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137-148. Example CAS Commands:
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]
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]
lim1
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)n1
first25 Table[N[a[n]],{n,1, 25}]
712 Chapter 10 Infinite Sequences and Series
The limit command does not work in this case, but the limit can be observed as 1.25.
Clear[minN, lim]
lim1.25
Do[{diff Abs[a[n] lim], If[diff .01,{minN n, Abort[]}]},{n, 2,1000}]
minN
10.2 INFINITE SERIES
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Thomas calculus early transcendentals 13th edition thomas solutions manual
- 1. 1 2 34 2 2 2 2 9 10 2 3 4 5 6 7 8 9 10 4 7 2 2 2 24 Thomas Calculus Early Transcendentals 13th Edition Thomas Solutions Manual Full clear download (no formatting errors) at: http://testbanklive.com/download/thomas-calculus-early-transcendentals-13th- edition-thomas-solutions-manual/ Thomas Calculus Early Transcendentals 13th Edition Thomas Test Bank Full clear download (no formatting errors) at: http://testbanklive.com/download/thomas-calculus-early-transcendentals-13th- edition-thomas-test-bank/ CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 SEQUENCES 11 12 1 13 2 14 31. a1 1 0, a2 2 4 , a3 3 9 , a 4 16 2. a 1 1,1! a 1 1 ,2! 2 a 1 1 ,3! 6 a 1 1 4! 24 (1)2 (1)3 1 (1)4 1 (1)5 13. a1 21 1, a2 41 3 , a3 61 5 , a4 81 7 4. a1 2 (1)1 1, a2 2 (1)2 3, a3 2 (1)3 1, a4 2 (1)4 3 2 1 22 1 23 1 24 15. a1 2 2 , a2 3 2 , a3 4 2 , a4 5 2 6. 1 21 1 22 1 3 2 23 1 7 3 24 1 15 4a 2 2 , a 22 4 , a 23 8 , a 24 16 7. a1 1, a2 1 1 3 , a3 3 1 7 , a4 7 1 15 , a5 15 1 31, a6 63 , a7 127 , a8 255 , a 511 ,256 2 2 a 1023 512 2 22 4 4 23 8 8 24 16 32 64 128 8. a1 1, a 1 ,2 1 a 2 1 ,3 6 1 a 6 1 ,4 24 1 a 24 1 ,5 120 a 1 ,720 a 1 ,5040 a 1 ,40,320 a 1 ,362,880 a 1 3,628,800 (1)2 (2) (1)3 (1) 1 4 1 5 1 1 1 (1) 2 1 (1) 4 19. a1 2, a2 2 1, a3 2 2 , a4 2 , a5 2 8 , a6 16 , a ,32 1 1 9 1 10a8 64 , a 128 , 1(2) a 256 2(1) 2 2 1 3 3 1 4 2 2 1 a 2 , 10. a1 2, a2 2 1, a3 3 3 , a4 4 2 , a5 5 5 , a6 3 , 7 7 1 9 2 a10 1a8 4 , a 9 , 5
- 2. 3 4 1 9 2 11. a1 1, a2 1, a3 1 1 2, a4 2 1 3, a5 3 2 5, a6 8, a7 13, a8 21, a9 34, a10 55 12. a1 2, a2 1, a 1 ,2 1 a 2 1 ,1 2 1 a5 2 1, 2 a6 2, a7 2, a8 1, a 1 ,2 a10 1 13. an (1)n1 , n 1, 2, 14. an (1)n , n 1, 2, 15. an (1)n1 n2 , n 1, 2, 16. a (1) , n 1, 2, n1n n2 701
