(Solution Manual) Thomas Calculus solution manual Complete with All Chapters

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Solution Manual

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INSTRUCTOR ’S SOLUTIONS MANUAL MULTIV ARIABLE 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 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 " #$%/g99" /g99 " /g99 /g99 #œœ œœ /g99 œœ /g99 œœ /g9911 2 13 2 14 3 1 4 39 41 6# ### 2. a 1, a , a , a " #$%"" #œœ œœ œœ œœ11 1 1 1 1! ! 2 3! 6 4! 24 3. a 1, a , a , a "# $ %/g99/g99 " /g99 /g99 #/g99 /g99 /g99 /g99"" "œœœœ /g99œœ œœ /g99(1 ) ( ) (1 ) (1 ) 1 4 13 6 1 5 8 17#$ % & 4. a 2 ( 1) 1, a 2 ( 1) 3, a 2 ( 1) 1, a 2 ( 1) 3 "#$%"#$%œ/g98 /g99 œ œ/g98 /g99 œ œ/g98 /g99 œ œ/g98 /g99 œ 5. a , a , a , a "#$% # ####"""" #œœ œœ œœ œœ2 222 22#$ &#$% % 6. a , a , a , a " #$%/g99 " " /g99/g99/g99 " ##œ œœœœœœœ22 1 3 2 1 7 2 1 5 24 28 1 6 2#$% #$ % 7. a 1, a 1 , a , a , a , a , "# $ % & '"" " " ## # # # #œ œ/g98œ œ/g98 œ œ/g98 œ œ /g98 œ œ3 3 7 7 15 15 31 63 4 4 8 8 16 32#$ % a, a, a, a ()*" ! œœœ œ127 255 511 1023 64 128 256 512 8. a 1, a , a , a , a , a , a , a , "# $ % & ' ( )"" " " " " "## # # œ œ œ œ œ œ œ œœœ œˆ‰ ˆ‰ ˆ ‰"" " ## 3 6 4 4 5 1 0 7 0 5040 40,32064 a, a *" !""œœ362,880 3,628,800 9. a 2, a 1, a , a , a , "# $ % &/g99/g99 ## # #"" " /g99/g99 /g99/g99œœ œœ œ /g99œ œ /g99œ œ(1 ) ( 2 ) (1 ) ( 1 ) 24 8(1 ) (1 )#$ % " #& " ˆ‰ ˆ‰4 a , a , a , a , a '( ) * " !""" " " ##œœ /g99œ /g99œ œ16 3 64 1 8 256 10. a 2, a 1, a , a , a , a , " # $ %&'/g99/g99 ##/g99/g99 ""œ/g99 œ œ/g99 œ œ/g99 œ œ/g99 œ œ/g99 œ/g991( 2 ) 2( 1 ) 33 4 5 5 322 34 †† ††ˆ‰ ˆ‰2 3" # a, a, a, a ()*" !""œ/g99 œ/g99 œ/g99 œ/g9922 749 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/g98œ œ/g98œ œ/g98œ œ œ œ œ œ 12. a 2, a 1, a , a , a 1, a 2, a 2, a 1, a , a "# $ % & ' () * " !"" "" #/g99 # ##/g99 /g99œ œ/g99 œ/g99 œ œ œ œ/g99 œ/g99 œ œ/g99 œ/g99 œˆ‰ ˆ ‰ ˆ‰"" ## "# 1 13. a ( 1) , n 1, 2, 14. a ( 1) , n 1, 2, n nn1 nœ/g99 œ á œ/g99 œ á/g98 15. a ( 1) n , n 1, 2, 16. a , n 1, 2, n nn1 () nœ/g99 œ á œ œ á/g98# /g99"n1 /g98 # 17. a , n 1, 2, 18. a , n 1, 2, n n2 2n 5 3n 2 nn 1œœ á œ œ án1 /g99 ab ab /g98 /g98/g99 19. a n 1, n 1, 2, 20. a n 4 , n 1, 2, n n œ /g99 œá œ /g99œá# 21. a 4n 3, n 1, 2, 22. a 4n 2 , n 1, 2, n n œ/g99 œ á œ/g99 œ á 23. a , n 1, 2, 24. a , n 1, 2, n n3n 2 n n! 5œœ á œ œ á/g983 n1 /g98 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 n1( 1 ) n( 1 ) nœœ á œ œ Ú Û œ á/g98/g99 # ##/g99/g98 /g99n1 n /g98 "" ##ˆ‰ 27. lim 2 (0.1) 2 converges (Theorem 5, #4)nÄ_/g98œ Ên 28. lim lim 1 1 convergesnnÄ_ Ä_n( ) ( 1 ) nn/g98/g99 " /g99nnœ/g98 œ Ê 29. lim lim lim 1 convergesnn nÄ_ Ä_ Ä_"/g99 /g99 /g98# #/g99 /g982n 2 1n22œœ œ /g99 ʈ‰ˆ‰" "n n 30. lim lim divergesnnÄ_ Ä_2n 13n2n 3/g98" /g99/g98 /g99ÈÈ Š‹ Š‹œœ /g99_Ê" "È Èn n 31. lim lim 5 convergesnnÄ_ Ä_"/g99 /g98/g99 /g985n n8 n5 1% %$" %œœ /g99ÊŠ‹ ˆ‰n 8 n 32. lim lim lim 0 convergesnn nÄ_ Ä_ Ä_n3 n3 n 5n 6 (n 3)(n 2) n/g98/g98 " /g98/g98 /g98 /g98 /g98 ## œœ œ Ê 33. lim lim lim (n 1) divergesnn nÄ_ Ä_ Ä_n2 n 1 n1 n1(n 1)(n 1)#/g99/g98 /g99/g99/g99/g99œœ /g99 œ_Ê 34 lim lim divergesnnÄ_ Ä_"/g99 /g99/g99 /g99n 70 4nn4 $ #œœ _ÊŠ‹Š‹" # #n 70 n 35. lim 1 ( 1) does not exist diverges 36. lim ( 1) 1 does not exist divergesnnÄ_ Ä_ab ˆ‰ /g98/g99 Ê /g99 /g99 Ên n " n 37. lim 1 lim 1 convergesnnÄ_ Ä_ˆ ‰ ˆ‰ ˆ ‰ ˆ‰n nn n n/g98" " " " " " ## # #/g99œ /g98 /g99œÊ 38. lim 2 3 6 converges 39. lim 0 convergesnnÄ_ Ä_ˆ‰ ˆ‰/g99 /g98 œÊ œÊ"" ## # /g99/g99" nnn1() n1/g98 40. lim lim 0 convergesnnÄ_ Ä_ˆ‰/g99œ œ Ê"##/g99" n ()n n 41. lim lim lim 2 convergesnn nÄ_ Ä_ Ä_É É Ê Š‹ È 2n 2n n1 n1 /g98/g98œœ œ Ê2 1/g98" n 42. lim lim divergesnnÄ_ Ä_"" (0.9) 90 nœœ _Ê ˆ‰n 43. lim sin sin lim sin 1 convergesnnÄ_ Ä_ˆ‰ ˆ‰ Š‹11 1 ## #""/g98œ /g98 œ œ Ênn 44. lim n cos (n ) lim (n )( 1) does not exist divergesnnÄ_ Ä_11 1 œ/g99 Ên 45. lim 0 because converges by the Sandwich Theorem for sequencesnÄ_sin n sin n nn n nœ/g99 Ÿ Ÿ Ê"" 46. lim 0 because 0 converges by the Sandwich Theorem for sequencesnÄ_sin n sin n## ## #" nn n œŸ Ÿ Ê 47. lim lim 0 converges (using l'Hopital's rule) ^ nnÄ_ Ä_n ln 2 ##" nnœœ 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) ^ nn n nÄ_ Ä_ Ä_ Ä_33 ln 3 n3 n 6 n 63( l n 3 ) 3( l n 3 )nn nn $##$œ œœœ _Ê 49. lim lim lim lim 0 convergesnn n nÄ_ Ä_ Ä_ Ä_ln (n ) n2n n11/g98" /g98/g98ÈÈ Š‹ Š‹œœ œ œ ʈ‰ Š‹" /g98 " #"n1 n2 n n ÈÈ 50. lim lim 1 convergesnnÄ_ Ä_ln n ln 2nœœ ʈ‰ ˆ‰" n 2 2n 51. lim 8 1 converges (Theorem 5, #3)nÄ_1n ÎœÊ 52. lim (0.03) 1 converges (Theorem 5, #3) nÄ_1n ÎœÊ 53. lim 1 e converges (Theorem 5, #5)nÄ_ˆ‰/g98œ Ê7 nn( 54. lim 1 lim 1 e converges (Theorem 5, #5)nnÄ_ Ä_ˆ‰ ’“ /g99œ /g98 œ Ê" /g99" /g99" nn() nn 55. lim 10n lim 10 n 1 1 1 converges (Theorem 5, #3 and #2)nnÄ_ Ä_Ènœœ œÊ1n 1nÎΆ† 56. lim n lim n 1 1 converges (Theorem 5, #2)nnÄ_ Ä_È ˆ‰Èn n ###œœ œ Ê 57. lim 1 converges (Theorem 5, #3 and #2)nÄ_ˆ‰3n11n lim 3 lim nÎ " œœ œ Ên nÄ_ Ä_1n 1nÎ Î 58. lim (n 4) lim x 1 converges; (let x n 4, then use Theorem 5, #2)n x Ä_ Ä_/g98œ œ Ê œ /g981n4 1 x ÎÐ /g98 Ñ Î 59. lim diverges (Theorem 5, #2)nÄ_ln n n lim n lim ln n 11n 1n ÎÎœœ œ _ Ên nÄ_ Ä__ 60. lim ln n ln (n 1) lim ln ln lim ln 1 0 convergesnn nÄ_ Ä_ Ä_cd ˆ‰ Š‹ /g99/g98 œ œ œœ Ênn n1 n1/g98/g98 61. lim 4 n lim 4 n 4 1 4 converges (Theorem 5, #2)nnÄ_ Ä_È Èn n nœœ œ Ê † 62. lim 3 lim 3 lim 3 3 9 1 9 converges (Theorem 5, #3)nn nÄ_ Ä_ Ä_Èn2n 1 21 n 1 n /g98 /g98Î # Îœœœ œ Êab†† 63. lim lim lim 0 and 0 lim 0 convergesnn n n Ä_ Ä_ Ä_ Ä_n! n! n! n nnn nn n n n2 3 (n 1)(n) n nn œŸ œ   Ê œ Ê"â /g99 â" †† †† †ˆ‰ 64. lim 0 converges (Theorem 5, #6)nÄ_(4 ) n!/g99nœÊ 65. lim lim diverges (Theorem 5, #6)nnÄ_ Ä_n! 106n (10 ) n n!œœ _Ê" Š‹' 66. lim lim diverges (Theorem 5, #6)nnÄ_ Ä_n! 23nn 6n n!œœ _ Ê" ˆ‰ 67. lim lim exp ln lim exp e convergesnn nÄ_ Ä_ Ä_ˆ‰ ˆ ‰ ˆ ‰ ˆ‰"" " /g99 ÎÐ Ñ/g99" nl n n n l n n1l n n ln 1 ln nœœ 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)nnÄ_ Ä_ˆ‰ ˆ‰ Š‹ /g98œ /g98 œœ Ê"" nnnn 69. lim lim exp n ln lim expnn nÄ_ Ä_ Ä_ˆ‰ ˆ ‰ ˆ‰ Š‹3n 3n 3n 1 3n 1n ln (3n 1) ln (3n 1) /g98" /g98" /g99/g99/g98/g99 /g99œœ " n lim exp lim exp exp e converges œœœ œ ÊnnÄ_ Ä_ /g58/g59Š‹ ˆ‰33 3n 1 3n 1 n/g98/g99 " #/g99 /g99Š‹6n 6 (3n 1)(3n 1) 9# /g98/g99#Î$ 70. lim lim exp n ln lim exp lim expnn n n Ä_ Ä_ Ä_ Ä_ˆ‰ ˆ ‰ ˆ‰ Š‹/g58/g59nn n1 n1n l n nl n ( n1 ) /g98/g98/g99/g98œœ œˆ‰Š‹""" /g98 " # nnn 1 n/g99 /g99 lim exp e converges œ/g99 œ ÊnÄ_Š‹n n(n 1)# /g98/g99" 71. lim lim x x lim exp ln x lim expnn n nÄ_ Ä_ Ä_ Ä_ˆ‰ ˆ‰ ˆ ‰ˆ‰ Š‹x 2n 1 n 1 n n 1 n1n 1n ln (2n 1)n /g98# /g98 #/g98Î"" " Î /g99/g98œœ œ x lim exp xe x, x 0 convergesœœ œ /g128 ÊnÄ_ˆ‰/g99 /g98! 2 2n 1 72. lim 1 lim exp n ln 1 lim exp lim expnn n nÄ_ Ä_ Ä_ Ä_ˆ‰ ˆ ‰ ˆ‰/g58/g59 – —/g99œ /g99œ œ"" nnn ##ln 1 1 Š‹ ‚ ˆ‰Š‹ Š ‹ Š‹/g99 /g99 /g99" # "$#" " #n n2 nn n lim exp e 1 convergesœœ œ ÊnÄ_ˆ‰/g99 /g99! 