 1.570791601 cos (1.570791601) 1.570796327 to 9 decimal places. After a few steps, the % # œ � œ œ 1 arc x and line segment cos x are nearly the same as the quarter circle. a b a b n 1 n 1 � � 135-146. Example CAS Commands: : (sequence functions may vary): Mathematica Clear[a, n] a[n_]; = n1 / n first25= Table[N[a[n]],{n, 1, 25}] Limit[a[n], n Ä 8] Mathematica: (sequence functions may vary): Clear[a, n] a[n_]; = n1 / n first25= Table[N[a[n]],{n, 1, 25}] Limit[a[n], n Ä 8] The last command (Limit) will not always work in Mathematica. You could also explore the limit by enlarging your table to more than the first 25 values. If you know the limit (1 in the above example), to determine how far to go to have all further terms within 0.01 of the limit, do the following. Clear[minN, lim] lim= 1 Do[{diff=Abs[a[n] � lim], If[diff < .01, {minN= n, Abort[]}]}, {n, 2, 1000}] minN For sequences that are given recursively, the following code is suggested. The portion of the command a[n_]:=a[n] stores the elements of the sequence and helps to streamline computation. Clear[a, n] a[1]= 1; a[n_]; = a[n]= a[n 1] (1/5) � � (n 1) � first25= Table[N[a[n]], {n, 1, 25}] The limit command does not work in this case, but the limit can be observed as 1.25. Clear[minN, lim] lim= 1.25 Do[{diff=Abs[a[n] � lim], If[diff < .01, {minN= n, Abort[]}]}, {n, 2, 1000}] minN 10.2 INFINITE SERIES 1. s lim s 3 n n a 1 r (1 r) 2 1 1 1 2 œ œ Ê œ œ a b ˆ ˆ ‰ ‰ ˆ ‰ ˆ ‰ � � � � � n 3 n 3 3 " " " n Ä _ 2. s lim s n n a 1 r (1 r) 11 1 1 1 œ œ Ê œ œ a b ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ � � � � � " n 9 9 100 100 100 n 100 100 " " " n Ä _ 3. s lim s n n a 1 r (1 r) 3 1 1 2 œ œ Ê œ œ a b ˆ ‰ ˆ ‰ ˆ ‰ � � � � � � " n n 3 " # " # # n Ä _ 4. s , a geometric series where r 1 divergence n 1 ( 2) œ 1 ( 2) � Ê � � � � n k k 5. s lim s " " " " " " " " " " " " � � # � � # # � � # # � # # (n 1)(n ) n 1 n 3 3 4 n 1 n n n n œ � Ê œ � � � � á � � œ � Ê œ ˆ ‰ ˆ ‰ ˆ ‰ n Ä _ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 578 Chapter 10 Infinite Sequences and Series 6. s 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 n(n 1) n n 1 2 2 3 3 4 n 1 n n n 1 n 1 n � � � � � œ � Ê œ � � � � � � á �