# 0811 s 1 44 1 also rn 1 34 1 f x dx n thus 0 rn f

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Unformatted text preview: n 0.1 . 1 dx x Hence, Sn ≤ 1 (c) an 40 < e40 0.5822 (Actually 0.577216.) ≥0 456 Chapter 9 Infinite Series xln n 77. n 2 (a) x 1ln n 1: n 1 : e (b) x 1, diverges 2 n n 2 1 e 2 ln n e n 2 n (c) Let x be given, x > 0. Put x xln n n p ln n e 2 n e n ⇔ ln x p p n 2 1 , diverges 2n ln n 2 n 2 p. 1 np 1 This series converges for p > 1 ⇒ x < . e 1 79. n 1 2n 1 n 1 Let f x 1 81. 2x 1 1 4 nn n p-series with p . 1 n5 4 1 5 4 Converges by Theorem 9.11 f is positive, continuous, and decreasing for x ≥ 1. 1 2x 1 1 dx ln 2x 1 1 Diverges by Theorem 9.10 83. n 0 2 3 n 85. n Geometric series with r 2 3 1 lim n→ Converges by Theorem 9.6 n n2 n n2 1 1 lim n→ Diverges by Theorem 9.9 1 87. n 1 lim 1 n→ 1 n 1 n n n e 0 Fails nth-Term Test Diverges by Theorem 9.9 89. n 2 1 n ln n Let f x 3 1 . x ln x 3 f is positive, continuous and decreasing for x ≥ 2. 2 1 dx x ln x 3 ln x 2 3 1 dx x ln x 2 Converges by Theorem 9.10. See Exercise 51. 2 2 1 2 ln x 2 2 1 2 ln 2 2 1 1 1 n2 1 0 Section 9.4 Section 9.4 1. (a) 1 n 3 1n 6 1 6 23 2 2 6 4 3 6 an 6 an = 3/2 n 6 23 2 3 2 . . .; S 1 3 5 an = 4 3 6 1 . . .; S 1 6 6 n 457 Comparisons of Series 6 n3 2 n Comparisons of Series n n2 0.5 6 1 1.5 6 2 4.5 6 1.5 . . .; S 1 6 an = 3/2 n +3 2 4.9 6 n n 2 + 0.5 1 n 3 2 (b) The first series is a p-series. It converges p (c) The magnitude of the terms of the other two series are less than the corresponding terms at the convergent p-series. Hence, the other two series converge. Sn n Σ k=1 12 6 4 2 > 1. 8 10 6 k 3/2 n Σ 10 (d) The smaller the magnitude of the terms, the smaller the magnitude of the terms of the sequence of partial sums. k=1 6 k k 2+ 0.5 8 6 4 n Σ 2 k=1 6 k 3/2 + 3 n 4 2 1 1 <2 n2 1 n Therefore, 3. 0 < 5. 1 1 > > 0 for n ≥ 2 n1 n Therefore, 1 n 1 n2 n 1 . n2 1 n ln n 1 > > 0. n1 n1 9. For n ≥ 3, n 1 n 1 . n 1 Note: n 1 n 1 0 3n 1 converges by comparison with the convergent geometric series 1 . n 2 n 1 1 > > 0. n2 n! 0 1n . 3 1 1 ≤n e n2 e Therefore, 13. 0 < 1 n! 0 n converges by comparison with the convergent p-series diverges by comparison with the divergent series n n Therefore, ln n n1 1 1 1 1 11. For n > 3, Therefore, n n 10 1 1 <n 3n 1 3 Therefore, diverges by comparison with the divergent p-series converges by comparison with the convergent p-series n 2 8 7. 0 < 1 1 6 1 e n2 0 converges by comparison with the convergent geometric series 1 2. 1n n 0 1n . e diverges by the Integral Test. 15. lim n→ n n2 1 1n lim n→ n2 n2 1 1 Therefore, 17. lim 1 n→ 1 n2 1 diverges by a limit comparison with the divergent p-series n 1 . 1n 1 lim n→ n n2 1 1 Therefore, n n n2 1n n 0 1 n2 1 diverges by a limit comparison with the divergent p-series n 1 . n 1 458 Chapter 9 Infinite Series 2n2 1 3n5 2n 1 19. lim n→ 1 n3 lim n→ 2n5 3n5 n3 2n n3 nn 2 21. lim n→ 1n 2 3 1 2n2 n 3n 1 1 2n 1 n 1 . n3 1 n 1 n n2 1 n2 23. lim n→ 3n 2n 1 1 lim n→ n2 n n2 1 1 n nn 3 2 diverges by a limit comparison with the divergent p-series converges by a limit comparison with the convergent p-series n n→ n2 n2 Therefore, Therefore, 5 lim 1 1 . n 1 25. lim nk 1 n→ Therefore, 1 lim n→ nk k n 1 1 Therefore, 1 n2 1n nk 1n nk 1 k 1 1n n 1 n converges by a limit comparison with the convergent p-series n diverges by a limit comparison with the divergent p-series n sin 1 n 1n 27. lim n→ lim n→ 1 n2 cos 1 n 1 n2 lim cos n→ 1 . 1n n 1 2. 1n 1 1 n p-series with p 1 n 1 1 2 29. n n n Diverges 1 Therefore, 1 n sin n 1 diverges by a limit comparison with the divergent p-series n 1 . 1n n n 13 1 31. n 33. 2 n 1 2n Direct comparison with n 1 1 3 n lim n→ n 2n an lim nan. By given conditions lim nan n→ 1 n n→ is finite and nonzero. Therefore, 37. lim n→ an n 1 diverges by a limit comparison with the p-series n 1 . 1n n Diverges; nth-Term Test Converges 3 n 35. 3 1 2 1 n2 1 2 Converges; Integral Test 0 39. 1 2 2 5 3 10 4 17 5 26 n ... n 1 n2 1 , which diverges since the degree of the numerator is only one less than the degree of the denominator. S ection 9.4 1 41. n 3 1n 43. lim n 1 n→ converges since the degree of the numerator is three less than the degree of the denominator. 45. 1 400 1 200 1 600 ... n 1 1 200n n 47. 1 5 0 3 1 216 1 n 1 200 n2 converges 49. Some series diverge or converge very slowly. You cannot decide convergence or divergence of a series by comparing the first few terms. 51. See Theorem 9.13, page 626. One example is n n 1.0 n4 5n4 3 1 1n1 diverges because lim n→ n1 1n 2 1 diverges ( p-series). n 2 Terms of ∞ Σ an 0.8 n=1 0.6 Terms of ∞ 2 Σ an 0.4 n=1 0.2 n 4 8 12 16 20 For 0 < an < 1, 0 < an2 < an < 1. Hence, the lower terms are those of 55. False. Let an 1 and bn n3 1 . 0 < an ≤ bn and both n2 n 57. True bn converges, lim bn n n→ 1 0. There exists N such that bn < 1 for n > N. Thus, anbn < an for n > N and an bn converges by comparison to the convergent n 1 an . series i 1 65. Suppose lim n→ an bn 0 and bn converges. From the definition of...
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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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