2n n1# 73. lim lim 0 converges (Theorem 5, #6)nnÄ_ Ä_36 3 6 2n ! n !nn n n† †/g99œœ Ê 74. lim lim lim 0 convergnn nÄ_ Ä_ Ä_ˆ‰ ˆ‰ ˆ‰ ˆ ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ ‰10 12 10 120 11 11 11 121nn n n 9 11 12 9 12 11 108 10 12 11 10 11 12 110n n nn nn n/g98/g98 /g98œœ œ Ê1es (Theorem 5, #4) 75. lim tanh n lim lim lim lim 1 convergesnn n n nÄ_ Ä_ Ä_ Ä_ Ä_œœ œ œ " œ Êee e 2 e ee e 1 2 en n 2n 2n n n 2n 2n/g99/g99 " /g98/g98/g99 /g99 76. lim sinh (ln n) lim lim divergesnn nÄ_ Ä_ Ä_œœ œ _Êee 2n ln n ln nn /g99 /g99 #/g99"ˆ‰ 77. lim lim lim lim convergesnn n nÄ_ Ä_ Ä_ Ä_n s i n 2n 1# "ˆ‰n /g99 #"œœ œœ Êsin cos cos ˆ‰ ˆ‰ Š‹ Š ‹ˆ‰ˆ‰ Š‹ ˆ‰" " " ## $"" #n n 22 2 n nn nn n 2n/g99/g99 /g98/g99 /g99 /g99# /g98 78. lim n 1 cos lim lim lim sin 0 convergesnn n nÄ_ Ä_ Ä_ Ä_ˆ‰ ˆ ‰ /g99œ œ œ œ Ê" " n nˆ‰ ˆ‰/g60‘ˆ‰ Š‹ Š‹"/g99cos sin " """ # " #n nn n n 79. lim n sin lim lim lim cos cos 0 1 convergesnn n nÄ_ Ä_ Ä_ Ä_È Š‹ Š‹1 1 n nsin cos È ÈŠ‹ Š‹ Š ‹œœ œ œ œ Ê11 1 nn 11 n2n32 2n32ÈÈ È/g99 /g99Î Î 80. lim 3 5 lim exp ln 3 5 lim exp lim expnn n nÄ_ Ä_ Ä_ Ä_ab ab ’“’ “ –—nn nn 1n 1n l n3 5 n1/g98œ /g98 œ œÎÎ /g98 abnn 3 l n3 5 l n5nn 35nn/g98 /g98 lim exp lim exp exp ln 5 5 œœ œ œnnÄ_ Ä_’“ ’ “ abŠ‹ ˆ‰ ˆ ‰ˆ‰3n 5n 33nn 55n3 5nln 3 ln 5 11ln 3 ln 5 /g98 /g98/g98/g98 81. lim tan n converges 82. lim tan n 0 0 convergesnnÄ_ Ä_/g99" /g99" # #"œÊ œ œÊ1 1 Èn† Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.1 Sequences 573 83. lim lim 0 converges (Theorem 5, #4)nnÄ_ Ä_ˆ‰ ˆ‰Š‹ Š‹"" " " 33nn 22n /g98œ /g98 œ ÊÈÈ n 84. lim n n lim exp lim exp e 1 convergesnn nÄ_ Ä_ Ä_È ’“ ˆ‰n# /g98 /g98 /g98!/g98œ œ œ œ Êln n n nn n2n 1 ab# # 85. lim lim lim lim 0 convergesnn n n Ä_ Ä_ Ä_ Ä_(ln n) 200 (ln n) 200 199 (ln n) nn n n200!# !! "** "*)œœ œ ᜠœ ʆ 86. lim lim lim lim lim 0 convergesnn n n nÄ_ Ä_ Ä_ Ä_ Ä_(ln n) 10(ln n) 80(ln n) nn n n3840&% $ ÈÈ È Èœ œœœ ᜠœ Ê–—Š‹ Š‹5(ln n) n n% " # È 87. lim n n n lim n n n lim lim nn n n Ä_ Ä_ Ä_ Ä_Š‹ Š‹ Š ‹ÈÈ/g99/g99 œ /g99/g99 œ œ## /g98/g99 /g98/g99 /g98/g99" /g98/g99nn nnn n nn nn 11ÈÈÈÉ# ## " n convergesœÊ" # 88. lim lim lim nn n Ä_ Ä_ Ä_"" /g99/g99 /g98 /g99/g99 /g98 /g99/g98 /g98/g99/g98 /g98 /g99/g98 /g98 /g99/g99 ÈÈ È ÈÈ ÈÈÈ ÈÈ n1 nn n1 nn n1 nnn1 nn n1 nn 1n ## ## # ### # #œœ Š‹ Š‹ lim 2 converges œœ /g99ÊnÄ_É É ˆ‰11 1/g99/g98 /g98 /g99/g99"" # "n n n 89. lim dx lim lim 0 converges (Theorem 5, #1)nn nÄ_ Ä_ Ä_"" " nx n nln n ' 1n œœ œ Ê 90. lim dx lim lim 1 if p 1 convergesnn nÄ_ Ä_ Ä_' 1n n 1"" " " " " /g99/g99 /g99 x1 px 1 pn p 1pp 1 p 1 œœ /g99 œ /g128 Ê ’“ ˆ‰/g99/g99 91. Since a converges lim a L lim a lim L L 1 L 72 L L 72 0 nn n 172 72 1a 1L2Êœ Ê œ Ê œ Ê /g98 œ Ê /g98 /g99 œnn nÄ_ Ä_ Ä_/g98 /g98/g98 nab L 9 or L 8; since a 0 for n 1 L 8Êœ /g99 œ /g128  Êœ n 92. Since a converges lim a L lim a lim L L L 2 L 6 L L 6 0 nn n 1a6 a2 L 2L6 2Êœ Ê œ Ê œ Ê /g98 œ /g98 Ê /g98 /g99 œnn nÄ_ Ä_ Ä_/g98/g98/g98/g98/g98 n nab L 3 or L 2; since a 0 for n 2 L 2Êœ /g99 œ /g128  Êœ n 93. Since a converges lim a L lim a lim 8 2a L 8 2L L 2L 8 0 L 2 nn n 1 n2Êœ Ê œ /g98 Ê œ /g98 Ê /g99 /g99 œ Ê œ /g99nn nÄ_ Ä_ Ä_/g98È È or L 4; since a 0 for n 3 L 4 œ/g128   Ê œ n 94. Since a converges lim a L lim a lim 8 2a L 8 2L L 2L 8 0 L 2 nn n 1 n2Êœ Ê œ /g98 Ê œ /g98 Ê /g99 /g99 œ Ê œ /g99nn nÄ_ Ä_ Ä_/g98È È or L 4; since a 0 for n 2 L 4 œ/g128   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 nn n 1 n2Êœ Ê œ Ê œ Ê /g99 œ Ê œ œnn nÄ_ Ä_ Ä_/g98È È a0 f o r n 1 L 5 n/g128  Ê œ 96. Since a converges lim a L lim a lim 12 a L 12 L L 25L 144 0 nn n 1 n2Êœ Ê œ /g99 Ê œ /g99 Ê /g99 /g98 œnn nÄ_ Ä_ Ä_/g98 ˆ‰ È Š‹ È L 9 or L 16; since 12 a 12 for n 1 L 9Êœ œ /g99 /g127  Êœ Èn 97. a 2 , n 1, a 2. Since a converges lim a L lim a lim 2 L 2 n1 1 n n n11 11 a aL /g98 /g98 œ/g98   œ Ê œ Ê œ /g98 Ê œ/g98 n n nn nÄ_ Ä_ Ä_Š‹ L 2L 1 0 L 1 2; since a 0 for n 1 L 1 2Ê /g99 /g99œÊ œ„ /g128  Ê œ/g982nÈÈ 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 n1 n 1 n n n1 n /g98 /g98 œ/g98   œ Ê œ Ê œ /g98 Ê œ/g98 È È È È nn nÄ_ Ä_ Ä_ L L 1 0 L ; since a 0 for n 1 LÊ /g99/g99 œÊœ /g128  Êœ2 15 15 22 n„/g98ÈÈ 99. 1, 1, 2, 4, 8, 16, 32, 1, 2 , 2 , 2 , 2 , 2 , 2 , x 1 and x 2 for n 2 ᜠá Ê œ œ  !"#$%&" nn2 /g99 100. (a) 1 2(1) 1, 3 2(2) 1; let f(a b) (a 2b) 2(a b) a 4ab 4b 2a 4ab 2b## ## # # # # # #/g99 œ /g99 /g99 œ ßœ/g98 /g99 /g98 œ/g98 /g98 /g99 /g99 /g99 2b a ; a 2b 1 f(a b) 2b a 1; a 2b 1 f(a b) 2b a 1 œ/g99 /g99œ /g99 Ê ß œ/g99 œ /g99œ Ê ß œ/g99 œ /g99## # # ## # # ## (b) r 2 2 r 2# /g98/g98 /g98 /g99 /g99 /g99 „ " " /g98/g98 /g98# /g99/g99# n n /g99œ /g99œ œ œ Ê œ „ ˆ‰ Ê Š‹a 2b a 4ab 4b 2a 4ab 2b a b (a b) (a b) y ya2 b## ## ## ###ab n n In the first and second fractions, y n. Let represent the (n 1)th fraction where 1 and b n 1 n /g99    /g99aa bb for n a positive integer 3. Now the nth fraction is and a b 2b 2n 2 n y n. Thus,  /g98     /g99   Ê  a2 b ab/g98 /g98n lim r 2.nÄ_nœÈ 101. (a) f(x) x 2; the sequence converges to 1.414213562 2 œ/g99 ¸# È (b) f(x) tan (x) 1; the sequence converges to 0.7853981635 œ/g99 ¸1 4 (c) f(x) e ; the sequence 1, 0, 1, 2, 3, 4, 5, diverges œ /g99/g99/g99/g99/g99áx 102. (a) lim nf lim lim f (0), where xnxxÄ_Ä! Ä!ˆ‰" " /g98/g99 w nx x nf( x) f(0 x) f(0)œœ œ œ ??/g98/g98?? ??? (b) lim n tan f (0) 1, f(x) tan xnÄ_/g99" w /g99" "" /g98ˆ‰n1 0œœœ œ # (c) lim n e 1 f (0) e 1, f(x) e 1nÄ_ab1n x Î/g99œ œœ œ/g99w! (d) lim n ln 1 f (0) 2, f(x) ln (1 2x)nÄ_ˆ‰/g98œ œ œ œ /g9822 n 1 2(0)w /g98 103. (a) If a 2n 1, then b 2n 2n 2n 2n, c 2n 2n œ /g98 œÚ ÛœÚ ÛœÚ /g98 /g98 Ûœ /g98 œÜ ÝœÜ /g98 /g98 Ýa4 n 4 n 1 a## # ## # # #/g98/g98 " " ## # 2n 2n 1 and a b (2n 1) 2n 2n 4n 4n 1 4n 8n 4n œ/g98 /g98 /g98 œ/g98 /g98 /g98 œ/g98 /g98 /g98/g98/g98# # # # # # %$# #ab 4n 8n 8n 4n 1 2n 2n 1 c . œ/g98/g98/g98 /g98 œ /g98 /g98œ%$# # # #ab (b) lim lim 1 or lim lim sin lim sin 1aa aa 2 Ä_ Ä_ Ä_ Ä_ ÄÎÚÛ ÚÛ ÜÝ ÜÝaa aa## ## ## ##œœ œ œ œ2n 2n 2n 2n 1# #/g98 /g98/g98)) )1 104. (a) lim (2n ) lim exp lim exp lim exp e 1;nn n nÄ_ Ä_ Ä_ Ä_112 n ln 2n 2n nÎ ! ##" abŠ‹œœ œ œ œ ˆ‰ ˆ ‰/g58/g5912 2n1 1 n! 2n , Stirlings approximation n! (2n ) for large values of n ¸Ê ¸ ¸ ˆ‰ ˆ‰ ÈÈnn n ee e12 n nn11Îab (b) n n! 40 15.76852702 14.71517765 50 19.48325423 18.393Èn n e 97206 60 23.19189561 22.07276647 105. (a) lim lim lim 0nn nÄ_ Ä_ Ä_ln n nc n c ncc 1 c œœ œˆ‰" n /g99" (b) For all 0, there exists an N e such that n e ln n ln n ln %/g128œ/g128 Ê /g128 /g99 Ê /g128/g99Ð ÑÎ /g99Ð ÑÎ " ln c ln c c ln c%% % %ˆ‰ n 0 lim 0nÊ/g128 Ê/g127 Ê /g99 /g127 Ê œÄ_c nn n"" " " %cc c %% ¸¸ 106. Let {a } and {b } be sequences both converging to L. Define {c } by c b and c a , where nn n2 n n2 n 1 n œœ /g99 n 1, 2, 3, . For all 0 there exists N such that when n N then a L and there exists N œá /g128 /g128 /g99 /g127 %% "" # kkn such that when n N then b L . If n 1 2max{N N }, then c L , so {c } converges to L. /g128/g99 /g127 /g128 /g98 ß /g99 /g127 #" # kk kk n nn %% Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.1 Sequences 575 107. lim n lim exp ln n lim exp e 1nn nÄ_ Ä_ Ä_1n nnÎ! ""œœ œ œ ˆ‰ ˆ ‰ 108. lim x lim exp ln x e 1, because x remains fixed while n gets largennÄ_ Ä_1n nÎ! "œœ œ ˆ‰ 109. Assume