# N n 1 1 n 1 n 25 1 n diverges by the nth term test 0

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Unformatted text preview: comparison to the convergent geometric series converges absolutely. n 1 2 0 n since 1 2n 1! < 1 for n > 0. 2n Therefore, the given series converges absolutely. cos n 1 0n 59. n n 1 n 0 n n 0 0 n 1 1 . 1n n1 1, therefore the series 1n converges conditionally. lim n→ cos n n 1 is a convergent p-series. n2 1 n 1 Therefore, the given series converges absolutely. diverges by a limit comparison to the divergent harmonic series, n n n 1 1 n 1 n2 cos n n2 1 61. The given series converges by the Alternating Series Test, but cos n n1 463 The given series converges by the Alternating Series Test, but does not converge absolutely since converges by comparison to the p-series Therefore, the given series converges absolutely. 51. 1 49. 2 1 n n Alternating Series 464 Chapter 9 Infinite Series 63. An alternating series is a series whose terms alternate in sign. See Theorem 9.14. 1n . n 67. False. Let an an converges and But, an 1 n 1 If p n an is absolutely convergent if an converges. is conditionally convergent if an diverges, but converges. 69. True. S100 Then 71. 65. 1 1 3 1 2 1 an converges. Since the next term 101 is negative, S100 is an overestimate of the sum. 1 np 73. Since an 1 n diverges. 0, then n 1 1 converges we have lim an 1 nn If p < 0, then n n→ 1 n 1 p n→ diverges. p < 1 np 0. Thus, there must exist an N > 0 such that aN < 1 for all n > N and it follows that an2 ≤ an for all n > N. Hence, by the Comparison Test, 1 If p > 0, then lim 1 1 100 1 diverges. n n an ... an an 0 and an2 1 np n an. 1 converges. Let an 1 n to see that the converse is false. Therefore, the series converges for p > 0. 75. n 1 converges, hence so does n2 1 n 1 . n4 1 77. (a) No, the series does not satisfy an 1 ≤ an for all n. For example, 1 9 1 < 8. (b) Yes, the series converges. S2n 1 2 1 3 1 2 ... 1 1 2 79. n 1 10 n3 2 10 n 1 1 12 1 , n3 2 convergent p-series 85. Convergent geometric series r 1 e or Integral Test 1 3n ... 1 2n ... 1 1 3n 1 3 1 2n , S2n → As n → 1 2n ... 1 1 1 3 ... 1 13 2 1 3n 3 2 1 . 2 81. Diverges by nth-Term Test lim an n→ 87. Converges (absolutely) by Alternating Series Test 83. Convergent geometric series r 7 <1 8 89. The first term of the series is zero, not one. You cannot regroup series terms arbitrarily. S ection 9.6 91. s 1 1 2 1 3 1 4 1 1 3 1 2 1 5 1 7 465 ... S The Ratio and Root Tests (i) s4n 1 2 1 s 2 2n 1 3 1 2 1 1 4 1 8 1 s 2 2n S lim S3n 1 6 1 6 ... 1 ... 4n 1 10 1 4n 4n 1 2 2 1 7 1 4 1 5 1 ... 4n 1 3 4n 1 1 2n S3n ln 2.) 1! 2! 1 s 2 2n 1 s 2 s 3 s 2 s. Section 9.6 The Ratio and Root Tests n 1nn 1n n 2! n 1nn 2! 1 3. Use the Principle of Mathematical Induction. When k 1 1 4n 1 1 ... 1 3 1 s4n n→ Thus, S 1. 1 11 1 0 since s > . 2 s n n 1 9 1 5 (In fact, s s n→ 1 4 1 6 Adding: s4n (ii) lim sn 1 4 5 . . . 2n 3 1 1, the formula is valid since 1 21 ! . Assume that 21 1! 2n ! 2n n! and show that 1 5 . . . 2n 3 1 2n 1 1 2n 1 2n 2n 1 2! . 1! n To do this, note that: 1 5 . . . 2n 3 1 5 . . . 2n 3 2n ! 2n n! 2n 1 2n ! 2n 1 2n n! 1 2n Induction hypothesis 2n 2n 2n ! 2n 1 2n 2n 1n! n 1 2n 2n 1 n 1 2 1 2 2! 1! The formula is valid for all n ≥ 1. 5. n n S1 1 3 4 n 3 ,S 42 Matches (d). 1 3 4 1.875 2 9 16 ... 3n n! 7. n S1 1 1 9 Matches (f). 9 33 2 ... 9. n S1 1 4n 5n 3 2, S2 Matches (a). n 4 2 3.31 8 7 2 ... 466 Chapter 9 Infinite Series 11. (a) Ratio Test: lim n→ (b) an 1 an lim n n→ 125 8 n2 5 8 n n 1 lim n n 5 10 15 20 9.2104 16.7598 18.8016 19.1878 5 < 1. Converges 8 5 8 25 Sn (c) 2 1 n n→ 19.2491 20 0 12 0 (d) The sum is approximately 19.26. (e) The more rapidly the terms of the series approach 0, the more rapidly the sequence of the partial sums approaches the sum of the series. n! 3n 0 13. n lim n→ 15. n an 1 an lim 3n n 1! 3n 1 lim n→ n lim n→ n! lim lim lim n1 2n 1 n 1 2n n→ 2n n→ n 0 lim n→ an 1 an n→ lim n→ 2 n 1 n! 2n 25. n 4n 0 n! lim n→ an 1 an 4n 1 n 1! lim n→ lim n→ 4 n 1 1 n! 4n 0 Therefore, by the Ratio Test, the series converges. 2 n2 2n 2 lim n→ n! n3n an 1 an n lim n n→ n→ Therefore, by the Ratio Test, the series converges. 2n2 n1 2 Therefore, by the Ratio Test, the series diverges. lim 0 2n 1 n1 n→ n→ n 2n 1 n 1! 1 3 4 3n 1 4n lim lim 23. lim an 1 an 1 2 1 n 2n n! n 2n n2 1 lim n Therefore, by the Ratio Test, the series converges. 21. 1 34 n3 4n Therefore, by the Ratio Test, the series converges. n n→ n lim n→ lim 19. an 1 an an 1 an n→ n 2n 1 n→ n 1 Therefore, by the Ratio Test, the series diverges. n 1 3 4 3 n→ 17. n 1! 1 3n 1 n3n n! n 3 Therefore, by the Ratio Test, the series diverges. S ection 9.6 The Ratio and Root Tests 467 3n 27. n n 0 n 1 an 1 an lim n→ 3n lim n n→ n→ n→ ln y 2 n n 1 2 e n lim n n n→ n 3 lim 1 n n→ n 2n 1 2 n 1 e 0 0 1n . Then, 2 ln n 1n 1n n→ 1 3n 1 n 2n lim 3 1 1 lim 1 n lim n 0 0 2 2 1 by L’Hôpital’s Rule 1 n2 n→ y 1 n→ n lim n ln ln y n n 1n , let y 2 n n To find lim 1 1 . e 1 Therefore, by the Ratio Test, the series converges. 4n 29....
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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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