- 3. n n n n n , n 4 n n n 2 2 2 702 Chapter 10 Infinite Sequences and Series 17. a 2n1 ,3(n2) n 1, 2, 18. a 2n5 ,n(n1) n 1, 2, 19. an n2 1, n 1, 2, 20. an n 4, n 1, 2, 21. an 4n 3, n 1, 2, 22. an 4n 2, n 1, 2, 23. a 3n2 ,n! n 1, 2, 24. 3 a , 5n1 n 1, 2, 25. an 1(1) n1 2 , n 1, 2, 26. an n 1 (1)n 1 2 2 2 n 2 n 1, 2, 27. lim 2 (0.1)n 2 converges (Theorem 5, #4) n n n n 28. lim ( 1) lim 1 ( 1) 1 converges n n n n 29. lim 12n lim12n 1 2n lim1 2 2 1 converges2n n n n 30. lim 2n1 2 n 1 lim n diverges n 13 n n 1 3 31. lim n 15n4 lim n4 8n3 n 1 5 n 18 5 converges n3 n3 132. lim n n2 5n6 lim (n3)(n2) lim n2 0 converges 33. lim n n2 2n1 n1 lim n (n1)(n1) n1 lim (n 1) diverges n 34. lim 1n 3 lim 1 n n diverges n 704n n 70 4 n 35. lim 1 (1) n does not exist diverges 36. lim (1)n 1 1 does not exist diverges n 37. n1 1 1 1 1 1 n n lim n 2n 1 n lim n 2 2n 1 n 2 converges 38. lim 2 1 3 1 6 converges 39. lim (1)n1 0 converges n 2n 2n n 2n1
- 4. 1 2 n n n n 40.lim 1 n 2 lim (1)n 2n 0 converges Section 10.1 Sequences 703 n n 41. lim 2n n1 lim 2n n1 lim 2 1 2 converges n n n n 42. lim 1 lim 10 n diverges n (0.9)n n 9 43. 1 1lim sin n 2 n sin lim n 2 n sin 1 converges 44. lim n n cos (n) lim (n)(1)n n does not exist diverges 45. lim sin n 0 because 1 sin n 1 converges by the Sandwich Theorem for sequences n n n n n 46. lim sin2 n n 0 because 0 sin2 n n 1 converges by the Sandwich Theorem for sequences n 2 2 2 47. lim n lim 1 0 converges (using l'Hoˆpital's rule) n 2n n 2n ln 2 3n 3n ln 3 3n (ln 3)2 3n (ln 3)3 48. lim n 3 lim n ln (n1) 3n2 lim n 1 lim6n n 6 2 diverges (using l'Hoˆpital's rule) 49. lim lim n1 lim 2 n lim n 0 converges n n n 1 2 n n n1 n 11 50. lim n ln n limln 2n n 1 n 1 converges2 2n 51. lim 81 n 1 converges (Theorem 5, #3) n 52. 53. lim (0.03)1 n 1 converges (Theorem 5, #3) n lim 1 7 n e7 converges (Theorem 5, #5) n n 54. lim 1 1 n lim 1 (1) e1 converges (Theorem 5, #5) n n n n 55. lim n n 10n lim 101 n n1 n 11 1 converges (Theorem 5, #3 and #2) n
- 5. 1 n n n! 9 n n n 704 Chapter 10 Infinite Sequences and Series 56. lim n n n2 lim n n 2 12 1 converges (Theorem 5, #2) n 57. lim lim 31 n 1 n 1 converges (Theorem 5, #3 and #2) 3 n n n 1 lim n 1 n 58. lim (n 4)1 (n4) lim x1 x 1 converges; (let x n 4, then use Theorem 5, #2) n x ln n lim ln n n 59. lim n n 1 n lim n 1 n 1 diverges (Theorem 5, #2) n 60. lim ln n ln (n 1) lim ln 1 ln lim 1 ln 1 0 converges n n n n n 61. lim n n 4n n lim 4 n n 41 4 converges (Theorem 5, #2) n 62. lim n 32n1 lim 32(1 n) lim 32 31 n 91 9 converges (Theorem 5, #3) n n n 63. lim n! lim 123 (n1)(n) lim 1 0 and n! 0 lim n! 0 converges n nn n nnn nn n n nn n nn 64. (4) lim n n! 0 converges (Theorem 5, #6) n! 165. lim 6n lim diverges (Theorem 5, #6) n n 10 n 106 n! 66. lim n! lim 1 diverges (Theorem 5, #6) n 2n 3n n 6n 67. 1 1 (ln n) 1 1 ln 1ln n 1lim n n lim exp n ln n ln n lim exp n ln n e converges 68. lim ln 1 1 n ln lim 1 1 n ln e 1 converges (Theorem 5, #5) n n n 69. 3n1n 3n1 ln (3n1)ln (3n1) 3 3 3n1 3n1 lim 3n1 lim exp n ln 3n1 lim exp 1 lim exp 1n n n n n 2 lim exp n 6n 2 (3n1)(3n1) exp 6 e2 3 converges n
