Chapter9

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Unformatted text preview: n n 03 lim 1 an 1 an 4n lim 3n 1 3n 1 1 lim 4 3n 3n 1 41 3 1 1 1 lim n→ n→ n→ n→ 4n 1 3n 1 3n 4 3 Therefore, by the Ratio Test, the series diverges. 31. n 0 lim n→ 3 1 n 1n! 5 . . . 2n an 1 an n→ 1 lim 1 1 n 1! 5 . . . 2n 1 2n 3 1 3 3 5 . . . 2n n! 1 lim n→ n 2n 1 3 1 2 Therefore, by the Ratio Test, the series converges. Note: The first few terms of this series are 33. n 1 lim n→ 1 1 n3 2 1 1 3 1 2! 35 3! 1 35. n an 1 an lim n→ lim n→ 1 n1 n3 2 1 32 1 lim n→ 3 5 7 . . .. 1 n4 an 1 an 1 lim n n→ 1 4 n4 1 32 n n 1 1 Ratio Test is inconclusive. n n 37. n 1 lim n 2n n 39. 1 an n lim n→ lim n→ 2n 1 1 1 2 n 2n 1 1 n n n n 2 2n n Therefore, by the Root Test, the series converges. n an lim n lim lim n→ 2n n n→ n→ 2n n 1 1 1 1 Since 2 > 1, the series diverges. n 2 lim n→ 4 n n 1 1 468 Chapter 9 41. n 2 Infinite Series 1n ln n n n lim an n→ lim 1n ln n n n n→ 1 ln n lim n→ 0 Therefore, by the Root Test, the series converges. 2n n 43. n n 1 1 n lim an n→ n lim 2 n→ n To find lim n n, let y n→ n n n 1 ln n n 1 and lim 2 n n 1 n→ n→ n 4n 1 45. n n lim lim 1n 1 21 1 n→ 0. 3. Therefore, by the Root Test, the series diverges. 1 n 47. n lim an n lim n→ 1 n . Then ln n n n→ e0 0, so y lim 2 n→ lim lim n→ Thus, ln y n lim n→ lim ln n n ln y n 1 n→ n n lim lim an n→ n 1 n lim n 4n n1 n 4 n→ 1 1 n2 1 n 1 n2 n→ 1 4 n→ 1 Since 4 < 1, the series converges. 1 n2 n 0 0 0<1 Hence, the series converges. Note: You can use L’Hôpital’s Rule to show lim n1 n 1. n→ n1 Let y ⇒ ln y n ln n n lim n→ lim n→ 1n 1 Hence, ln y → 0 ⇒ y 49. n 2 n ln n n 0. n1 n → 1. n lim an n n1 n ln n n→ n→ n ln n 3 n 1 nn 3 n This is convergent p-series. 1 5 5 1 0 1 n3 2 1 n n 1 an n n lim n→ < 1 5 n 5 n an 0 Therefore, by the Alternating Series Test, the series converges (conditional convergence). Since 0 < 1, the series converges by the Root Test. 53. 1 51. n lim lim n→ ln n n 2n 55. n 1n lim n→ 1 2n n 1 2 0 This diverges by the nth-Term Test for Divergence. S ection 9.6 1 n 3n 2n 57. n 1 2 n 3 2 Since r 1 n 3n 3 2n 1 2 1 19 n 3 2 n 59. n > 1, this is a divergent geometric series. The Ratio and Root Tests 469 10n 3 n2n 1 10n lim n→ 3 n2n 1 2n lim n→ 10n 3 n 10 Therefore, the series converges by a Limit Comparison Test with the geometric series n 61. n cos n n 12 63. n cos n 1 ≤n 2n 2 n 1 0 n7n 1 n! an 1 an lim n→ Therefore, the series 1n . 2 lim n n→ 1 7n n 1! 1 n! n7n 1 n 3n n! 65. n 1n . 2 0 1 1 an 1 an lim n→ 3n lim n n→ 1! n! 3n 1 lim n→ 3 n 1 0 Therefore, by the Ratio Test, the series converges. (Absolutely) 67. n 1 lim n→ 5 3n . . . 