the hypotheses of the theorem and let be a positive number. For all there exists a N such that %% " when n N then a L a L L a , and there exists a N such that when /g128/g99 /g127 Ê /g99 /g127 /g99 /g127 Ê /g99 /g127 " # kknn n %% % % n N then c L c L c L . If n max{N N }, then /g128/g99 /g127 Ê /g99 /g127 /g99 /g127 Ê /g127 /g98 /g128 ß # "# kknn n %% % % L abcL bL l i m bL . /g99/g127 Ÿ Ÿ /g127 /g98 Ê /g99 /g127 Ê œ%% % nnn n n kknÄ_ 110. Let . We have f continuous at L there exists so that x L f(x) f(L) . Also, a L there %$ $ %/g128! Ê /g99 /g127 Ê /g99 /g127 Ä Ê kk k k n exists N so that for n N a L . Thus for n N, f(a ) f(L) f(a ) f(L). /g128/g99 /g127 /g128 /g99 /g127 Ê Äkk k knn n $% 111. a a 3n 3n 4n 4 3n 6n n 2 n1 n3(n 1) 1 (n 1) 1 n 1 n n 13n 1 3n 4 3n 1/g98/g98/g98 /g98/g98 /g98 /g98 # /g98/g98/g98 /g98 ##  Ê /g128Ê/g128Ê /g98 /g98 /g98 /g128 /g98 /g98 /g98 4 2; the steps are reversible so the sequence is nondecreasing; 3 3n 1 3n 3 Ê/g128 /g127Ê /g98 /g127 /g983n n1/g98" /g98 1 3; the steps are reversible so the sequence is bounded above by 3 Ê/g127 112. a a n1 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)! /g98/g98 /g98 /g98/g98 /g98/g98 /g98 /g98/g98 /g98 /g98 /g98 /g98 /g98  Ê /g128Ê/g128Ê/g128 (2n 5)(2n 4) n 2; the steps are reversible so the sequence is nondecreasing; the sequence is not Ê/g98 /g98 /g128 /g98 bounded since (2n 3)(2n 2) (n 2) can become as large as we please(2n 3)! (n 1)!/g98 /g98œ /g98 /g98â/g98 113. a a 2 3 n 1 which is true for n 5; the steps are n1 n23 2 3 23 (n 1)! n! 2 3 n!(n 1)! /g98 /g98/g98ŸÊ Ÿ Ê Ÿ Ê Ÿ /g98  n1n1 nn n1n1 nn/g98/g98 /g98/g98† reversible so the sequence is decreasing after a , but it is not nondecreasing for all its terms; a 6, a 18,&" # œœ a 36, a 54, a 64.8 the sequence is bounded from above by 64.8 $%&œœœ œ Ê324 5 114. a a 2 2 ; the steps are n1 n22 2 2 2 n1 n n n1 n ( n 1 ) /g98 /g98# # /g98 # # /g98 #"" " " " Ê /g99 /g99   /g99 /g99Ê /g99   /g99Ê   /g99 n1 n n1 n n1 /g98/g98 /g98 reversible so the sequence is nondecreasing; 2 2 the sequence is bounded from above /g99/g99 Ÿ Ê2 n" #n 115. a 1 converges because 0 by Example 1; also it is a nondecreasing sequence bounded above by 1 nœ/g99 Ä""nn 116. a n diverges because n and 0 by Example 1, so the sequence is unbounded nœ/g99 Ä _ Ä""nn 117. a 1 and 0 ; since 0 (by Example 1) 0, the sequence converges; also it is n21 2n nœœ /g99 /g127 /g127 Ä Ê Än nn n n/g99" " " " " ## # a nondecreasing sequence bounded above by 1 118. a ; the sequence converges to by Theorem 5, #4 n21 2 33 3nœœ/g99 !n nn/g99"ˆ‰ 119. a ( 1) 1 diverges because a 0 for n odd, while for n even a 2 1 converges to 2; it nn nn n1 n nœ/g99 /g98 œ œ /g98 ab ˆ‰ ˆ ‰/g98 " 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 nn 1 n n œ ßßß á ß œ ßßß á ß /g98   Ÿ /g98 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 nn 112 n n2(n 1) n1 Í   Í /g98 /g98 /g98   /g98 /g98 Í /g98   /g98/g98 "/g98 /g98 /g98##ÈÈ ÈÈÈÈ ÈÈ ÈÈ and 2 ; thus the sequence is nonincreasing and bounded below by 2 it converges12 n n/g98È È ÊÈÈ 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 nn 1n1 n1 nn 1 n(n 1) Í  Í /g98 /g98   /g98 Í     /g98/g98 /g98 /g98/g98 " /g98## nonincreasing and bounded below by 1 it converges Ê 123. 4 so a a 4 4 1 and43 3 3 3 3 3 3 44 4 4 4 4 4nn n n n1 nn 1n1 n n/g98/g98/g98/g98" /g98œ/g98   Í /g98  /g98 Í   Í   ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ 4 4; thus the sequence is nonincreasing and bounded below by 4 it converges /g98  ʈ‰3 4n 124. a 1, a 2 3, a 2(2 3) 3 2 2 3, a 2 2 2 3 3 2 2 1 3, "# $ %## $ $œ œ /g99 œ /g99 /g99 œ /g99 /g99" œ /g99 /g99" /g99 œ /g99 /g99 ab a b ab ab22†† a 2 2 2 1 3 3 2 2 1 3, , a 2 2 1 3 2 3 2 3 &$ $ % % /g99" /g99" /g99" /g99œ/g99 /g99/g99 œ /g99 /g99á œ/g99 /g99 œ/g99 /g98cd a b a bab nnn n n 1† 2 ( 1 3 ) 3 23 ; aa 23 2 3 2 2 1 2 œ /g99/g98œ /g99/g98   Í /g99/g98  /g99 /g98Í /g99  /g99 Í Ÿn1 n n n1 n n1nn 1/g99 /g98/g98/g98 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 /g127/g127 /g128Ê/g128 Ê /g99 /g128MM 1M 1M/g99/g99 n M nM n M(n 1) M. Ê/g128/g98 Ê/g128 /g98 Ê /g128n n1/g98 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. #" "# Ÿœ 127. The sequence a 1 is the sequence , , , , . This sequence is bounded above by , n() 33 3œ/g98 á/g99" # #### #""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 /g128Ê /g99/g127 /g128Ê /g99/g127 kk k kmn%% ## a a a L L a a L L a whenever m N and n N. kk k k k k k kmn m n m n /g99 œ /g99/g98/g99 Ÿ /g99 /g98 /g99 /g127/g98œ /g128 /g128%% ##% 129. Given an 0, by definition of convergence there corresponds an N such that for all n N, %/g128/g128 La a n d La . N o w LL LaaL La aL 2 .kk kk k k k k kk kk"# # " # " # " /g99/g127 /g99/g127 /g99 œ /g99 /g98 /g99 Ÿ /g99/g98/g99 /g127 /g98 œ nn n n n n %% %% % L L 2 says that the difference between two fixed values is smaller than any positive number 2 .kk#"/g99/g127 % % The only nonnegative number smaller than every positive number is 0, so L L 0 or L L . kk"# " #/g99œ 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. Cons ider the two subsequences a and a , where a L , kn in kn ÐÑ ÐÑ ÐÑ "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 in kn in ÐÑ ÐÑ ÐÑ #" # " # Ä Á /g99 Ä/g99/g128 /g128 ¸¸ kk a a . So by Exercise 128, the sequence a is not convergent and hence diverges. Ê/g99 /g127 Ö × kkmn n % 131. a L given an 0 there corresponds an N such that 2k N a L . Similarly, 2k 2k ÄÍ /g128 /g128 Ê /g99 /g127 %% "" cd kk a L 2k 1 N a L . Let N max{N N }. Then n N a L whether 2k 1 2k 1 n /g98# /g98 " # ÄÍ /g98 /g128 Ê /g99 /g127 œ ß /g128Ê /g99 /g127 cd k k kk %% n is even or odd, and hence a L. nÄ 132. Assume a 0. This implies that given an 0 there corresponds an N such that n N a 0 n n Ä/g128 /g128 Ê /g99/g127 %% kk a a a 0 a 0. On the other hand, assume a 0. This implies that Ê/g127 Ê /g127 Ê /g99 /g127 Ê Ä Ä kk k k k k kk kk kk kk nn n n n %% % given an 0 there corresponds an N such that for n N, a 0 a a %% % %/g128/g128 /g99 /g127 Ê /g127 Ê /g127 kk k k k kkk kknn n a 0 a 0. Ê/g99 /g127 ÊÄkknn % 133. (a) f(x) x a f (x) 2x x x x œ/g99 Ê œ Ê œ/g99 Ê œ œ œ#w/g98/g98/g99/g99 /g99 /g98 ##/g98 n1 n n1xa 2 x xa xa x2 x 2xx ## # # nn n n n nn nna x abˆ‰ (b) x 2, x 1.75, x 1.732142857, x 1.73205081, x 1.732050808; we are finding the positive "# $ % & œœ œ œ œ number where x 3 0; that is, where x 3, x 0, or where x 3 .