- 6. x n2n1 n n n 2 n 1 n 2 3 2 n 5 n Section 10.1 Sequences 705 n ln nln (n1) 1 1 70. n n n n1 lim n n1 lim exp n nln n1 lim exp n lim exp1 n 1 n 2 lim exp n n2 n(n1) e1 converges 71. n 1/n lim 2n1 lim x 1 1/n 2n1 x lim exp 1 ln 1 x lim exp ln(2n1) n x lim exp 2 n n n n n 2n 1 xe0 x, x 0 converges n ln 1 1 2 1 1 72. lim 1 1 lim exp n ln 1 1 lim exp lim exp lim exp 2n n2 n3 n2 n n 2 n n 2 n 1 n 1 n n2 1 n n 2 e0 1 converges 73. 3 6 36lim 2 n n! lim n! 0 converges (Theorem 5, #6) n n 74. lim 10 n 11 lim 12 n 10 n 11 11 lim 120 n 121 0 converges (Theorem 5, #4) n 9 n 11 n n 12 n 9 n 12 n 11 n n 108 n 10 12 11 10 11 12 110 1 en en e2n 1 2e2n 75. lim tanh n lim n n lim 2n lim 2n lim 1 1 converges n n e e n e 1 n 2e n ln n ln n 1 76. lim sinh (ln n) lim e e 2 lim n n 2 diverges n n n n2 sin 1 sin 1 cos1 1 cos1 n n n n n 177. lim 2n1 lim lim 2 1 2 2 lim 2 2 2 converges n n n n n n2 n2 n 3 n 1cos 1 sin 1 1 78. lim n 1 cos 1 lim n lim n 2 lim sin 0 converges n n n 1 n 1 n n n 79. lim sin 1 lim n lim n 2n lim cos 1 cos 0 1 sin 1 cos 1 1 n n n n 1 n 1 n n converges n 2n3 2 80. n n 1 n n n 1 n ln3n 5n 3n ln 35n ln 5 3n 5n lim 3 5 lim exp ln 3 5 lim exp n lim exp 1n n n n n n 3 ln3ln5 n 3 ln 3ln5 lim exp 5 lim exp 5 exp(ln5) 5 n 3 n 1 5 n 3 n 1
- 7. n 1 706 Chapter10 Infinite Sequences and Series 81. lim tan1 n converges 82. lim 1 tan1 n 0 0 converges n 2 n n 2 n n n 83. lim 1 1 lim 1 1 0 converges (Theorem 5, #4) n 3 2n n 3 2 n 2 lnn2 n 2n1 084. lim n n lim exp n lim exp n2 n e 1 converges n n n 85. lim n (ln n)200 n lim n 200 (ln n)199 n lim n 200199 (ln n)198 n lim n 200! n 0 converges 5ln n4 5 n 4 3 86. lim ln n lim lim 10ln n lim 80ln n lim 3840 0 converges n n n 1 2 n n n n n n n 87. lim n n2 n lim n n2 n n n2 n lim n lim 1 1 converges n n n n2 n n n n2 n n 1 11 2 n 1 1 n2 1n2 n n2 1 n2 n 1 1 11 n 2 n 88. lim 2 2 lim 2 2 2 2 lim 1n lim 1 1 2 n n 1 n n n n 1 n n n 1 n n n n n converges 89. lim 1 1 dx lim ln n lim 1 0 converges (Theorem 5, #1) n n 1 x n n n n 90. n lim dx lim n lim 1 1 1 1 1 if p 1 converges 1 1 1 1 1 1 1 n xp n 1p x p n p np p 91. Since an converges lim a L lim a 1 lim 72 L 72 L(1 L) 72 n n n n n 1an 1L L2 L 72 0 L 9 or L 8; since an 0 for n 1 L 8 an 6 L692. Since an converges lim an L lim an1 lim a 2 L L2 L(L 2) L 6 n n n n L2 L 6 0 L 3 or L 2; since an 0 for n 2 L 2 93. Since an converges lim an L lim an1 lim 8 2an L 8 2L L2 2L 8 0 n n n L 2 or L 4; since an 0 for n 3 L 4 94. Since an converges lim an L lim an1 lim 8 2an L 8 2L L2 2L 8 0 n n n
