# Lim n 1 n3 lim n 2n5 3n5 n3 2n n3 nn 2 21 lim n 1n 2

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Unformatted text preview: limit of a sequence, there exists M > 0 such that an bn 1 and n3 1 n an2. 1 converge. n2 1 59. True 61. Since 459 diverges. 5n4 1 209 1 1 204 1 201 lim n→ n3 Therefore, diverges, (harmonic) 53. n3 5n4 3 Comparisons of Series 0 <1 whenever n > M. Hence, an < bn for n > M. From the Comparison Test, an converges. 63. 1 and n2 1 n2 1 n3 1 both converge, and hence so does n3 1 . n5 1 and 460 Chapter 9 67. (a) Let Infinite Series 1 an lim n→ an bn and 3, n 1 lim 1 n1 1 n2 n→ 3 lim n→ 1 By Exercise 65, n 1 n 1 3 69. Since lim an 0, the terms of n→ sufficiently large n. Since lim n→ sin an an 1 and converges, so does 1 , converges. n2 bn n2 n 1 0 3 lim n→ converges. 1 lim lim n 1 1 n n n n→ 1 bn n→ 1 n n , converges. 1 n 0 converges. 71. The series diverges. For n > 1, sin an are positive for n < 2n n1 an n <2 1 1 > n1 n 2 sin an . 1n Since n , and n 1 1. an bn n By Exercise 65, nn Section 9.5 1 n an (b) Let > 1 2n 1 diverges, so does 2n 1 nn 1 n. Alternating Series 6 n2 1 3. n 3 n! 1 5. n 10 n2n 1 S1 6 S1 3 S1 5 S2 7.5 S2 4.5 S2 6.25 S3 8.1667 S3 5.0 Matches (e). Matches (a). Matches (d). 7. n 1 1n 1 2n 1 (a) 0.7854 4 n 2 3 4 5 6 7 8 9 10 Sn (b) 1 1 0.6667 0.8667 0.7238 0.8349 0.7440 0.8209 0.7543 0.8131 0.7605 (c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases. 1.1 (d) The distance in part (c) is always less than the magnitude of the next term of the series. 0 11 0.6 1n n2 9. n 1 (a) 1 2 12 0.8225 n 2 3 4 5 6 7 8 9 10 Sn (b) 1 1 0.75 0.8611 0.7986 0.8386 0.8108 0.8312 0.8156 0.8280 0.8180 (c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases. 1.1 (d) The distance in part (c) is always less than the magnitude of the next term in the series. 0 11 0.6 S ection 9.5 1n n 11. n 1 an 1 13. n 1 < 1 n n 1 n lim n→ n 1 1 n 1. Thus, lim an n→ < 1 2n an 1 0 1 1n n 17. 1 1 Converges by Theorem 9.14 1 2 n2 1 1 2n 2n n→ 1 n n2 1 n lim n→ 1 lim 0 n2 1 an an Converges by Theorem 9.14 15. 1n 1 2n 1 1 1 Alternating Series an 0. Diverges by the nth-Term Test 1 < n1 1 1 n lim n→ 1 n an 0 Converges by Theorem 9.14 1n 1n 1 ln n 1 19. n 1 n ln n lim n→ 1 1 21. sin n 1 1n 1 lim n→ lim n 2n 1 1 2 n n 1 1 n n 1 1 Diverges by the nth-Term Test 1 n→ 1 Diverges by the nth-Term Test cos n 23. n n 1 1 n 1 n! 25. 1 n Diverges by the nth-Term Test 0 n 27. n 1 an 1 lim n→ n 1 n! 1! < 1 n! an an 1 1 n < 0 Converges by Theorem 9.14 lim n→ n n 2 n n 2 n1 1 2 n for n ≥ 2 2 0 Converges by Theorem 9.14 29. n an 1 1 3 1 n 1n! 5 . . . 2n 1 1 3 lim an n→ lim n→ lim n→ n 1! 5 . . . 2n 1 2n n! 5 . . . 2n 1 1 3 1 1 3 n! 5 . . . 2n 1 2 3. . .n 3 5 . . . 2n 1 lim 2 n→ 1 3 3 4 5 1 5... n 7 2n 3 Converges by Theorem 9.14 1 2n 1 0 1 n 2n 1 1 an n 2n 1 < an 1 461 462 Chapter 9 n n 1 en 31. 1 1 Infinite Series n 2ex Let f x 1n e2n 2 n e e2x 1 1 1 6 2en 1 n lim e2n n→ lim 1 n→ 2en lim 2e2n 2.4325 3 49 S6 ≤ a7 S 0.0612 0.0612 2.3713 ≤ S ≤ 2.4937 Thus, f x is decreasing. Therefore, an 2en 1 0.0612 ≤ S ≤ 2.4325 R6 2ex e2x 1 < 0. e2x 1 2 fx 1 2.4325 . Then 1n n2 3 33. S6 n→ 1 1 en < an , and 0. The series converges by Theorem 9.14. 5 2 35. S6 n 1 n! 0 n n 0.002778 ≤ S ≤ 0.7333 0.7333 R6 2 6! S6 ≤ a7 S 1 n! 37. 0.7333 0 n (a) By Theorem 9.15, 0.002778 RN ≤ aN 0.002778 0.7305 ≤ S ≤ 0.7361 1 1 N 1! < 0.001. This inequality is valid when N 6. (b) We may approximate the series by 6 n n 1 n! 0 1 1 2 1 1 6 1 24 1 120 1 720 0.368. (7 terms. Note that the sum begins with n 1 39. n 2n 0 41. 1! n RN ≤ aN 1 1 1 2N 1! This inequality is valid when N RN ≤ aN < 0.001. 1 0 2n n 1 6 1 1! 1 120 1 1 N 1 < 0.001. This inequality is valid when N 2. 1000. (b) We may approximate the series by (b) We may approximate the series by 2 1 1 (a) By Theorem 9.15, (a) By Theorem 9.15, n 1n n n 0.) 1000 1n n 0.842. n (3 terms. Note that the sum begins with n 1 1 1 1 2 1 3 1 4 ... 1 1000 0.693. 0.) (1000 terms) 43. n 1 1n n3 1 n 1 1 ⇒N Use 10 terms. N 1 3 1 n 3 2n 1 1 By Theorem 9.15, By Theorem 9.15, RN ≤ aN 1 45. 1 3 RN ≤ aN < 0.001 > 1000 ⇒ N 1 > 10. 1 2N 1 1 3 1 This inequality is valid when N < 0.001. 7. Use 7 terms. S ection 9.5 1 47. n 1 1 1 n 1 n n 1 2 1 . n2 1 n n 1 n n 1 lim n→ n 1 1 n2 n 1 is a divergent p-series. Therefore, the series converges conditionally. 1 2 n n n 2 n3 1 nn 1 n n 2 n3 1 1n ln n 2 The given series converges by the Alternating Series Test but does not converge absolutely since the series Therefore, the series diverges by the nth-Term Test. 55. 1 n 1 n n2 1 n 1 53. 2 n 1 diverges by comparison to the harmonic series ln n 2 1 . Therefore, the series converges conditionally. n 1 1 57. n 0 2n n 0 2n n n 1! 1 converges by a limit comparison to the convergent p-series 1 . Therefore, the given series n2 2 1! is convergent by...
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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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