##/g99œ œ /g128 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, "# $ œœ /g98 œ œ /g98 /g98 œ x 1.570791601 cos (1.570791601) 1.570796327 to 9 decimal places. After a few steps, the % #œ/g98 œœ1 arc x and line segment cos x are nearly the same as the quarter circle. ab abn1 n1 /g99/g99 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 8] Ä Mathematica : (sequence functions may vary): Clear[a, n] a[n_]; = n1 / n first25= Table[N[a[n]],{n, 1, 25}] Limit[a[n], n 8] Ä 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}] /g99 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) /g99/g98 (n 1)/g99 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}] /g99 minN 10.2 INFINITE SERIES 1. s lim s 3 nna1 r (1 r)21 1 1 2œœ Ê œ œab ˆ‰ ˆ‰ ˆ‰ ˆ‰/g99 /g99/g99 /g99/g99n 3n 33" ""nÄ_ 2. s lim s nna1 r (1 r) 111 1 1 œœ Ê œœab ˆ‰ ˆ ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰/g99 /g99/g99 /g99/g99"n 99 100 100 100n 100 100" ""nÄ_ 3. s lim s nna1 r (1 r) 31 1 2œœ Ê œ œab ˆ‰ ˆ‰ ˆ‰/g99 /g99/g99/g99 /g99/g99"nn 3" # "# # nÄ_ 4. s , a geometric series where r 1 divergence n1 ( 2) 1 ( 2)œ/g128 Ê/g99/g99 /g99/g99nkk 5. s lim s"" " " " " " " " " " " /g98 /g98# /g98 /g98# # /g98 /g98# # /g98# # ( n 1 ) ( n ) n 1 n 3 3 4 n 1 n n n n œ /g99 Êœ/g99/g98/g99/g98 á /g98 /g99 œ /g99 Ê œ ˆ‰ ˆ‰ ˆ ‰nÄ_ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.578 Chapter 10 Infinite Sequences and Series 6. s 5 55 5 5 5 55 55 5 5 5 5 5 n ( n 1 ) nn 1 2 23 34 n 1n nn 1 n 1 n /g98/g98 /g99 /g98 /g98œ /g99 Êœ/g99/g98/g99/g98/g99/g98 á /g98 /g99/g98/g99 œ /g99 ˆ ‰ ˆ ‰ ˆ ‰ˆ‰ ˆ‰ lim s 5ÊœnÄ_n 7. 1 , the sum of this geometric series is /g99 /g98/g99/g98 á œ œ"" " " " /g99/g99 /g98 41 66 4 5 1 1 4 ˆ‰ ˆ ‰"" 44 8. , the sum of this geometric series is "" " " /g99 # 16 64 256 1 1 /g98/g98 /g98 á œˆ‰ ˆ‰" "16 4 9. , the sum of this geometric series is 77 7 7 41 66 4 3 1 /g98/g98/g98 á œˆ‰ ˆ‰7 4 4/g99" 10. 5 , the sum of this geometric series is 4 /g99 /g98/g99/g98 á œ55 5 5 41 66 4 1 /g99/g99ˆ‰" 4 11. (5 1) , is the sum of two geometric series; the sum is /g98 /g98/g98/g98/g98/g98/g98 /g98 á ˆ‰ ˆ‰ ˆ ‰555 34 98 7 ##"" " 1053 23 1 1 /g99/g99" ## ˆ‰ ˆ‰"" #/g98œ /g98 œ 3 12. (5 1) , is the difference of two geometric series; the sum is /g99 /g98/g99/g98/g99/g98/g99 /g98 á ˆ‰ ˆ‰ ˆ ‰555 34 98 7 ##"" " 1053 17 1 1 /g99/g99" ## ˆ‰ ˆ‰""#/g99œ /g99 œ 3 13. (1 1) , is the sum of two geometric series; the sum is /g98 /g98/g99/g98/g98 /g98/g99 /g98 á ˆ‰ ˆ ‰ ˆ ‰11 1 54 2 58 1 5 ##"" " 215 17 1 1 66 /g99/g98" ˆ‰ ˆ‰""#/g98œ /g98 œ 5 14. 2 2 1 ; the sum of this geometric series is 2 /g98/g98 /g98 /g98 á œ /g98/g98 /g98 /g98 á œ48 1 6 24 8 10 52 51 2 5 52 51 2 5 3 1 ˆ‰ Š‹" /g99ˆ‰2 5 15. Series is geometric with r 1 Converges to œÊ /g127Ê œ22 15 55 3 1¹¹/g9925 16. Series is geometric with r 3 3 1 Diverges œ/g99 Ê /g99 /g128 Ê ¹¹ 17. Series is geometric with r 1 Converges to œÊ /g127Ê œ11 1 88 7 1¹¹18 18 /g99 18. Series is geometric with r 1 Converges to œ/g99 Ê /g99 /g127 Ê œ/g9922 2 33 5 1¹¹/g99 /g99/g9923 23ˆ‰ 19. 0.23 20. 0.234 œœ œ œ œ œ !! ˆ‰ ˆ‰_ _ œ œ n0 n023 23 234 234 100 10 99 1000 10 999n n 1 1 " " /g99 /g99# $Š‹ Š‹ ˆ‰Š‹23 234 100 1000 1000" "100 21. 0.7 22. 0.d œœ œ œœ œ !! ˆ‰ ˆ‰_ _ œ œ n0 n077 dd 10 10 9 10 10 9n n 1 1 " " /g99 /g99Š‹ Š‹ Š‹ Š‹7 d 10 10 10 10" " 23. 0.06 œœ œ œ!ˆ‰ ˆ‰ ˆ‰_ œn016 6 10 10 10 90 15n 1 "" /g99Š‹ Š‹6 100 10" 24. 1.414 1 1 1 œ/g98 œ/g98 œ/g98 œ ! ˆ‰_ œn0414 414 413 1000 10 999 999n 1 "" /g99$Š‹ Š‹414 1000 1000" Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.2 Infinite Series 579 25. 1.24123 œ/g98 œ/g98 œ/g98 œ/g98 œ œ124 123 124 124 123 124 123 100 10 10 100 100 10 10 100 99,900 99,900 33,300n 1 123,999 41,333 ! ˆ‰_ œn0&$ & #" /g99/g99Š‹ Š‹123 10 10& " $ 26. 3.142857 3 3 3 œ/g98 œ/g98 œ/g98 œ œ ! ˆ‰_ œn0142,857 142,857 3,142,854 116,402 10 10 10 1 999,999 37,037n 1'' '" /g99/g99Š‹ Š‹142,857 10 10' " ' 27. 1 0 diverges lim lim nnn1 n1 0 1Ä_ Ä_/g98œ œÁÊ 28. 1 0 diverges lim lim lim lim nn n nnn 1 n2 n3 n 5 n6 2 n5 2nn 2 n 1 2 Ä_ Ä_ Ä_ Ä_/g98 /g98/g98 /g98 /g98 /g98/g98/g98 ab ab abœ œ œ œÁÊ2 2 29. 0 test inconclusive lim n1 n4Ä_/g98œÊ 30. 0 test inconclusive lim lim nnn1 n3 2 nÄ_ Ä_/g982œœ Ê 31. cos cos 0 1 0 diverges lim n1 nÄ_œ œÁÊ 32. 1 0 diverges lim lim lim lim nn n nee e 1 en e1 e 1Ä_ Ä_ Ä_ Ä_/g98/g98nn n nn nœ œ œ œÁÊ 33. ln 0 diverges lim n1nÄ_ œ/g99 _ Á Ê 34. cos n does not exist diverges lim nÄ_1œÊ 35. s 1 1 lim s k k 22 33 4 k 1 kk k 1 k 1œ/g99/g98/g99/g98/g99/g98 á /g98 /g99/g98/g99 œ /g99 ʈ ‰ ˆ ‰ ˆ ‰ˆ‰ ˆ‰"" "" " " "" " " /g99/g98 /g98kÄ_ lim 1 1, series converges to 1œ/g99 œkÄ_ˆ‰" /g98k 1 36. s 3 lim s k k33 33 3 3 3 3 3 3 314 49 91 6 k k k 1 k 1 k 1 œ/g99/g98/g99/g98/g99 /g98 á /g98 /g99 /g98 /g99 œ /g99 ʈ‰ ˆ‰ ˆ ‰ Š‹ Š‹ab ab ab /g99/g98 /g9822 2 2 2kÄ_ lim 3 3, series converges to 3 œ/g99 œkÄ_Š‹3 k 1ab/g982 37. s ln 2 ln 1 ln 3 ln 2 ln 4 ln 3 ln k ln k 1 ln k 1 ln k kœ/g99/g98/g99/g98/g99/g98 á /g98/g99 /g99 /g98/g98 /g99Š‹ Š‹ Š‹ Š ‹ Š ‹ÈÈ È È ÈÈ ÈÈ È È ln k 1 ln 1 ln k 1 lim s lim ln k 1 ; series divergesœ/g98 /g99œ/g98 Ê œ /g98 œ _ÈÈ È È kkÄ_ Ä_k 38. s tan 1 tan 0 tan 2 tan 1 tan 3 tan 2 tan k tan k 1 tan k 1 tan k kœ/g99/g98/g99/g98/g99/g98 á /g98/g99/g99 /g98 /g98 /g99ab ab ab a b a b ab ab tan k 1 tan 0 tan k 1 lim sk lim tan k 1 does not exist; series divergesœ/g98 /g99 œ/g98 Ê œ /g98 œ ab ab abkkÄ_ Ä_ 39. s cos cos cos cos cos cos k11 11 1111 11 1123 34 45 œ/g99/g98/g99/g98/g99/g98 ሉ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰/g99/g99 /g99/g99 /g99/g99 cos cos cos cos cos /g98 /g99/g98/g99œ /g99ˆ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ‰ ˆ‰ ˆ‰/g99/g99 /g99 /g99 /g99 /g98 /g98 /g98# /g98#11 1 1 1 11 1 1 1kk 1 k 1k 3 k1 lim s lim cos , series converges to Ê œ /g99 œ/g99œkkÄ_ Ä_k 3k 3 2 6 61 1’“ ˆ‰11 1 1 1 /g99 /g98# 40. s 5 4 6 5 7 6 k 3 k 2 k 4 k 3 kœ/g99/g98/g99/g98/g99/g98 á /g98/g98 /g99 /g98 /g98/g98 /g99 /g98Š‹ Š‹ Š‹ Š ‹ Š ‹ÈÈ È È ÈÈ ÈÈ ÈÈ k 4 2 lim s lim k 4 2 ; series divergesœ/g98 /g99 Ê œ /g98 /g99 œ _ÈÈ ’“kkÄ_ Ä_k Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.580 Chapter 10 Infinite Sequences and Series 41. s 14 (4n 3)(4n 1) 4n 3 4n 1 5 5 9 9 13 4k 7 4k 3 k /g99/g98 /g99 /g98 /g99/g99" " "" "" " " "œ /g99 Êœ/g99/g98/g99/g98/g99 /g98 á /g98 /g99 ˆ‰ ˆ‰ ˆ ‰ ˆ ‰ 1 lim s lim 1 1 /g98/g99œ /g99Ê œ /g99œˆ‰ ˆ ‰"" " " /g99/g98 /g98 /g98 4k 3 4k 1 4k 1 4k 1 kkkÄ_ Ä_ 42. A(2n 1) B(2n 1) 6 (2A 2B)n (A B) 66A B (2n 1)(2n 1) 2n 1 2n 1 (2n 1)(2n 1)A(2n 1) B(2n 1) /g99/g98 /g99 /g98 /g99/g98/g98/g98 /g99œ/g98œ Ê/g98 /g98 /g99 œ Ê /g98 /g98 /g99 œ 2A 6 A 3 and B 3. Hence, 3 2A 2B 0 A B 0 A B6 AB6ÊÊ Ê œÊœ œ /g99 œ /g99/g98œ /g98 œ /g99œ /g99 œœœ!! ˆ‰kk n1 n1œœ6 (2n 1)(2n 1) n 1 n 1/g99 /g98 #/g99 #/g98"" 3 3 1 the sum is œ /g99/g98/g99/g98/g99/g98 á /g99 /g98 /g99 œ /g99 ÊŠ‹ ˆ‰"""""" " " " " # /g99 /g98 /g99 #/g98 #/g98 133557 ( k 1 ) 12 k 1 k 1 k 1 lim 3 1 3kÄ_ˆ‰/g99œ" #/g98k 1 43. 40n A B C D (2n 1) (2n 1) (2n 1) (2n 1) (2n 1) (2n 1) (2n 1) (2n 1)A(2n 1)(2n 1) B(2n 1) C(2n 1)(2n 1) D(2n 1) /g99 /g98 /g99/g99 /g98/g98 /g99 /g98/g99/g98 /g98/g98 /g98/g98/g99 /g98/g99 ## # # #### # #œ/g98 /g98/g98 œ A(2n 1)(2n 1) B(2n 1) C(2n 1)(2n 1) D(2n 1) 40nÊ /g99 /g98/g98 /g98/g98 /g98 /g99/g98 /g99œ## ## A 8 n4 n2 n 1B 4 n4 n 1C 8 n4 n2 n 1D 4 n4 n 14 0 nÊ /g98 /g99/g99/g98 /g98/g98/g98 /g99 /g99/g98œ /g99/g98œ ab a b ab a b$ ##$ # # (8A 8C)n (4A 4B 4C 4D)n ( 2A 4B 2C 4D)n ( A B C D) 40nÊ /g98 /g98 /g98/g99/g98 /g98 /g99 /g98/g99/g99 /g98 /g99 /g98 /g98 /g98œ$# 8A 8C 0 8A 8C 0 4A 4B 4C 4D 0 A B C D 0 2A 4B 2C 4D 40 A 2 A B C D 0ÊÊ/g98œ /g98œ /g98/g99/g98œ /g98/g99 /g98œ /g99 /g98/g99/g99œ /g99 /g98 /g99 /g98/g98/g98œÚÚÝÝÝÝ ÛÛÝÝÝÝÜÜBC2 D2 0 2 B2 D2 0 A BC D 0 4B 20 B 5 B D 0 /g99/g99 œ /g99 œ /g99/g98 /g98 /g98 œÊÊ œ Ê œ/g98œœ and D 5 C 0 and A 0. Hence, A C 0 A5C50œ/g99 Ê Ê œ œ/g98œ /g99/g98/g98 /g99œœ!’