- 8. 2 2 y n n f (0) 2 n L 2 or L 4; since an 0 for n 2 L 4 Section 10.1 Sequences 707 95. Since an converges lim an L lim an1 lim 5an L 5L L2 5L 0 L 0 or L 5; n since an 0 for n 1 L 5 n n 96. Since an converges lim an L lim an1 lim 12 an L 12 L L2 25L 144 0 n n n L 9 or L 16; since 12 an an 12 for n 1 L 9 n n1 an L 97. an1 2 1 , n 1, a1 2. Since an converges lim a n L lim a n lim 2 1 n L 2 1 L2 2L 1 0 L 1 2; since an 0 for n 1 L 1 2 98. an1 1 an , n 1, a1 1. Since an converges lim an L lim an1 lim 1 an L 1 L 2 15 n 15 n n L L 1 0 L 2 ; since an 0 for n 1 L 2 99. 1, 1, 2, 4, 8, 16, 32, … 1, 20 , 21 , 22 , 23 , 24 , 25 , x1 1 and xn 2n2 for n 2 100. (a) 12 2(1)2 1, 32 2(2)2 1; let f (a, b) a 2b 2 2a b 2 a2 4ab 4b2 2a2 4ab 2b2 2b2 a2 ; a2 2b2 1 f (a, b) 2b2 a2 1; a2 2b2 1 f (a, b) 2b2 a2 1 (b) r2 2 a2b 2 2 a2 4ab4b2 2a2 4ab2b2 a 2b 1 r 2 1 n ab ab 2 ab 2 2 n yn In the first and second fractions, y n. Let a b represent the (n 1) th fraction where a 1 and b n 1b for n a positive integer 3. Now the nth fraction is a2b a b and a b 2b 2n 2 n yn n. Thus, lim n rn 2. 101. (a) f (x) x2 2; the sequence converges to 1.414213562 2 (b) f (x) tan (x) 1; the sequence converges to 0.7853981635 4 (c) f (x) ex ; the sequence 1, 0, 1, 2, 3, 4, 5, … diverges 102. (a) lim n f 1 lim f x lim f 0xf (0) f (0), where x 1 n n x0 x x0 x n (b) lim n tan 1 1 1 2 1, f (x) tan1 x n 1 0 (c) lim n e1/n 1 f (0) e0 1, n f (x) ex 1 (d) lim n ln 1 2 f (0) 2 2, f (x) ln (1 2x) n n 12(0)
- 9. n n n! n e 4015.76852702 14.71517765 50 19.48325423 18.39397206 60 23.19189561 22.07276647 2 e e c1 c 708 Chapter 10 Infinite Sequences and Series a2 4n2 4n1 2 1 2 a2 2 1103. (a) If a 2n 1, then b 2 2 2n 2n 2 2n 2n, c 2n2 2n 2 2n2 2n 1 and a 2 b 2 2n 1 2 2n 2 2n 2 4n 2 4n 1 4n 4 8n 3 4n 2 4n4 8n3 8n2 4n 1 2n2 2n 1 2 c2 . (b) lim a2 2 lim a2 2n2 2n 1 or 2n2 2n1 lim a2 2 lim sin lim sin 1 a2 a 2 a a 2 a 2 104. (a) n 1 (2n) ln 2n 2 n 1 e0 n n lim 2 lim exp 2n lim exp 2 lim exp 2n 1; n! e 2n, n n n n (b) Stirling’s approximation n n! n 2n 1 (2n) n for large values of n 105. (a) lim ln n c lim 1 n lim 1 0 n n n cn n cn (b) For all 0, there exists an N e(ln ) c such that n e(ln ) c ln n ln ln nc ln 1 nc 1 1 1 0 lim c 1 0 nc nc n nc 106. Let {an} and {bn} be sequences both converging to L. Define {cn} by c2n bn and c2n1 an , where n 1, 2, 3, . For all 0 there exists N1 such that when n N1 then an L and there exists N2 such that when n N2 then bn L . If n 1 2max{N1, N2}, then cn L , so {cn} converges to L. 107. lim n1 n lim exp 1 ln n lim exp 1 e0 1 n n n n n 108. lim x1 n lim exp 1 ln x e0 1, because x remains fixed while n gets large n n n 109. Assume the hypotheses of the theorem and let be a positive number. For all there exists an N1 such that when n N1 then an L an L L an , and there exists an N2 such that when n N2 then cn L cn L cn L . If n max{N1, N2}, then L an bn cn L bn L lim bn L. n 110. Let 0. We have f continuous at L there exists so that x L f (x) f (L) . Also, an L there exists N so that for n N, f (an ) f (L). an L . Thus for n N, f (an ) f (L)
