MTH 201 Chapter 10 Solutions, Fall 2015

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 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. 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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 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 œ Ê 52. lim (0.03) 1 converges (Theorem 5, #3) n Ä _ 1 n 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 1 n 1 n 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 œ œ œ _ Ê n n _ 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 � � 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. 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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 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 � 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 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 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}] � 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 � � � � œ � ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ lim s 5 Ê œ n Ä _ n 7. 1 , the sum of this geometric series is � � � � á œ œ " " " " " � � � 4 16 64 5 1 1 4 ˆ ‰ ˆ ‰ 4 4 8. , the sum of this geometric series is " " " " � # 16 64 256 1 1 � � � á œ ˆ ‰ ˆ ‰ 16 4 9. , the sum of this geometric series is 7 7 7 7 4 16 64 3 1 � � � á œ ˆ ‰ ˆ ‰ 7 4 4 � 10. 5 , the sum of this geometric series is 4 � � � � á œ 5 5 5 5 4 16 64 1 � � ˆ ‰ 4 11. (5 1) , is the sum of two geometric series; the sum is � � � � � � � � á ˆ ‰ ˆ ‰ ˆ ‰ 5 5 5 3 4 9 8 7 # # " " " 10 5 3 23 1 1 � � " # # ˆ ‰ ˆ ‰ � œ � œ 3 12. (5 1) , is the difference of two geometric series; the sum is � � � � � � � � á ˆ ‰ ˆ ‰ ˆ ‰ 5 5 5 3 4 9 8 7 # # " " " 10 5 3 17 1 1 � � " # # ˆ ‰ ˆ ‰ � œ � œ 3 13. (1 1) , is the sum of two geometric series; the sum is � � � � � � � � á ˆ ‰ ˆ ‰ ˆ ‰ 1 1 1 5 4 25 8 1 5 # # " " " 2 1 5 17 1 1 6 6 � � " ˆ ‰ ˆ ‰ � œ � œ 5 14. 2 2 1 ; the sum of this geometric series is 2 � � � � á œ � � � � á œ 4 8 16 2 4 8 10 5 25 125 5 25 125 3 1 ˆ ‰ Š ‹ " � ˆ ‰ 2 5 15. Series is geometric with r 1 Converges to œ Ê � Ê œ 2 2 1 5 5 5 3 1 ¹ ¹ � 2 5 16. Series is geometric with r 3 3 1 Diverges œ � Ê � � Ê ¹ ¹ 17. Series is geometric with r 1 Converges to œ Ê � Ê œ 1 1 1 8 8 7 1 ¹ ¹ 1 8 1 8 � 18. Series is geometric with r 1 Converges to œ � Ê � � Ê œ � 2 2 2 3 3 5 1 ¹ ¹ � � � 2 3 2 3 ˆ ‰ 19. 0.23 20. 0.234 œ œ œ œ œ œ ! ! ˆ ‰ ˆ ‰ _ _ œ œ n 0 n 0 23 23 234 234 100 10 99 1000 10 999 n n 1 1 " " � � Š ‹ Š ‹ ˆ ‰ Š ‹ 23 234 100 1000 1000 100 21. 0.7 22. 