“k n1œ40n (n 1 )( 2 n 1 )#/g99 /g98## 5 5 œ /g99 œ /g99 /g98 /g99/g98/g99 á /g99 /g98 /g99!’“ Š ‹k n1œ" " """ " " " " " #/g99 #/g98 # # /g99 /g98 #/g99 #/g98 ( n 1) ( n 1) 1 9 9 5 5 (2(k 1) 1) ( k 1) ( k 1)## ### 5 1 the sum is lim 5 1 5 œ/g99 Ê /g99 œ Š‹ Š‹"" /g98/g98(2k 1) (2k 1)##nÄ_ 44. s 12n 1 n (n 1) n (n 1) 4 4 9 9 16 (k 1) k k (k 1) k/g98 "" " " " " " """" /g98/g98 /g99/g98## # # ## # # œ/g99 Êœ/g99/g98/g99/g98/g99 /g98 á /g98 /g99 /g98 /g99 ˆ‰ ˆ‰ ˆ ‰ ’“ ’“ lim s lim 1 1 Êœ/g99 œkkÄ_ Ä_k (k 1)’“" /g98# 45. s 1 1 k22 4 33 k 1 k k k 1 k 1œ /g99/g98/g99/g98/g99/g98 á /g98 /g98/g98/g99 œ /g99Š ‹ Š‹ Š‹ Š ‹ Š ‹"" "" " " "" " " /g99 /g98/g98 ÈÈ ÈÈ È ÈÈÈ È È lim s lim 1 1 Êœ/g99 œkkÄ_ Ä_kk 1Š‹" /g98È 46. s kœ /g99/g98/g99/g98/g99/g98 á /g98 /g99/g98/g99 œ /g99ˆ ‰ ˆ‰ ˆ‰ ˆ ‰ ˆ ‰" " "" "" " " " " " " # # ## ## # # ## # #" Î# "Î# "Î$ "Î$ "Î% ÎÐ /g99 Ñ Î Î ÎÐ /g98 Ñ ÎÐ /g98 Ñ 1k1 1 k 1 k 1k1 1k1 lim sÊ œ/g99œ /g99kÄ_k 1"" "## 47. s k ln 3 ln ln 4 ln 3 ln 5 ln 4 ln (k 1) ln k ln (k 2) ln (k 1)œ/g99/g98/g99/g98/g99/g98 á /g98 /g99/g98 /g99ˆ‰ ˆ‰ ˆ‰ Š‹ Š ‹"" "" "" " " " " #/g98 /g98 /g98 lim s œ/g99 /g98 Ê œ/g99"" " #/g98 # ln ln (k 2) ln kkÄ_ 48. s tan (1) tan (2) tan (2) tan (3) tan (k 1) tan (k) kœ/g99/g98/g99/g98 á /g98 /g99 /g99cd cd c d/g99" /g99" /g99" /g99" /g99" /g99" tan (k) tan (k 1) tan (1) tan (k 1) lim s tan (1) /g98 /g99 /g98 œ /g99 /g98 Ê œ /g99œ/g99œ /g99cd/g99" /g99" /g99" /g99" /g99" ##kÄ_k 4411 1 1 49. convergent geometric series with sum 2 2" /g99 /g99 1 2 21 Š‹È È " È2œœ /g98 È 50. divergent geometric series with r 2 1 51. convergent geometric series with sum 1 kkÈœ/g128 œŠ‹ Š‹3 # " #1 /g99/g99 Copyright © 2010 Pearson Education Inc. 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Section 10.2 Infinite Series 581 52. lim ( 1) n 0 diverges 53. lim cos (n ) lim ( 1) 0 divergesnn n Ä_ Ä_ Ä_/g99Á Ê œ /g99 Á Ên1 n /g981 54. cos (n ) ( 1) convergent geometric series with sum 1œ/g99 Ê œn 1 5 6" /g99Š‹/g99" 5 55. convergent geometric series with sum " /g99/g991 e e1Š‹" #eœ# # 56. lim ln 0 divergesnÄ_" 3nœ/g99 _Á Ê 57. convergent geometric series with sum 222 0 1 8 2 1 999/g99Š‹" 10/g99 œ/g99œ 58. convergent geometric series with sum " /g99/g991 x x 1Š‹" xœ 59. difference of two geometric series with sum 3"" /g99/g99##1 1 33 Š‹ Š‹2 33/g99œ /g99 œ" 60. lim 1 lim 1 e 0 divergesnnÄ_ Ä_ˆ‰ ˆ ‰/g99œ /g98 œÁ Ê"/g99 " /g99" nnnn 61. lim 0 diverges 62. lim lim lim n divergesnn n n Ä_ Ä_ Ä_ Ä_n! nn n n 1000 n! 1 nnnœ_Á Ê œ /g128 œ_ ʆ †â #â 63. ; both and are geometric series, and both conve !! ! ! ! ! ! ˆ‰ ˆ‰ ˆ‰ ˆ‰__ _ _ _ _ _ œœ œ œ œ œ œn1 n1 n1 n1 n1 n1 n123 2 3 1 3 1 3 44 4 2 4 2 4nn n n nn n n nn n/g98œ/g98œ /g98 œ rge since r 1 and r 1, respectivley 1 and 3 œÊ /g127 œÊ /g127 Ê œ œ œ œÊ11 33 1 3 22 44 2 4nn 11¹¹ ¹¹ !!ˆ‰ ˆ‰__ œœn1 n113 24 1324/g99/g99 1 3 4 by Theorem 8, part (1) !_ œn123 4nn n/g98œ/g98œ 64. 1 0 diverges by n term test for divergence lim lim lim nn n24 1 34 1th Ä_ Ä_ Ä_/g98/g98/g98" /g98" /g98" /g98"nn nn21n 42n 33nn 44nn œ œ œ œÁʈ‰ ˆ‰ 65. ln ln (n) ln (n 1) s ln (1) ln (2) ln (2) ln (3) ln (3) ln (4) !! ˆ‰ c d cd cd cd__ œœn1 n1n n 1/g98œ /g99 /g98 Ê œ/g99/g98/g99/g98/g99/g98 á k ln (k 1) ln (k) ln (k) ln (k 1) ln (k 1) lim s , diverges/g98 /g99 /g99/g98/g99 /g98 œ /g99 /g98 Ê œ /g99 _ Êcd cdkÄ_k 66. lim a lim ln ln 0 divergesnnÄ_ Ä_nn 2n 1œœ Á Ê ˆ‰ˆ ‰/g98#" 67. convergent geometric series with sum " /g99 /g99 1 e ˆ‰e 1œ1 1 68. divergent geometric series with r 1 kkœ¸ /g128e 23.141 22.4591 1e 69. ( 1) x ( x) ; a 1, r x; converges to for x 1 !! kk__ œœn0 n0/g99œ /g99œ œ /g99 œ /g127nn n "" /g99/g99 /g981 ( x) 1 x 70. ( 1) x x ; a 1, r x ; converges to for x 1!! ab k k__ œœn0 n0/g99œ /g99 œ œ /g99 /g127n2 n n 1 x## " /g98# Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.582 Chapter 10 Infinite Sequences and Series 71. a 3, r ; converges to for 1 1 or 1 x 3 œœ œ /g99 /g127/g127 /g99 /g127 /g127x 1 3 6 x 1 3 x/g99/g99 " #/g99 #/g99Š‹x /g99" # 72. ; a , r ; converges to !! ˆ‰ ˆ‰__ œœn0 n0(1 ) 3 sin x 3 sin x 3 sin xnn 1 /g99 #/g98 # /g98 # /g98"" /g99 " " /g99 " /g99nœœ œˆ‰ Š‹" # /g99" /g983 sin x for all x since for all x œœ Ÿ Ÿ3 sin x 3 sin x 2(4 sin x) 8 2 sin x 4 3 sin x/g98/g98 " " " /g98/g98 /g98 #ˆ‰ 73. a 1, r 2x; converges to for 2x 1 or x œœ /g127 /g127"" /g99#1 2xkk k k 74. a 1, r ; converges to for 1 or x 1. œœ /g99 œ /g127 /g128"" /g99/g98 xx 1 x1 x1 ## ## Š‹/g99" #x¸¸ kk 75. a 1, r (x 1) ; converges to for x 1 1 or 2 x 0 œ œ /g99/g98 œ /g98 /g127 /g99 /g127/g127n 1 (x 1) x"" /g98/g98 # /g98kk 76. a 1, r ; converges to for 1 or 1 x 5 œœ œ /g127 /g127 /g1273 x 2 3 x 1 x 1/g99" /g99 #/g99 #/g99Š‹3 x /g99 #¸¸ 77. a 1, r sin x; converges to for x (2k 1) , k an integer œœ Á /g98" /g99#1 sin x1 78. a 1, r ln x; converges to for ln x 1 or e x e œœ /g127 /g127 /g127" /g99/g99" 1 ln xkk 79. (a) (b) (c) !!!___ œ/g99 œ œn 2 n0 n5"" " /g98/g98 /g98/g98 /g99/g99 #(n 4)(n 5) (n 2)(n 3) (n 3)(n ) 80. (a) (b) (c) !!!___ œ/g99 œ œn 1 n3 n2 055 5 (n 2)(n 3) (n 2)(n 1) (n 19)(n 18) /g98/g98 /g99/g99 /g99 /g99 81. (a) one example is 1""" " #/g99/g98/g98/g98 /g98 á œ œ481 61Š‹ Š‹" # "# (b) one example is 3 /g99/g99/g99/g99 /g99 á œ œ /g99333 3 481 61#/g99 /g99Š‹ Š‹3 # "# (c) one example is 1 1 0. /g99/g99/g99/g99 /g99 á œ/g99 œ""" " #/g99481 61Š‹ Š‹" # "# 82. The series k is a geometric series whose sum is k where k can be any positive or negative number. !ˆ‰_ œn01 2n1 1/g98 /g99Š‹ Š‹k # " #œ 83. Let a b . Then a b 1, while (1) diverges. nn n nnn a bœœ œ œ œ œ ˆ‰ ˆ‰ !!! ! ! Š‹"" ##___ _ _ œœœ œ œ n1 n1 n1 n1 n1n n 84. Let a b . Then a b 1, while a b AB. nn n n n nnn n 43œœ œ œ œ œ œÁ ˆ‰ ˆ‰ ˆ‰ !!! ! ! ab"" " " ##___ _ _ œœœ œ œ n1 n1 n1 n1 n1 85. Let a and b . Then A a , B b 1 and 1 . nn n n 43 b Bnn n a Aœœ œ œ œ œ œ œ Áˆ‰ ˆ‰ ˆ‰ !! ! ! Š‹"" " " # #__ _ _ œœ œ œn1 n1 n1 n1n n 86. Yes: diverges. The reasoning: a converges a 0 diverges by the !! !Š‹ Š‹" "" a aa nnn nnÊ Ä Ê Ä_Ê nth-Term Test. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.3 The Integral Test 583 87. Since the sum of a finite number of terms is finite, adding or subtracting a finite number of terms from a series that diverges does not change the divergence of the series. 88. Let A a a a and lim A A. Assume a b converges to S. Let nn nn n œ/g98/g98 á /g98 œ /g98 "#nÄ_!ab S( ab ) ( ab ) ( ab ) S( aa a ) ( bb b ) nn n n n n œ/g98/g98/g98/g98 á /g98/g98 Êœ/g98 /g98 á /g98/g98/g98 /g98 á /g98 "" ## "# "# b b b S A lim b b b S A b converges. ThisÊ/g98 /g98 á /g98 œ /g99 Ê /g98 /g98 á /g98œ /g99 Ê "# "# nnn n nnÄ_ab ! contradicts the assumption that b diverges; therefore, a b diverges. !! ab nn n/g98 89. (a) 5 1 r r ; 2 2 222 3 3 3 1r 5 5 5 5/g99#œÊœ /g99 Ê œ /g98 /g98 /g98 á ˆ‰ ˆ‰ (b) 5 1 r r ; Š‹13 2 1 r 10 10 2 10 10 1013 3 13 13 3 13 3 13 3 /g99# # ##$œ Ê œ /g99 Ê œ /g99 /g99 /g98/g99/g98 á ˆ‰ ˆ‰ ˆ‰ 90. 