- 10. 5 n1 n n n 3 n Section 10.1Sequences 709 111. 3(n1)1 3n1 3n4 3n1 2 2 an1 an (n1)1 n1 n2 n1 3n 3n 4n 4 3n 6n n 2 4 2; the steps are reversible so the sequence is nondecreasing; 3n1 3 3n 1 3n 3 1 3;n1 the steps are reversible so the sequence is bounded above by 3 112. 2(n1)3! (2n3)! (2n5)! (2n3)! (2n5)! (n2)! an1 an (n1)1! (n1)! (n2)! (n1)! (2n3)! (n1)! (2n 5)(2n 4) n 2; the steps are reversible so the sequence is nondecreasing; the sequence is not bounded since (2n3)! (n1)! (2n 3)(2n 2) (n 2) can become as large as we please 2n1 3n1 2n 3n 2n1 3n1 (n1)! 113. an1 an (n1)! n! 2 n 3 n n! 2 3 n 1 which is true for n 5; the steps are reversible so the sequence is decreasing after a5, but it is not nondecreasing for all its terms; a1 6, a2 18, a3 36, a4 54, a 324 64.8 the sequence is bounded from above by 64.85 114. a a 2 2 1 2 2 1 2 2 1 1 2 1 ; the steps are reversible so then1 2n1 n 2n n n1 2n1 2n n(n1) 2n1 sequence is nondecreasing; 2 2 1 2 the sequence is bounded from aboven 2n 115. a 1 1 n converges because 1 0 by Example 1; also it is a nondecreasing sequence bounded above by 1n 116. a n 1 n diverges because n and 1 0 by Example 1, so the sequence is unboundedn 117. a 2n 1 1 1 and 0 1 1 ; since 1 0 (by Example 1) 1 0, the sequence converges; also it isn 2n 2n 2n n n 2n a nondecreasing sequence bounded above by 1 118. an 2n 1 3n 2 n 1 ; the sequence converges to 0 by Theorem 5, #4 3n 119. an (1)n 1n1 diverges because an 0 for n odd, while for n even an 21 1 converges to 2; itn n diverges by definition of divergence 120. xn max {cos 1, cos 2, cos 3, , cos n} and xn1 max {cos 1, cos 2, cos 3,, cos (n 1)} xn so the sequence is nondecreasing and bounded above by 1 the sequence converges. with xn 1 121. 12n 12(n1) 2 2 12nan an1 n1 n 1 2n 2n n 2n 2n n 1 n and 2; n thus the sequence is nonincreasing and bounded below by 2 it converges 122. n1 (n1)1 2 2 n1an an1 n n1 n 2n 1 n 2n 1 0 and n 1; thus the sequence is nonincreasing
- 11. n n m n and boundedbelow by 1 it converges 710 Chapter 10 Infinite Sequences and Series 123. 4n1 3n 4 3 n so 3 n 3 n1 3 n 3 n1 3 3 n 4n 4 an an1 4 4 4 4 4 4 1 4 and 4 4 4; thus the sequence is nonincreasing and bounded below by 4 it converges 124. a1 1, a2 2 3, a3 2(2 3) 3 22 22 13, a4 222 22 13 3 23 23 13, a5 2 23 23 13 3 24 24 13, , a 2n1 2n1 1 3 2n1 3 2n1 3n 2n1 (1 3) 3 2n 3; an an1 2n 3 2n1 3 2n 2n1 1 2 so the sequence is nonincreasing but not bounded below and therefore diverges 125. For a given , choose N to be any integer greater than 1/ . Then for n N , sinn 0 sinn 1 1 . n n n N 126. For a given , choose N to be any integer greater than 1/ e. Then for n N , 1 1 1 1 1 . n2 n2 N2 M M127. Let 0 M 1 and let N be an integer greater than 1M . Then n N n 1M n nM M n M nM n M (n 1) n M. n1 128. Since M1 is a least upper bound and M2 is an upper bound, M1 M2. Since M2 is a least upper bound and M1 is an upper bound, M2 M1. We conclude that M1 M2 so the least upper bound is unique. 129. The sequence a 1 (1) is the sequence 1 , 3 , 1 , 3 , . This sequence is bounded above by 3 , but itn 2 2 2 2 2 2 clearly does not converge, by definition of convergence. 130. Let L be the limit of the convergent sequence {a }. Then by definition of convergence, for 2 there corresponds an N such that for all m and n, m N a L 2 and n N a L . Now2 am an am L L an am L L an 2 2 whenever m N and n N. 131. Given an 0, by definition of convergence there corresponds an N such that for all n N, L1 an and L2 an . Now L2 L1 L2 an an L1 L2 an an L1 2 . L2 L1 2 says that the difference between two fixed values is smaller than any positive number 2 . The only nonnegative number smaller than every positive number is 0, so L1 L2 0 or L1 L2. 132. 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 ak(n) and ai(n), where ak(n) L1, ai(n) L2 and L1 L2. Thus ak(n) ai(n) L1 L2 0. So there does not exist N such that for all m, n N am an . So by Exercise 128, the sequence {an} is not convergent and hence diverges.
- 12. n n nn xn Section 10.1 Sequences 711 133. a2k L given an 0 there corresponds an N1 such that 2k N1 a2k L . Similarly, a2k1 L 2k 1 N2 a2k1 L . Let N max{N1, N2}. Then n N an L whether n is even or odd, and hence an L. 134. Assume an 0. This implies that given an 0 there corresponds an N such that n N an 0 an an an 0 an 0. On the other hand, assume an 0. This implies that given an 0 there corresponds an N such that for n N, an 0. an 0 an an an 0 x2 a 2x2 x2 a x2 a xn a 135. (a) f (x) x2 a f (x) 2x xn1 xn xn1 2xn 2xn 2xn 2 (b) x1 2, x2 1.75, x3 1.732142857, x4 1.73205081, x5 1.732050808; we are finding the positive number where x2 3 0; that is, where x2 3, x 0, or where x 3. 136. x1 1, x2 1 cos(1) 1.540302306, x3 1.540302306 cos(1 cos(1)) 1.570791601, x4 1.570791601 cos(1.570791601) 1.570796327 2 to 9 decimal places. After a few steps, the arc
- 13. 3 1 n 1 11 2 33 n n 2 xn1 and line segment cos xn1 are nearly the same as the quarter circle. 