0.d œ œ œ œ œ œ ! ! ˆ ‰ ˆ ‰ _ _ œ œ n 0 n 0 7 7 d d 10 10 9 10 10 9 n n 1 1 " " � � Š ‹ Š ‹ Š ‹ Š ‹ 7 d 10 10 10 10 23. 0.06 œ œ œ œ ! ˆ ‰ ˆ ‰ ˆ ‰ _ œ n 0 1 6 6 10 10 10 90 15 n 1 " " � Š ‹ Š ‹ 6 100 10 24. 1.414 1 1 1 œ � œ � œ � œ ! ˆ ‰ _ œ n 0 414 414 413 1000 10 999 999 n 1 " " � Š ‹ Š ‹ 414 1000 1000 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.2 Infinite Series 579 25. 1.24123 œ � œ � œ � œ � œ œ 124 123 124 124 123 124 123 100 10 10 100 100 10 10 100 99,900 99,900 33,300 n 1 123,999 41,333 ! ˆ ‰ _ œ n 0 " � � Š ‹ Š ‹ 123 10 10 26. 3.142857 3 3 3 œ � œ � œ � œ œ ! ˆ ‰ _ œ n 0 142,857 142,857 3,142,854 116,402 10 10 10 1 999,999 37,037 n 1 " � � Š ‹ Š ‹ 142,857 10 10 27. 1 0 diverges lim lim n n n 1 n 10 1 Ä_ Ä_ � œ œ Á Ê 28. 1 0 diverges lim lim lim lim n n n n n n 1 n 2 n 3 n 5n 6 2n 5 2 n n 2n 1 2 Ä_ Ä_ Ä_ Ä_ � � � � � � � � a b a ba b œ œ œ œ Á Ê 2 2 29. 0 test inconclusive lim n 1 n 4 Ä_ � œ Ê 30. 0 test inconclusive lim lim n n n 1 n 3 2n Ä_ Ä_ 2 � œ œ Ê 31. cos cos 0 1 0 diverges lim n 1 n Ä_ œ œ Á Ê 32. 1 0 diverges lim lim lim lim n n n n e e e 1 e n e 1 e 1 Ä_ Ä_ Ä_ Ä_ � � n n n n n n œ œ œ œ Á Ê 33. ln 0 diverges lim n 1 n Ä_ œ �_ Á Ê 34. cos n does not exist diverges lim nÄ_ 1 œ Ê 35. s 1 1 lim s k k 2 2 3 3 4 k 1 k k k 1 k 1 œ � � � � � � á � � � � œ � Ê ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ " " " " " " " " " " � � � k Ä _ lim 1 1, series converges to 1 œ � œ k Ä _ ˆ ‰ " � k 1 36. s 3 lim s k k 3 3 3 3 3 3 3 3 3 3 3 1 4 4 9 9 16 k k k 1 k 1 k 1 œ � � � � � � á � � � � œ � Ê ˆ ‰ ˆ ‰ ˆ ‰ Š ‹ Š ‹ a b a b a b � � � 2 2 2 2 2 k Ä _ lim 3 3, series converges to 3 œ � œ k Ä _ Š ‹ 3 ak 1 b � 2 37. s ln 2 ln 1 ln 3 ln 2 ln 4 ln 3 ln k ln k 1 ln k 1 ln k k œ � � � � � � á � � � � � � Š ‹ Š ‹ Š ‹ Š ‹ Š ‹ È È È È È È È È È È ln k 1 ln 1 ln k 1 lim s lim ln k 1 ; series diverges œ � � œ � Ê œ � œ _ È È È È k k Ä _ 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 k œ � � � � � � á � � � � � � a b a b a b a b a b a b a b tan k 1 tan 0 tan k 1 lim sk lim tan k 1 does not exist; series diverges œ � � œ � Ê œ � œ a b a b a b k k Ä _ Ä _ 39. s cos cos cos cos cos cos k 1 1 1 1 1 1 1 1 1 1 1 1 2 3 3 4 4 5 œ � � � � � � á ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ � � � � � � cos cos cos cos cos � � � � œ � ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ � � � � � � � � # � # 1 1 1 1 1 1 1 1 1 1 k k 1 k 1 k 3 k 1 lim s lim cos , series converges to Ê œ � œ � œ k k Ä _ Ä _ k 3 k 3 2 6 6 1 1 ’ “ ˆ ‰ 1 1 1 1 1 � � # 40. s 5 4 6 5 7 6 k 3 k 2 k 4 k 3 k œ � � � � � � á � � � � � � � � Š ‹ Š ‹ Š ‹ Š ‹ Š ‹ È È È È È È È È È È k 4 2 lim s lim k 4 2 ; series diverges œ � � Ê œ � � œ _ È ’È “ k k Ä _ Ä _ k Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 580 Chapter 10 Infinite Sequences and Series 41. s 1 4 (4n 3)(4n 1) 4n 3 4n 1 5 5 9 9 13 4k 7 4k 3 k � � � � � � " " " " " " " " " œ � Ê œ � � � � � � á � � ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ 1 lim s lim 1 1 � � œ � Ê œ � œ ˆ ‰ ˆ ‰ " " " " � � � � 4k 3 4k 1 4k 1 4k 1 k k k Ä _ Ä _ 42. A(2n 1) B(2n 1) 6 (2A 2B)n (A B) 6 6 A B (2n 1)(2n 1) 2n 1 2n 1 (2n 1)(2n 1) A(2n 1) B(2n 1) � � � � � � � � � œ � œ Ê � � � œ Ê � � � œ 2A 6 A 3 and B 3. Hence, 3 2A 2B 0 A B 0 A B 6 A B 6 Ê Ê Ê œ Ê œ œ � œ � � œ � œ � œ � œ œ œ ! !ˆ ‰ k k n 1 n 1 œ œ 6 (2n 1)(2n 1) n 1 n 1 � � # � # � " " 3 3 1 the sum is œ � � � � � � á � � � œ � Ê Š ‹ ˆ ‰ " " " " " " " " " " # � � � # � # � 1 3 3 5 5 7 (k 1) 1 2k 1 k 1 k 1 lim 3 1 3 k Ä _ ˆ ‰ � œ " # � k 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) � � � � � � � � � � � � � � � � � œ � � � œ A(2n 1)(2n 1) B(2n 1) C(2n 1)(2n 1) D(2n 1) 40n Ê � � � � � � � � � œ # # # # A 8n 4n 2n 1 B 4n 4n 1 C 8n 4n 2n 1 D 4n 4n 1 40n Ê � � � � � � � � � � œ � � œ a b a b a b a b $ # # $ # # (8A 8C)n (4A 4B 4C 4D)n ( 2A 4B 2C 4D)n ( A B C D) 40n Ê � � � � � � � � � � � � � � � œ $ # 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 0 Ê Ê � œ � œ � � � œ � � � œ � � � � œ � � � � � � œ Ú Ú Ý Ý Ý Ý Û Û Ý Ý Ý Ý Ü Ü B C 2D 20 2B 2D 20 A B C D 0 4B 20 B 5 B D 0 � � œ � œ � � � � œ Ê Ê œ Ê œ � œ œ and D 5 C 0 and A 0. Hence, A C 0 A 5 C 5 0 œ � Ê Ê œ œ � œ � � � � œ œ ! ’ “ k n 1 œ 40n ( n 1) (2n 1) # � � 5 5 œ � œ � � � � � á � � � !’ “ Š ‹ k n 1 œ " " " " " " " " " " # � # � # # � � # � # � ( 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 � Ê � œ Š ‹ Š ‹ " " � � (2k 1) (2k 1) n Ä _ 44. s 1 2n 1 n (n 1) n (n 1) 4 4 9 9 16 (k 1) k k (k 1) k � " " " " " " " " " " " � � � � œ � Ê œ � � � � � � á � � � � ˆ ‰ ˆ ‰ ˆ ‰ ’ “ ’ “ lim s lim 1 1 Ê œ � œ k k Ä _ Ä _ k (k 1) ’ “ " � 45. s 1 1 k 2 2 4 3 3 k 1 k k k 1 k 1 œ � � � � � � á � � � � œ � Š ‹ Š ‹ Š ‹ Š ‹ Š ‹ " " " " " " " " " " � � � È È È È È È È È È È lim s lim 1 1 Ê œ � œ k k Ä _ Ä _ k k 1 Š ‹ " � È 46. s k œ � � � � � � á � � � � œ � ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ " " " " " " " " " " " " # # # # # # # # # # # # 1 k 1 1 k 1 k 1 k 1 1 k 1 Ê lim s œ � œ � k Ä _ k 1 " " " # # 47. sk ln 3 ln ln 4 ln 3 ln 5 ln 4 ln (k 1) ln k ln (k 2) ln (k 1) œ � � � � � � á � � � � ˆ ‰ ˆ ‰ ˆ ‰ Š ‹ Š ‹ " " " " " " " " " " # � � � lim s œ � � Ê œ � " " " # �