1 e e 9 1 e e b ln /g98/g98 /g98 á œ œÊœ /g99 Ê œÊœb2 b b b 1e 99 988 "" /g99b ˆ‰ 91. s 1 2r r 2r r 2r r 2r , n 0, 1, n2n 2n 1œ /g98/g98/g98 /g98/g98 /g98 á /g98 /g98 œ á#$ %& /g98 s 1 rr r 2 r 2 r2 r 2 r l i m sÊ œ /g98 /g98 /g98 á/g98 /g98 /g98 /g98 /g98 á/g98 Ê œ /g98 n n2n 2n 1 1r 1r2rab a b#% $ & /g98 " /g99/g99 nÄ_## , if r 1 or r 1 œ/g127 /g12712 r 1r/g98 /g99# #kk k k 92. L s /g99œ /g99 œ naa r 1r 1r 1ra1 r /g99/g99 /g99/g99abn n 93. area 2 2 (1) 4 2 1 8 m œ /g98 /g98 /g98 /g98á œ /g98 /g98 /g98 /g98á œ œ## ###"" # /g99Š‹ Š ‹È È24 1" # 94. (a) L 3, L 3 , L 3 , , L 3 lim L lim 3 "# $#œœ œ á œ Ê œ œ _ ˆ‰ ˆ‰ ˆ‰ ˆ‰44 4 4 33 3 3nnn1 n1 /g99/g99 nnÄ_ Ä_ (b) Using the fact that the area of an equilatera l triangle of side length s is s , we see that A ,ÈÈ33 442"œ AA3 , AA3 4 , #" $#"" ##œ/g98 œ /g98 œ/g98 œ /g98/g98 Š‹ Š‹ ˆ‰ ˆ ‰ abÈ ÈÈ È ÈÈÈ 3 33 3 333 43 4 1 43 4 1 2 722 2 A A 3 4 , A A 3 4 , . . . , %$""œ/g98 œ/g98 ab ab Š‹ Š‹ ˆ‰ ˆ‰23 33 43 4322 54ÈÈ 34 A 3 4 3 3 4 3 3 . nnn n k2 k2 k2œ/g98 œ/g98 œ/g98ÈÈ È È 33 3 3 44 3 4 9 4k2 k k1 k1 4 9!! !ab ab Š‹ ˆ‰ ˆ ‰ ÈÈŒ/g57 œœ œ/g99/g99 $ ""/g99/g99 2k k1/g99 $ /g99 lim A lim 3 3 3 3 3 3 1nnÄ_ Ä_nn k2œ /g98 œ/g98 œ/g98 œ /g98 Œ/g57 Œ /g57ÈÈ ÈŒ/g57! ˆ‰ ˆ ‰ÈÈ È È33 3 3 44 4 2 0 4 541 3 9 1œk k1/g99 $ /g991 36 4 9/g99 A œœÈ3 45 588ˆ‰" 10.3 THE INTEGRAL TEST 1. f x is positive, continuous, and decreasing for x 1; dx lim dx lim ab ’“ œ  œ œ /g9911 1 1 xx x x2 22'' 11b 1b _ bbÄ_ Ä_ lim 1 1 dx converges convergesœ/g99 /g98 œ Ê ÊbÄ_ˆ‰ ! 11 1 bx nn1' 1_ 22 œ_ 2. f x is positive, continuous, and decreasing for x 1; dx lim dx lim xab ’“ œ  œ œ1 11 5 x xx 40.8 0.2 0.2 0.2'' 11b 1b _ bbÄ_ Ä_ lim b dx diverges diverges œ/g99 œ _Ê ÊbÄ_ˆ‰ ! 55 1 1 44 x n0.8 n1' 1_ 0.2 0.2 œ_ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.584 Chapter 10 Infinite Sequences and Series 3. f x is positive, continuous, and decreasing for x 1; dx lim dx lim tanab ’“ œ  œ œ1 11 1 x x4 x4 x4 2 21 2 22 /g98 /g98/g98/g99 '' 11b 1b _ bbÄ_ Ä_ lim tan tan tan dx converges converges œ/g99 œ /g99 Ê ÊbÄ_ˆ‰ ! 1b 11 11 1 1 22 22 4 22 x 4 n 411 1 n1/g99/g99 /g99 /g98/g98œ_1 ' 1_ 22 4. f x is positive, continuous, and decreasing for x 1; dx lim dx lim ln x 4ab ’“ œ   œ œ l/g98l1 11 x4 x4 x4/g98 /g98/g98'' 11b 1b _ bbÄ_ Ä_ lim ln b 4 ln 5 dx diverges diverges œ l/g98l /g99 œ _ Ê ÊbÄ_ab ! ' 1_11 x4 n4n1/g98/g98œ_ 5. f x e is positive, continuous, and decreasing for x 1; e dx lim e dx lim eab ’“ œ  œ œ /g99/g99 /g99/g99 /g99 2x 2x 2x 2x 1 2'' 11b 1b _ bbÄ_ Ä_ lim e dx converges e converges œ/g99 /g98 œ Ê ÊbÄ_ˆ‰ ! 11 1 2e 2e 2e2x 2n n12b 22 ' 1_/g99/g99 œ_ 6. f x is positive, continuous, and decreasing for x 2; dx lim dx lim ab ’“ œ  œ œ /g991 11 1 xl n x xl n x xl n x ln x ab ab ab2 22 '' 22b 2b _ bbÄ_ Ä_ lim dx converges converges œ/g99 /g98 œ Ê ÊbÄ_ˆ‰ ! 11 1 1 1 ln b ln 2 ln 2 xl n x nl n nn2' 2_ ab ab22 œ_ 7. f x is positive and continuous for x 1, f x 0 for x 2, thus f is decreasing for x 3; ab ab œ  œ /g127 /g128  x4 x x4 x42 22 2 /g98w /g99 /g98ab dx lim dx lim ln x 4 lim ln b 4 ln 13 dx '' ' 33 3b 3b _ _xx 1 1 1 x x4 x4 2 2 2 x422 22 2 /g98/g98 /g98œœ /g98 œ /g98 /g99 œ _Êbb bÄ_ Ä_ Ä_’“ab aba b ˆ‰ diverges diverges diverges ÊÊ œ /g98/g98 !! ! n3 n1 n3nn 1 2 n n4 n4 5 8 n4œœ œ__ _ /g98/g98 /g9822 2 8. f x is positive and continuous for x 2, f x 0 for x e, thus f is decreasing for x 3; ab ab œ  œ /g127 /g128  ln x 2 ln x xx2 2 2w /g99 dx lim dx lim 2 ln x lim 2 ln b 2 ln 3 dx '' ' 33 3b 3b _ _ln x ln x ln x xx x22 2œœœ /g99 œ _ÊbbbÄ_ Ä_ Ä_’“ab a b ab ab diverges diverges diverges ÊÊ œ /g98 !! ! n3 n2 n3ln n ln n ln 4 ln n nn 2 nœœ œ__ _22 2 9. f x is positive and continuous for x 1, f x 0 for x 6, thus f is decreasing for x 7; ab ab œ  œ /g127 /g128  x e3 exx 62 x3 x3 Î Îw /g99/g99ab dx lim dx lim lim '' 77b 7b _x x 3x 18x 54 3b 18b 54 327 e e eee e e22 2 2 x3 x3 x3 x3 x3 b3 73 Î Î ÎÎÎ Î Î œœ /g99 /g99 /g99 œ /g98 œbb b Ä_ Ä_ Ä_’“ Š‹/g99/g99 /g99 lim lim dx converges converges œ/g98 œ /g98 œ Ê ÊbbÄ_ Ä_Š‹ ˆ‰ !36 b1 8 ee e e ee e327 54 327 327 x n n7ab/g99/g99 /g99 œ_ b3 73 b3 73 73 x3 n322 ÎÎ Î Î ÎÎ Î ' 7_ converges Ê œ/g98/g98 /g98/g98/g98 /g98!! n1n 1 4 9 16 25 36 n ee e e e e een7 œ__ œ2 2 n3 13 23 43 53 n3 12 ÎÎ Î Î Î Î 10. f x is continuous for x 2, f is positive for x 4, and f x 0 for x 7, thus f is ab ab œœ   /g128 œ /g127 /g128x4 x4 7x x2 x 1 x1 x1/g99/g99 /g99 /g99/g98 /g99 /g99w 22 3ab ab decreasing for x 8; dx lim dx dx lim dx dx  œ /g99 œ /g99'' ' ' ' 88 8 8 8bb b b _x4 x1 3 1 3 x1 x1 x1 x1 x1/g99/g99 /g99/g99 /g99 /g99 /g99 ab ab ab ab22 2 2bbÄ_ Ä_”• ” • lim ln x 1 lim ln b 1 ln 7 dx diverges œ l/g99l /g98 œ l/g99l /g98 /g99 /g99 œ _ ÊbbÄ_ Ä_’“ ˆ‰33 3 x4 x1 b1 7 x1 /g99/g99/g99 /g998b 8'_ ab2 diverges 2 0 diverges Ê Ê œ /g99 /g99 /g98 /g98/g98/g98/g98 !! ! n8 n2 n8n4 n4 1 1 2 3 n4 n2 n 1 n2 n 1 4 1 6 2 5 3 6 n2 n 1œœ œ__ _/g99/g99 /g99 /g99/g98 /g99/g98 /g99/g9822 2 11. converges; a geometric series with r 1 12. converges; a geometric series with r 1 œ/g127 œ /g127" " 10 e 13. diverges; by the nth-Term Test for Divergence, lim 1 0nÄ_n n1/g98œÁ Copyright © 2010 Pearson Education Inc. 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Section 10.3 The Integral Test 585 14. diverges by the Integral Test; dx 5 ln (n 1) 5 ln 2 dx '' 11n55 x1 x1/g98/g98œ/g98 /g99Ê Ä __ 15. diverges; 3 , which is a divergent p-series (p ) !!__ œœn1 n13 nnÈÈœœ"" # 16. converges; 2 , which is a convergent p-series (p ) !!__ œœn1 n1/g99" #23 nn n Èœ/g99 œ$ Î# 17. converges; a geometric series with r 1 œ/g127" 8 18. diverges; 8 and since diverges, 8 diverges !! ! !__ _ _ œœ œ œn1 n1 n1 n1/g99"81 1 nn n nœ/g99 /g99 19. diverges by the Integral Test: dx ln n ln 2 dx '' 22nln x ln x xxœ/g99 Ê Ä _"## ab_ 20. diverges by the Integral Test: dx; te dt lim 2te 4et ln x dt dx e dt'' 2l n2__ln x xdx x tt2 t2 t2 ÈÔ× ÕØ/g60‘œ œ œÄœ /g99ÎÎ Î bÄ_b ln 2 lim 2e (b 2) 2e (ln 2 2)œ/g99 /g99 /g99 œ _bÄ_/g60‘b2 l n 2 2 ÎÐ ÑÎ 21. converges; a geometric series with r 1 œ/g1272 3 22. diverges; lim lim lim 0nn nÄ_ Ä_ Ä_55 ln 5 l n 5 5 43 4 l n 4 l n 4 4n nn nn/g98œœ œ _Á ˆ‰ ˆ ‰ 23. diverges; 2 , which diverges by the Integral Test !!