1k 1 1 8 1 8 tk 2k 1 4k 2k 4k 1k 2 4k 2 k 2k 2 54 2

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Unformatted text preview: m 1 n 2 2 n→ 1 2 0 n 15 04 n 1 4 15 1 4 15 ,S 41 S0 n n 1 n 1nn n 1 4 ... 9 4 1 23. nn 1 n 1 25. 2 n 0 3 4 1 n 15 4 54 n lim Sn n 1 2 1 2 1 lim 1 n 1 3 1 3 ... 9 4 45 ,S 16 2 21 16 2.95, . . . n 17 1 3 94 34 1 2 3 1 4 . .. 17 ,... 6 1 2 1 4 1 5 n 17 31 17 2 33 1 . . ., S n 12 1 1 n 1 1 1 0.9 n 3 4 nn 2 3 n 1 21 1 2 n 0 Geometric series with r <1 0.9 < 1 Converges by Theorem 9.6 1 n 1 n 3 1 4 1 1 3 21 (b) 1 16 94 14 1 1 2 17 03 27. 2 Sn 0 n 6 1 1 4 5 4 n 17 ,S 31 1 4 Converges by Theorem 9.6 n 1 1 Matches (f). The series is geometric: 3 1 1 n→ Geometric series with r 29. (a) 9 4 1 4 17 03 S0 1 n→ 1 1 0 21. 3.05, . . . 15 4 14 1 9 1 4 9 ,S 41 n 1 16 Matches graph (a). Analytically, the series is geometric: 15 04 n2 n2 Matches graph (c). Analytically, the series is geometric: 1 4 45 ,S 16 2 n 91 04 4 n 19. 1 Diverges by Theorem 9.9 S0 Diverges by Theorem 9.9 2 1n lim 0 Diverges by Theorem 9.9 Diverges by Theorem 9.6 445 n2 13. 1 n lim n→ Series and Convergence 1 2 1 5 1 n 1 2 1 3 1 1 1 3 n 1 4 1 7 ... 1 2 11 3 1 6 n 3 3.667 n 5 10 20 50 100 Sn 2.7976 3.1643 3.3936 3.5513 (c) 5 3.6078 (d) The terms of the series decrease in magnitude slowly. Thus, the sequence of partial sums approaches the sum slowly. 0 11 0 34 9 446 Chapter 9 31. (a) n 2 0.9 n Infinite Series 1 2 0.9 1 (b) n 2 n 1 0 (c) 20 0.9 n 5 10 20 50 100 Sn 8.1902 13.0264 17.5685 19.8969 22 19.9995 0 11 0 (d) The terms of the series decrease in magnitude slowly. Thus, the sequence of partial sums approaches the sum slowly. n 10 0.25 33. (a) n 1 1 (b) 10 0.25 1 40 3 13.3333 (c) n 5 10 20 50 100 Sn 13.3203 13.3333 13.3333 13.3333 15 13.3333 11 0 0 (d) The terms of the series decrease in magnitude rapidly. Thus, the sequence of partial sums approaches the sum rapidly. 1 35. n 2 2n 1 n 12 n1 2 1 2 1 3 1 n 1 0 1 2 39. n 3 45. n 8 1n 0 2 1 n 1 3 1 2n 1 2 1 3 1 4 1 8 n 1 n n 1 1 4 1 5 n 8 2 3 13 1 2 1 2 41. n 1 1 1 1 2 n 1 n 1 6 ... 3 4 1 12 n 1 2 1 2 1 1 2 37. 12 n1 0 1 3 1 3 n 1 4 1 1 9 4 1 4 ... 2 3 12 47. n 1 5 0 n 1 3n 1 2n n 2 0.8415 < 1. The series sin 1 n sin 1 n n sin 1 1 sin 1 n 51. (a) 0.4 n 41 0 10 10 a 1 r 1 1 0 is geometric with r sin 1 sin 1 4 10 1 10 and r 4 9 1 1 12 3 2 0 1 1 3 1 1 10 n 1 13 1 2 5.3080. n 1 10 n n n 4 sin 1 < 1. Thus, 53. (a) 0.81 4 10 0 1 2 0 1 2 1 n (b) Geometric series with a S n n 1 10 43. 