137-148. Example CAS Commands: 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] 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] lim1 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)n1 first25 Table[N[a[n]],{n,1, 25}] 712 Chapter 10 Infinite Sequences and Series The limit command does not work in this case, but the limit can be observed as 1.25. Clear[minN, lim] lim1.25 Do[{diff Abs[a[n] lim], If[diff .01,{minN n, Abort[]}]},{n, 2,1000}] minN 10.2 INFINITE SERIES n a1rn 21 1 3 21. sn (1r) 11 lim n sn 3 1 3 a1 rn 9 1 1 9 100 100 100 12. sn (1r) 1 1 100 lim n sn 1 100 3. sn a 1rn (1r) 11 n 2 11 lim n sn 1 2 2 4. s 1(2) , a geometric series where1(2) r 1 divergence 5. 1 1 1 s 1 1 1 1 1 1 1 1 s 1 (n1)(n2) n1 n2 n 2 3 3 4 n1 n2 2 n2 lim n n 6. 5 5 5 s 5 5 5 5 5 5 5 5 5 5 s
- 14. 4 4 1 4 4 4 4 n(n1) nn1 n 5 2 2 3 3 4 n1 n n n1 5 n1 lim n 5 n 7. 1 1 1 1 1 41 4 16 64 , the sum of this geometric series is 11 11 5 8. 1 1 1 16 1 16 64 256 , the sum of this geometric series is 11 12 9. 7 7 7 7 7 4 16 64 , the sum of this geometric series is 11 3 10. 5 5 5 55 4 16 64 , the sum of this geometric series is 11 4 11. 5 1 5 1 5 1(5 1) 2 3 4 9 8 27 , is the sum of two geometric series; the sum is 5 1 3 23
- 15. 2 3 2 3 25 23 3 7 6 6 n n n 11 11 10 2 2 12. 5 1 5 1 5 1 Section 10.2 Infinite Series 713 (5 1) 2 3 4 9 8 27 , is the difference of two geometric series; the sum is 5 1 3 17 11 11 10 2 2 13. 1 1 1 1 1 1(1 1) 2 5 4 25 8 125 , is the sum of two geometric series; the sum is 1 1 5 17 11 11 2 6 6 14. 4 8 16 2 4 8 1 102 5 25 125 2 1 5 25 125 ; the sum of this geometric series is 2 12 3 5 15. Series is geometric with r 2 2 1 Converges to 1 5 5 5 12 3 5 16. Series is geometric with r 3 3 1 Diverges 1 17. Series is geometric with r 1 1 1 Converges to 8 1 8 8 11 7 8 2 18. Series is geometric with r 2 2 1 Converges to 3 2 19. 0.23 3 3 n 20. 12 5 n 234 23 100 1 102 100 23 1 1 99 0.234 234 1 1000 234 n0 100 n0 1000 103 1 1 1000 999 21. 7 1 n 10 7 d 1 n d d0.7 n0 10 10 1 1 9 10 22. 0.d 10 10 n0 10 1 1 9 10 23. 0.06 1 n0 6 1 6 100 10 6 1 10 10 10 1 1 90 15 24. n 414 1.414 1 414 1 1 1000 1 414 1413 n0 1000 103 1 1 1000 999 999 25. 1.24123 124 123 1 123 124 105 124 123 124 123 123,999 41,333 100 105 103 100 1 1 100 105 102 100 99,900 99,900 33,300 n0 103 26. 3.142857 3 142,857 1 142,857 3 10 3 142,857 3,142,854 116,402 n0 106 106 1 1 10 106 1 999,999 37,037
- 16. n n n (k1) 23 3 4 4 5 k 3 714 Chapter 10 Infinite Sequences and Series 27. lim n lim 1 1 0 diverges n n10 n 1 n(n1) n2 n 2n1 228. lim n (n2)(n3) lim n2 5n6 lim 2n5 lim 2 1 0 diverges 29. lim 1 0 test inconclusive n n4 30. lim n n lim n2 3 n 1 0 test inconclusive2n 31. lim cos 1 cos0 1 0 diverges n n n n n 32. lim e lim e lim e lim 1 1 0 diverges n en n n en 1 n en n 1 33. lim ln 1 0 diverges n n 34. lim cos n does not exist diverges n 35. s 1 1 1 1 1 1 1 1 1 1 s 1 k 1 2 2 3 3 4 k1 k k k1 1 k1 lim k k lim 1 k k1 1, series converges to 1 3 3 3 3 3 3 3 3 3 3 336. sk 1 4 4 9 9 16 2 2 2 k k (k1) 2 3 (k1)2 3lim k sk lim 