__ œœn0 n0/g99" /g98/g982 n1 n 1œ/g99 24. diverges by the Integral Test: ln (2n 1) as n ' 1ndx 2x 1/g99#"œ /g99 Ä_ Ä_ 25. diverges; lim a lim lim 0nn nÄ_ Ä_ Ä_n22 ln 2 n1 1œœ œ _ Ánn /g98 26. diverges by the Integral Test: ; ln n 1 ln 2 as nux du'' 12nn 1dx du xx 1dx xu ÈȈ‰È/g98–—È ˆ‰Èœ/g98 " œÄ œ /g98 /g99 Ä_ Ä_È /g98 27. diverges; lim lim lim 0nn nÄ_ Ä_ Ä_ÈÈ Š‹ Š‹nn ln nœœ œ _Á" "2n nÈ # 28. diverges; lim a lim 1 e 0nnÄ_ Ä_n nnœ/g98 œ Á ˆ‰" 29. diverges; a geometric series with r 1.44 1 œ¸ /g128" #ln 30. converges; a geometric series with r 0.91 1 œ¸ /g127" ln 3 31. converges by the Integral Test: dx; duu ln x du dx'' 3l n3__ Š‹ ÈÈ" x (ln x) (ln x) 1xuu 1# # /g99"" /g99”•œ œÄ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.586 Chapter 10 Infinite Sequences and Series lim sec u lim sec b sec (ln 3) lim cos sec (ln 3)œœ /g99 œ /g99bb b Ä_ Ä_ Ä_cd c d kk /g60‘ ˆ‰/g99" /g99" /g99" /g99" /g99" "b ln 3 b cos (0) sec (ln 3) sec (ln 3) 1.1439œ/g99 œ /g99 ¸/g99" /g99" /g99" #1 32. converges by the Integral Test: dx dx; duu ln x du dx'' ' 11 0__ _"" /g98/g98 /g98 " x 1 ln x 1 (ln x) 1 u xabŠ‹ ## # œÄœ œ" x”• lim tan u lim tan b tan 0 0œœ /g99 œ /g99 œbbÄ_ Ä_cd a b/g99" /g99" /g99" ##b 011 33. diverges by the nth-Term Test for divergence; lim n sin lim lim 1 0nn x0 Ä_ Ä_ Ĉ‰" nxsin xœœ œ Ásinˆ‰ ˆ‰" "n n 34. diverges by the nth-Term Test for divergence; lim n tan lim lim nn nÄ_ Ä_ Ä_ˆ‰"nœœtan sec ˆ‰ ˆ‰Š‹ ˆ‰ Š‹" "" ## " " #n nn n n/g99 /g99 lim sec sec 0 1 0œœ œ ÁnÄ_##"ˆ‰n 35. converges by the Integral Test: dx; du lim tan uue du e dx'' 1ex x__e 1e 1ux 2x/g98/g98" /g99"”• cdœ œÄœ #nÄ_b e lim tan b tan e tan e 0.35œ/g99 œ /g99 ¸bÄ_ab/g99" /g99" /g99" #1 36. converges by the Integral Test: dx; du duue du e dx dx du'' ' 1e e__ _22 2 2 1e u ( 1u ) u u1x x u/g98/g98 /g98"xÔ× Õ؈‰œ œ œÄœ /g99 lim 2 ln lim 2 ln 2 ln 2 ln 1 2 ln 2 ln 0.63œ œ /g99œ /g99œ /g99¸bbÄ_ Ä_/g60 ‘ ˆ‰ ˆ‰ ˆ‰ ˆ‰ub e e e u1 b1 e1 e1 e1/g98/g98 /g98 /g98 /g98b e 37. converges by the Integral Test: dx; 8u du 4u 4ut a n x du'' 1422 4_ Î ÎÎ Î8 tan x 3 1x 4 1 6 4 dx 1x/g99 " ## # # #/g98/g99" /g98#”• cd Š‹œ œÄœ œ /g99 œ 1111 11 1 38. diverges by the Integral Test: dx; lim ln u lim (ln b ln 2)ux 1 du 2x dx'' 12__xd u x1 4#/g98# # ##"" "”•/g60‘œ/g98 œÄœ œ /g99 œ _bbÄ_ Ä_b 2 39. converges by the Integral Test: sech x dx 2 lim dx 2 lim tan e '' 11bb 1_ œœbbÄ_ Ä_e 1exx x/g98/g99" ab# cd 2 lim tan e tan e 2 tan e 0.71œ/g99 œ /g99 ¸bÄ_ab/g99" /g99" /g99" b1 40. converges by the Integral Test: sech x dx lim sech x dx lim tanh x lim (tanh b tanh 1) '' 11bb 1_##œœ œ /g99bb bÄ_ Ä_ Ä_cd 1 tanh 1 0.76œ/g99 ¸ 41. dx lim a ln x 2 ln x 4 lim ln ln ; ' 1_ˆ‰ ˆ ‰ cdkkkka 3 x2 x4 b4 5(b 2) /g98/g98 /g98" /g98/g99œ /g98 /g99 /g98 œ /g99bbÄ_ Ä_b 1a a lim a lim (b 2) the series converges to ln if a 1 and diverges to if, a 1 1, a 1 bbÄ_ Ä_(b 2) b4 3a1 5 /g98 /g98/g99aœ/g98 œ Ê œ __/g128 œœˆ‰ a 1. If a 1, the terms of the series eventually become negative and the Integral Test does not apply. From /g128/g127 that point on, however, the series behaves like a negative multiple of the harmonic series, and so it diverges. 42. dx lim ln lim ln ln ; lim ' 3b 3_ˆ‰ ˆ ‰ ’“¹¹"/g99 /g99 /g99 " /g99/g98 /g98 /g98 /g98 x1 x1 ( x1 ) ( b1 ) 4 ( b1 )2 a x1 b1 2 b/g99œ œ /g99bb bÄ_ Ä_ Ä_2a 2a 2a 2a lim the series converges to ln ln 2 if a and diverges to if1, a , aœœ Ê œ œ_œ _/g127 bÄ_" " #/g98 # #" # "#a(b 1)4 2a 1 /g99/g74ˆ‰ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.3 The Integral Test 587 if a . If a , the terms of the series eventually become negative and the Integral Test does not apply. /g127/g128"" ## From that point on, however, the series behaves like a negative multiple of the harmonic series, and so it diverges. 43. (a) (b) There are (13)(365)(24)(60)(60) 10 seconds in 13 billion years; by part (a) s 1 ln n where ab* nŸ/g98 n (13)(365)(24)(60)(60) 10 s 1 ln (13)(365)(24)(60)(60) 10 œÊ Ÿ /g98 ab a b ab** n 1 ln (13) ln (365) ln (24) 2 ln (60) 9 ln (10) 41.55 œ/g98 /g98 /g98 /g98 /g98 ¸ 44. No, because and diverges !! !__ _ œœ œn1 n1 n1"" " " nx x n nœ 45. Yes. If a is a divergent series of positive numbers, then a also diverges and a . !! ! ˆ‰ ˆ ‰_ _ œ œ n1 n1n nnaa "## #_ n1 œœ/g127nn There is no “smallest" divergent series of positive num bers: for any divergent series a of positive numbers !_ œn1n has smaller terms and still diverges.!ˆ‰_ œn1an # 46. No, if is a convergent series of positive numbers, then 2 a 2a also converges, and 2a a . !! !_ __ œ œœ n1 n1 n1 a n nn n n œ  There is no “largest" convergent series of positive numbers. 47. (a) Both integrals can represent the area under the curve f x , and the sum s can be considered an abœ1 x150 È/g98 approximation of either integral using rectangles with x 1. The sum s is an overestimate of the ?œœ 501 n1!50 n1œ È/g98 integral dx. The sum s represents a left-hand sum (that is, the we are choosing the left-hand endpoint of ' 1511 x150 È/g98 each subinterval for c ) and because f is a decreasing function, the value of f is a maximum at the left-hand endpoint of i each sub interval. The area of each rectangle overestimates the true area, thus dx . In a similar ' 15111 x1 n1ÈÈ/g98/g98/g127!50 n1œ manner, s underestimates the integral dx. In this case, the sum s represents a right-hand sum and because 50 501 x1' 050 È/g98 f is a decreasing function, the value of f is aminimum at the right-hand endpoint of each subinterval. The area of each rectangle underestimates the true area, thus dx. Evaluating the integrals we find dx !50 n1œ 11 1 n1 x1 x1 ÈÈ È /g98/g98 /g98/g127'' 0 150 51 2 x 1 2 52 2 2 11.6 and dx 2 x 1 2 51 2 1 12.3. Thus,œ/g98 œ/g99 ¸ œ/g98 œ/g99 ¸ ’“ ’“ ÈÈ ÈÈ ÈÈ 1 051 50 050' 1 x1È/g98 11.6 12.3. /g127/g127!50 n1œ 1 n1È/g98 (b) s 1000 dx 2 x 1 2 n 1 2 2 1000 n 500 2 2 251414.2 nn1 x1n 2 /g128Ê œ /g98 œ/g98 /g99/g128Ê /g128/g98 /g99 ¸ ' 11 11 /g98 /g98 È/g98’“ÈÈ ÈÈ Š‹ n 251415. Ê  Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.588 Chapter 10 Infinite Sequences and Series 48. (a) Since we are using s to estimate , the error is given by . We can consider this sum as 3011 1 nn nœ!! !30 n1 n1 n3 1œœ œ__ 44 4 an estimate of the area under the curve f x when x 30 using rectangles with x 1 and c is the right-hand endpoint of abœ  œ1 x i 4 ? each subinterval. Since f is a decreasing function, the value of f is a minimum at the right-hand endpoint of each subinterval, thus dx dx 1.23 10 . ! ’“ Š‹_ œn3 1 11 1 1 1 1 nx x 3 x 3 bbb bb b 33 05 44 4 3 3 3 /g127œ œ /g99 œ /g99 /g98 ¸ ‚'' 30 3030_ Ä_ Ä_ Ä_/g99lim lim limab Thus the error 1.23 10 /g127‚Þ/g995 (b) We want S s 0.000001 dx 0.000001 dx dx /g99 /g127 Ê/g127 Êœ œ /g99 n11 1 1 xx x 3 xbbb b'' ' nn nn__ Ä_ Ä_44 4 3 lim lim’“ 0.000001 n 69.336 n 70. œ/g99 /g98 œ /g127 Ê /g128 ¸ Ê  lim b1 1 1 1000000 3b 3n 3n 3Ä_ˆ‰ É33 33 49. We want S s 0.01 dx 0.01 dx dx /g99 /g127 Ê/g127 Êœ œ /g99 œ /g99 /g98 n11 1 1 1 1 xx x 2 x 2 b 2 nbbbb b'' ' nn nn__ Ä_ Ä_ Ä_33 3 2 2 2 lim lim lim ’“ ˆ‰ 0.01 n 50 7.071 n 8 S s 1.195 œ/g127 Ê /g128 ¸ Ê   Ê ¸ œ ¸1 1 2n n 8 2 3È !8 n1œ 50. We want S s 0.1 dx 0.1 dx tan /g99/g127 Ê /g127 Ê œ n11 1 x x4 x4 2 2bbb1b'' nnn_ /g98/g98Ä_ Ä_/g99 22 lim lim’“ ˆ‰ tan tan tan 0.1 n 2tan 0.2 9.867 n 10 S s œ/g99 œ /g99 /g127 Ê/g128 /g99 ¸ Ê  Ê¸ lim b1b 1n 1n 22 22 4 22 211 110Ä_/g99/g99 /g99ˆ‰ ˆ ‰ ˆ ‰ ˆ‰ ˆ‰11 0.57œ¸!10 n1œ 1 n42/g98 51. S s 0.00001 dx 0.00001 dx dx /g99 /g127 Ê/g127 Êœ œ /g99 œ /g99 /g98 n11 1 1 0 1 0 1 0 xx x x b nbb bb b'' ' nn nn__ Ä_ Ä_ Ä_1.1 1.1 1.1 0.1 0.1 0.1 lim lim lim ’“ ˆ‰ 0.00001 n 1000000 n 10œ /g127 Ê/g128 Ê/g12810 n10 60 0.1 52. S s 0.01 dx 0.01 dx dx /g99 /g127 Ê/g127 Êœ œ /g99 n11 1 1 xl n x xl n x xl n x 2l n x bbb b'' ' nn nn__ Ä_ Ä_ ab ab ab ab33 3 2 lim lim’“ 0.01 n e 