1 49. Note that sin 1 8 81 1 0 100 100 n (b) Geometric series with a S a 1 r 1 81 100 1 100 81 100 1 100 and r 81 99 9 11 10 9 S ection 9.2 31 40 100 0 55. (a) 0.075 n n 57. n 3 40 (b) Geometric series with a 3 40 99 100 a S 1 r n 1 1 n n n 3n 2n 1 61. lim n→ 3n 2n 1 2 1 3 1 2 1 4 1 3 n 3 2 1 10 0 Diverges by Theorem 9.9 63. 1 1 n 10 10n 1 lim 5 66 447 n 10 10n 1 n→ 1 1 59. 1 1 100 and r 1 Series and Convergence 1 5 1 4 4 2n 0 1 6 4 n 0 1 2 ... 3 , converges 2 n 65. 1.075 n 1 2 Geometric series with r 0 1 2 1 n 0 Geometric series with r 1.075 Diverges by Theorem 9.6 Converges by Theorem 9.6 Diverges by Theorem 9.9 n ln n 67. Since n > ln n , the terms an do not approach 0 as n → n 2 69. For k 0, k n lim 1 . Hence, the series n→ 2 n→ 0 n a ar ... ar 2 . . ., a ar n 0 n an 77. diverges. n 1 x 2 n x 2n 0 x 2 x 2n x 21 0 x 2 1 x2 1 x 1 n x x x 2 x , x <2 x n 1 x n x 1 1 n 0 Geometric series: converges for x fx x2 22 x 0. n→ 1 n n an converges 1 an diverges because lim an n 81. x < 1 or x < 2 2 1 n n→ n Geometric series: converges for n lim n→ is a geometric series with ratio r. When 0 < r < 1, the series converges to a 1 r . The series diverges if r ≥ 1. 1 n lim an 0 fx n 1 ar n n 0. 73. See definition on page 606. 75. The series given by xn n 12 1 arctan n diverges. Hence, 79. k n 1 n 0 n→ 1 nk k 0. 0, lim 1 n 71. lim arctan n 1 ek Hence, n lim n→ n diverges. ln n For k k n n 1 1 1 1 <1⇒0<x<2 n 0 1 x 1 x 2 1 , x 0<x<2 448 Chapter 9 83. 1 nxn n Infinite Series x 0 n n n 0 Geometric series: converges for x <1⇒ x <1⇒ fx x n 1 n 1 0 1 x 85. x 0 n Geometric series: converges if 1<x<1 ⇒ x > 1 ⇒x < 1<x<1 , fx n 87. (a) Yes, the new series will still diverge. x (c) y1 ... 1 xn n 1 0 1 1x n 1 x x 1 , x > 1 or x < 1 1 1 3 x f x y2 x <1 , S3 1 x S5 1 x S5 S3 x2 y3 (b) 1 1 or x > 1 (b) Yes, the new series will converge. 89. (a) x is the common ratio. x2 0 1 x 1 <1 x x2 x3 x4 − 1.5 1.5 0 91. f x 3 0.5x 0.5 1 1 n S 10,000 < n2 6 −2 n n 10 1± n −1 3 12 1 < 0.0001 1 n 1 2 0 1 nn 0 < n2 Horizontal asymptote: y 3 93. y=6 7 10,000 12 6 41 2 10,000 Choosing the positive value for n we have n 99.5012. The first term that is less than 0.0001 is n 100. The horizontal asymptote is the sum of the series. f n is the nth partial sum. 1 8 n < 0.0001 10,000 < 8n This inequality is true when n at a faster rate. n 1 95. 8000 0.9 i 8000 1 i 0 0.9 n 1 0.9 1 n 1 1 97. 100 0.75 i D2 D3 16 0.81 2 16 0.81 11 22 2 32 0.81 down up n P2 0.81 16 11 22 2 32 0.81 2 n D 16 32 0.81 16 32 0.81 32 0.81 n 0 152.42 feet n ... 2 16 1 32 0.81 11 02 2 n 1 1 8 12 12 1...
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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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