2n 7 an 1 an n→ 3 lim 3 1 3n 7 . . . 2n 5 1 3 1 2n 3 5 7 . . . 2n 3n 1 lim n→ 3 2n 3 0 Therefore, by the Ratio Test, the series converges. 69. (a) and (c) n n5n 1 n! 71. (a) and (b) are the same. n n 5 n 0 n n 4n 1 n 1 5n 1! 25 2! 73. Replace n with n n1 n1 04 7 n Therefore, by the Ratio Test, the series converges. cos n 2n converges by comparison with the geometric series n lim n→ 2 1. 1 35 3! 3 45 4! 4 ... 75. Since 310 210 10! 9 k 1 2k 1.59 3k k! 10 5, use 9 terms. 0.7769 0 470 Chapter 9 an 1 an 77. lim n→ Infinite Series 4n lim 1 3n an n→ lim n→ 4n 3n 2 an 79. lim n→ an 1 an n→ 1 lim lim 1 n lim 1 n→ 83. lim n→ an 1 an 85. n 3 1 ln n n lim n→ lim lim n→ 87. lim n→ lim 1 ln n n n→ lim n n→ 1 ln n an 1 an lim n→ lim 0 Therefore, by the Root Test, the series converges. n→ lim n 1 n n lim n n→ 1 x 1 x 1 <1⇒ x 3 x 3 x <1⇒ 3 1! lim n→ n! x 2 1 lim n n→ 1<x ⇒ 2x 3n 1 2x 3n n For the series to converge, x 1<1 1 x 2 n 1 n x 2 The series converges only at x 0. 2 < x < 0. n For x n For x 1 1 2, n Answer: 1 , converges. n 0, 1 n 1 n n n 1 1 <1 2 3<x<3 a 91. lim n 1 n→ an 1 xn n n→ 1 1 3, the series diverges. Answer: 1 n 2n 3, the series diverges. For x x 1 1 2. . .n n 1 3 . . . 2n 1 2n 1 2. . .n 1 3 . . . 2n 1 For the series to converge: For x a 89. lim n 1 n→ an 0<1 The series converges by the Ratio Test. n an 1 n n→ 1 0, so the series diverges. n→ n an The series converges by the Ratio Test. The Ratio Test is inconclusive. But, lim an 1 an sin n n→ 1 n an an n→ sin n lim n→ 4 >1 3 1 2 The series diverges by the Ratio Test. 81. lim an 1 an 1 , diverges. n 1 2<x≤0 93. See Theorem 9.17, page 597. 95. No. Let an n The series n 97. The series converges absolutely. See Theorem 9.17. 1n 1 . 10,000 1 diverges. 10,000 3 < x < 3. S ection 9.6 99. First, let n→ an lim r<1 n an n→ an diverges. Rn n 1 0 converges, we can apply the Comparison Test to conclude that an n 1 an converges. converges which in turn implies that n 1 101. Ratio Test: lim n→ an 1 an lim n ln n p 1 ln n 1 n n→ 1, inconclusive. p Root Test: n lim an n→ lim n1 n lim 1 n ln n n n→ p ln ln n . lim ln y n→ n ln y lim n→ p ln n 1 n Thus, lim n→ 103. For n lim p n→ 1. Furthermore, let y n→ n1 n ln n lim n→ n1 n ⇒ pn pn n→ 1. 1, inconclusive. pn k an ≤ an ≤ an ⇒ 1, 2, 3, . . . , Taking limits as k → pn p ln l...
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This note was uploaded on 01/11/2013 for the course MATH 111 taught by Professor Man during the Spring '13 term at University of Washington-Tacoma Campus.

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