1177.405 n 1178œ/g99 /g98 œ/g127 Ê /g128 ¸ Ê  lim b11 1 2l n b 2l n n 2l n n50 Ä_Š‹ab ab abÈ 22 2 53. Let A and B 2 a , where {a } is a nonincreasing sequence of positive terms converging to nk n kk 2 œœ!!nn k1 k1œœ a abk 0. Note that {A } and {B } are nondecreasing sequences of positive terms. Now, nn B2 a4 a8 a 2 a 2 a 2 a2 a 2 a2 a2 a2 a nn2 œ/g98/g98/g98 á /g98 œ/g98 /g98 /g98 /g98/g98/g98 /g98 á #%) # %% )))) abn ab a b 2a 2a 2a 2a 2a 2a 2a 2a 2a 2a 2a /g98 /g98 /g98 á /g98 Ÿ/g98/g98 /g98 /g98 /g98/g98/g98 /g98 áðóóóóóóóóóóóóóóñóóóóóóóóóóóóóóòˆ‰ ab a b ab ab ab 22 2nn n "# $% &'() 2 t e r m sn1/g99 2a 2a 2a 2A 2 a . Therefore if a converges, /g98/g98 /g98 á /g98 œ Ÿˆ‰ !!ab a b ab ab 22 1 22 kkn1 n1 nn /g99/g99 /g98_ œk1 then {B } is bounded above 2 a converges. Conversely, n Ê!k 2 abk Aa aa aaaa aa2 a4 a 2 a aBa 2 a . nn nnk2 2 œ/g98 /g98 /g98 /g98/g98/g98 /g98 á /g98/g127/g98 /g98 /g98 á /g98 œ/g98 /g127/g98" #$ %&'( " # % " " ab a b !ab abn k_ œk1 Therefore, if 2 a converges, then {A } is bounded above and hence converges. !_ œk1k 2 n abk 54. (a) a 2 a 2 , which diverges ab ab ab 22 2 ln 2 2 n(ln 2) n(ln 2) ln nnnnnnn n nœœ Ê œ œ"" " " " ## ††!! !__ _ œœ œ n2 n2 n2 diverges.Ê!_ œn2" n ln n Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.3 The Integral Test 589 (b) a 2 a 2 , a geometric series that ab abab22nn 2n nnnp np p 1 p 1 n œÊ œ œ œ"" " " ## #!!! ! ˆ‰___ _ œœœ œn1 n1 n1 n1† /g99/g99 converges if 1 or p 1, but diverges if p 1." #p1 /g99/g127/g128 Ÿ 55. (a) ; u du lim lim b (ln 2)u ln x du'' 2l n2pb ln 2__ /g99 dx u x(ln x) p 1 1 p dx xp1 p1 pp1 ”• ’“ Š ‹ cdœ œÄœ œ /g99bbÄ_ Ä_/g99 /g98 /g99/g98 /g99" /g99/g98 /g99/g98 the improper integral converges if p 1 and diverges if p 1.(ln 2) , p 1 , pœÊ /g128 /g127/g128 _/g127 "œ" /g99p1/g99 /g98p1 For p 1: lim ln (ln x) lim ln (ln b) ln (ln 2) , so the improper integral diverges if œœ œ /g99 œ _' 2b 2_dx x ln xbbÄ_ Ä_cd c d p1 . œ (b) Since the series and the integral converge or diverge together, converges if and only if p 1. !_ œn2" n(ln n)p /g128 56. (a) p 1 the series diverges œÊ (b) p 1.01 the series converges œÊ (c) ; p 1 the series diverges !!__ œœn2 n2"" " n ln n 3 n(ln n)ab$œœ Ê (d) p 3 the series converges œÊ 57. (a) From Fig. 10.11(a) in the text with f(x) and a , we have dx 1 œ œ Ÿ/g98/g98/g98 á /g98"" " " " " # xk x 3 n k ' 1n1 /g98 1 f(x) dx ln (n 1) 1 1 ln n 0 ln (n 1) ln n Ÿ/g98 Ê /g98 Ÿ/g98/g98/g98 á /g98Ÿ/g98 Ê Ÿ /g98 /g99 ' 1n"" " # 3n 1 ln n 1. Therefore the sequence 1 ln n is bounded above by Ÿ /g98/g98/g98 á /g98 /g99 Ÿ /g98/g98/g98 á /g98 /g99 ˆ‰ ˜ ™ˆ‰"" " "" " # # 3n 3n 1 and below by 0. (b) From the graph in Fig. 10.11(b) with f(x) , dx ln (n 1) ln n œ/g127 œ /g98 /g99"" " /g98 xn 1 x' nn1 /g98 0 ln (n 1) ln n 1 ln (n 1) 1 ln n . Ê /g128 /g99 /g98 /g99 œ /g98/g98/g98 á /g98 /g99 /g98 /g99 /g98/g98/g98 á /g98/g99"" " " " " " /g98# /g98 # n1 3 n 1 3 ncd ˆ‰ ˆ ‰ If we define a 1 ln n, then 0 a a a a {a } is a decreasing sequence of nn 1 n n1 n n 3nœ/g98œ/g98/g99 /g128 /g99 Ê /g127 Ê"" "# /g98/g98 nonnegative terms. 58. e e for x 1, and e dx lim e lim e e e e dx converges by/g99/g99 /g99 /g99 /g99 xx x b 1# #Ÿ  œ /g99 œ /g99 /g98 œ Ê '' 1 1x 1x_ _ /g99 /g99/g99 bbÄ_ Ä_cd ˆ‰b " the Comparison Test for improper integrals e 1 e converges by the Integral Test. Êœ /g98!!__ œœn0 n1/g99/g99nn## 59. (a) s 1.97531986; dx lim x dx lim lim and 1010 n x 2 2b 242 2421x 1 1 1 3œœ œ œ/g99 œ/g99 /g98 œ ! ’“ ˆ‰ n1b 11 œ" /g99 33 22 '' 11 11b _ bb bÄ_ Ä_ Ä_/g99 dx lim x dx lim lim '' 10 10b _1x 1 1 1 x 2 2b 200 2003 3 22œœ /g99 œ /g99 /g98 œbb bÄ_ Ä_ Ä_/g99’“ ˆ‰/g99b10 1.97531986 s 1.97531986 1.20166 s 1.20253 Ê /g98 /g127/g127 /g98 Ê /g127/g12711 242 200 (b) s 1.202095; error 0.000435 œ¸ œ Ÿ œ!_"/g98 /g99 n1œn2 21.20166 1.20253 1.20253 1.20166 3 60. (a) s 1.082036583; dx lim x dx lim lim and 1010 n x 3 3b 3993 39931x 1 1 1 4œœ œ œ/g99 œ/g99 /g98 œ ! ’“ ˆ‰ n1b11 œ " /g99 44 33 '' 11 11b _ bb bÄ_ Ä_ Ä_/g99 dx lim x dx lim lim '' 10 10b _1x 1 1 1 x 3 3b 3000 30004 4 33œœ /g99 œ /g99 /g98 œbb bÄ_ Ä_ Ä_/g99’“ ˆ‰/g99b10 1.082036583 s 1.082036583 1.08229 s 1.08237 Ê /g98 /g127/g127 /g98 Ê /g127/g12711 3993 3000 (b) s 1.08233; error 0.00004 œ¸ œ Ÿ œ!_"/g98 /g99 n1œn2 21.08229 1.08237 1.08237 1.08229 4 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.590 Chapter 10 Infinite Sequences and Series 10.4 COMPARISON TESTS 1. Compare with , which is a convergent p-series, since p 2 1. Both series have nonnegative terms for n 1. For !_ œn1" n2 œ/g128   n 1, we have n n 30 . Then by Comparison Test, converges.  Ÿ /g98 Ê  22 11 1 nn 3 0 n 3 022 2 /g98/g98!_ œn1 2. Compare with , which is a convergent p-series, since p 3 1. Both series have nonnegative terms for n 1. For !_ œn1" n3 œ/g128   n 1, we have n n 2 . Then by Comparison Test,   Ÿ /g98 Ê  Ê  Ê   44 11 nn 1n n 1 n 1 nn 2nn 2nn 2 n 2 n 244 44 34 4 4 /g98/g98/g98 /g98 /g98/g99/g99 !_ œn1 converges. 3. Compare with , which is a divergent p-series, since p 1. Both series have nonnegative terms for n 2. For !_ œn2"" # ÈnœŸ   n 2, we have n 1 n . Then by Comparison Test, diverges.  /g99 Ÿ Ê   ÈÈ ! 11 1 n1 n n1 ÈÈ È /g99/g99_ œn2 4. Compare with , which is a divergent p-series, since p 1 1. Both series have nonnegative terms for n 2. For !_ œn2" nœŸ   n 2, we have n n n . Thus diverges.   /g99 ŸÊ  Ê  œ Ê    22 11 nn 1 n 2n 1 n 2 n nn n nn n n nn nn n n22 22 22 2 /g99/g99 /g99 /g99 /g99/g98/g98 !_ œn2 5. Compare with , which is a convergent p-series, since p 1. Both series have nonnegative terms for n 1. !_ œn1" n32 32 Î œ/g128   For n 1, we have 0 cos n 1 . Then by Comparison Test, converges.  Ÿ Ÿ Ê Ÿ2 cos n 1 cos n nn n2 2 32 32 32 ÎÎ Î!_ œn1 6. Compare with , which is a convergent geometric series, since r 1. Both series have nonnegative terms for ! ¹¹_ œn1" 331 n llœ /g127 n 1. For n 1, we have n 3 3 . Then by Comparison Test, converges.    †   Ê Ÿnn 11 1 n3 3 n3††nn n!_ œn1 7. Compare with . The series is a convergent p-series, since p 1, and the series !! !__ _ œœ œ n1 n1 n1È È 5 5 nn n13 232 32 32 ÎÎ Î œ/g128 5 converges by Theorem 8 part 3. Both series have nonnegative terms for n 1. For n 1, we haveœ    È!_ œn11 n32 Î n n 4n 4n n 4n n 4n 5n n 4n 5n 20 5 n 4 5.3 4 3 4 43 44 4 43 4 4 n4 n n4Ÿ ÊŸÊ /g98Ÿ /g98œÊ /g98Ÿ/g98 œ /g98 Ê Ÿ ab43 4/g98 /g98 5 Then by Comparison Test, converges. ÊŸ Ê Ÿ ÊŸ œnn 4 n4 n4 n n4 n n4n4 5 n4 5 n4 5 n3 44 3 4 3 4 32ab È /g98 /g98/g98 /g98 /g98/g98/g98 /g98ÉÉ É !Î_ œn1 8. Compare with , which is a divergent p-series, since p 1. Both series have nonnegative terms for n 1. For !_ œn1"" # ÈnœŸ   n 1, we have n 1 2 n 2 2 n 1 3 n 2 n 1 3n 3 2 n n n 3    Ê  Ê /g98 Ê /g98    Ê /g98  ÈÈÈ È È ˆ‰ n2 n n n n3 1 Ê /g98 /g98  /g98Ê  Ê  Ê  Ê  22 nn 2 n 1 n 1 n 1 n3 n3 n n3 n n3 nn2n 1 11 1 È Ê Éˆ‰ È ˆ ‰ ˆ ‰ ÈÈ È/g98/g98 /g98 /g98 /g98/g98 /g98/g98/g98/g98 22 2222 . Then by Comparison Test, diverges. Ê ÈÈ ÈÈ Èn1 n1 n3 n31 n/g98/g98 /g98/g982 2!_ œn1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.