3 2 3, series converges to 3 k (k1) 37. sk ln 2 ln 1 ln 3 ln 2 ln 4 ln 3 ln k ln k 1 ln k 1 ln k ln k 1 ln 1 ln k 1 lim k sk lim ln k k 1 ; series diverges 38. sk tan1 tan 0 tan 2 tan1 tan3 tan 2 tan k tank 1 tan k 1 tan k tank 1 tan0 tank 1 lim k sk lim tank 1 does not exist; series diverges k 39. sk cos 1 1 cos 1 1 cos 1 1 cos 1 1 cos 1 1 cos 1 1 cos 1 1 cos 1 1 k1 cos 1 1 k1 cos 1 1 k2 cos3 1 1 k 2 1 1 lim k sk lim cos k k2 3 2 6 , series converges to 6
- 17. k k Section 10.2Infinite Series 715 40. sk 5 4 6 5 7 6 k 3 k 2 k 4 k 3 k 4 2 lim sk lim k 4 2 ; series diverges k k 41. 4 1 1 1 1 1 1 1 1 1 1 1 1 (4n3)(4n1) 4n3 4n1 sk 1 5 5 9 9 13 4k7 4k3 4k3 4k1 1 4k1 1lim k sk lim 1 4k1 1 42. 6 A B A(2n1)B(2n1) (2n1)(2n1) 2n1 2n1 (2n1)(2n1) A(2n 1) B(2n 1) 6 (2A 2B)n (A B) 6 2A 2B 0 A B 0 k k 2A 6 A 3 and B 3. Hence, 6 3 1 1 A B 6 A B 6 (2n1)(2n1) n1 n1 2n1 2n1 1 1 1 1 1 1 1 1 1 1 13 1 3 3 5 5 7 2(k1)1 2k1 2k1 3 1 2k1 the sum is lim 3 1 2k1 3 40n A B C D A(2n1)(2n1)2 B(2n1)2 C(2n1)(2n1)2 D(2n1)2 43. (2n1) 2 (2n1) 2 (2n1) (2n1) 2 (2n1) (2n1) 2 (2n1)2 (2n1)2 A(2n 1)(2n 1)2 B(2n 1)2 C(2n 1)(2n 1)2 D(2n 1)2 40n A(8n3 4n2 2n 1) B(4n2 4n 1) C(8n3 4n2 2n 1) D(4n2 4n 1) 40n (8A 8C)n3 (4A 4B 4C 4D)n2 (2A 4B 2C 4D)n (A B C D) 40n 8A 8C 0 4A 4B 4C 4D 0 A C 0 A B C D 0 B D 0 2A 4B 2C 4D 40 A 2B C 2D 20 4B 20 B 5 2B 2D 20 A B C D 0 A B C D 0 and D 5 A C 0 C 0 and A 0. Hence, A 5 C 5 0 k k 40n 5 1 1 5 1 1 1 1 1 1 1 1 n1 (2n1)2 (2n1)2 n1 (2n1)2 (2n1)2 1 9 9 25 25 2(k1)1 2 (2k1)2 (2k1)2 1 the sum is lim 5 1 1 55 1 (2k1)2 (2k1)2 n 2n1 1 1 1 1 1 1 1 1 1 1 1 44. 2 2 2 2 k 2 2 2 2 n (n1) n (n1) s 1 1 4 4 9 9 16 (k1) k k (k1) lim sk lim 1 (k1)2 1 k k 45. sk 1 1 1 1 1 1 1 1 1 1 1 1 lim 2 2 3 3 4 sk lim 1 1 1 k1 k k k1 k1 46. k sk 2 k 1 2 1 2 k1 1 3 1 3 1 4 1 ( 1) 1 1 1 ( 1) 2 1 ( 1) 1 1 1 1 1 1 1 1 1 1 1 1 lim k 2 2 2 2 2 2 k 2 k
- 18. k s 1 11 2 1 2 2 k 2 k 2 k Thomas Calculus Early Transcendentals 13th Edition Thomas Solutions Manual Full clear download (no formatting errors) at: http://testbanklive.com/download/thomas-calculus-early-transcendentals-13th- edition-thomas-solutions-manual/ Thomas Calculus Early Transcendentals 13th Edition Thomas Test Bank Full clear download (no formatting errors) at: http://testbanklive.com/download/thomas-calculus-early-transcendentals-13th- edition-thomas-test-bank/ People also search: thomas calculus early transcendentals 13th edition pdf free thomas' calculus early transcendentals 13th edition solutions thomas calculus early transcendentals pdf thomas calculus early transcendentals 14th edition pdf thomas calculus 14th edition thomas' calculus early transcendentals 12th edition thomas calculus 13th edition ebook thomas calculus 13th edition amazon