1n n 3 1 1 let f x x 1 x 1 dx e x f is positive

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Unformatted text preview: ≥ 1. 1 1 n1 11. 1 , diverges 1 xk fx 2 Hence, the series diverges by Theorem 9.10. k for x > xk ck c xk 1 c2 <0 1. 1 xk 1 ck xk dx 1 ln x k k c 1 Diverges by Theorem 9.10 21. Let f x 1x ,fn x an. The function f is not positive for x ≥ 1. 1 n3 1 25. n Let f x n 1 1 dx x3 1 2x 2 1 1 2 Converges by Theorem 9.10 an. 1 1 n 1 Let f x f is positive, continuous, and decreasing for x ≥ 1. sin x ,fn x The function f is not decreasing for x ≥ 1. 27. 1 . x3 2 23. Let f x x 1 < 0 for x ≥ 1. 2x3 2 ,f x f is positive, continuous, and decreasing for x ≥ 1. 1 1 x dx 2x , diverges 1 Hence, the series diverges by Theorem 9.10. S ection 9.3 29. n 1 1 n 5 1 n1 5 1 n n Divergent p-series 1 5 with p 37. n 1 1 n1 2 1 31. 33. n Divergent p-series <1 1 2 with p 2 n3 4 n 1 1 n3 2 1 2 n n 3 2 with p 1 n Convergent p-series >1 with p 2 n 41. 2 453 1 n1.04 1 35. Convergent p-series <1 39. The Integral Test and p-Series n 1 2 nn n S1 2 S1 2 S1 3.1892 S2 2.6732 S2 2.7071 S3 4.0666 S3 3.0293 S3 3.0920 S4 4.7740 S4 3.2560 S4 2 n3 2 2 S2 1 1.04 > 1 3.3420 Matches (c), diverges Matches (d), converges Note: The partial sums for 39 and 41 are very similar because 2 3 2. 43. (a) Matches (b), converges Note: The partial sums for 39 and 41 are very similar because 2 3 2. n 5 10 20 50 Sn 3.7488 3.75 3.75 3.75 8 100 3.75 The partial sums approach the sum 3.75 very rapidly. 0 11 0 (b) n 5 10 20 50 Sn 1.4636 1.5498 1.5962 1.6251 1.635 2 The partial sums approach the sum slower than the series in part (a). 6 1.6449 an and 0 11 0 45. Let f be positive, continuous, and decreasing for x ≥ 1 and an f n . Then, n 5 100 47. Your friend is not correct. The series n f x dx 1 1 n 1 1 1 x 1 1 Since n 1 1 > n x dx 1 n ln n 2 51. n If p If p dx diverges, the series 2x 1 1 n 1 10,000 y 1.0 0.5 x 1 2 3 4 5 n ... 2 p 1, then the series diverges by the Integral Test. 1, 1 x ln x p dx Converges for 1 diverges. n 1 10,001 is the harmonic series, starting with the 10,000th term, and hence diverges. either both converge or both diverge (Theorem 9.10). See Example 1, page 618. 49. The area under the rectangles is greater than the area under the graph of y 1 x, x ≥ 1. 10,000 ln x 2 p p 1 dx x 1 < 0 or p > 1 ln x p p 1 1 . 2 454 Chapter 9 Infinite Series n 53. n 1 n2 1 If p p n 1, n x fx x2 1 1 fx n2 1 1 p, diverges (see Example 1). Let p 1 2p 1 x2 . 1 x2 p 1 For a fixed p > 0, p x x2 1 1 1, f x is eventually negative. f is positive, continuous, and eventually decreasing. 1 dx p x2 p 1 1 2 2p 1 For p > 1, this integral converges. For 0 < p < 1, it diverges. 55. n 1 n ln n 2 p 2 1 , diverges. n ln n n 61. S6 SN an n an n 1 an > 0. 1 n N 2 1 , converges. n ln n 2 1 24 1 R6 ≤ N S converges for p > 1. p Hence, 59. Since f is positive, continuous, and decreasing for x ≥ 1 and an f n , we have, RN 1 n ln n 2 57. n Hence, n converges for p > 1. 6 1 dx x4 n an ≤ aN SN n N 1 64 1.0811 0.0015 6 0.0015 1.0826 f x dx 1 N 1 ≤ 1 3x3 1 54 1 ≤ 1.0811 n4 1 1.0811 ≤ S 1 44 1 Also, RN 1 34 1 f x dx. N Thus, 0 ≤ RN ≤ f x dx. N 63. S10 1 2 1 5 1 10 1 R10 ≤ 10 x2 1 n 1 e 1 x2 dx ne n 1 ≤ 0.9818 4 e16 4 0.4049 ≤ 1 50 1 37 arctan x 1 3 e9 xe 1 26 n2 1 65 1 82 1 101 arctan 10 2 10 n2 2 e4 R4 ≤ dx 1 0.9818 ≤ 65. S4 1 17 0.0997 0.9818 0.0997 1.0815 67. 0 ≤ RN ≤ 0.4049 N 1 e 2 e x2 ≤ 0.4049 4 5.6 16 5.6 2 10 8 10 8 1 < 0.003 N3 N 3 > 333.33 N > 6.93 N≥7 1 dx x4 1 3x3 N 1 < 0.001 3N 3 S ection 9.3 69. RN ≤ 5x e 1 e 5 dx N 5N e 5x 1 71. RN ≤ < 0.001 5 N The Integral Test and p-Series x2 N 1 455 arctan x dx N 1 < 0.005 e5N arctan N < 0.001 2 e5N > 200 arctan N < 5N > ln 200 arctan N > 1.5698 N > tan 1.5698 ln 200 5 N> 1.5698 N ≥ 1004 N > 1.0597 N≥2 73. (a) n 2 1 . This is a convergent p-series with p n1.1 6 (b) n 1 dx x ln x 1 n1.1 2 6 n 2 n 2 1 is a divergent series. Use the Integral Test. n ln n 1 is positive, continuous, and decreasing for x ≥ 2. x ln x fx 2 1.1 > 1. ln ln x 2 1 21.1 1 n ln n 1 31.1 1 2 ln 2 1 41.1 1 51.1 1 3 ln 3 1 61.1 1 4 ln 4 0.4665 1 5 ln 5 0.2987 1 6 ln 6 0.2176 0.7213 0.3034 0.1703 0.1803 0.1393 0.1243 0.0930 For n ≥ 4, the terms of the convergent series seem to be larger than those of the divergent series! 1 1 < n1.1 n ln n (c) n ln n < n1.1 ln n < n0.1 This inequality holds when n ≥ 3.5 75. (a) Let f x Sn 1 x. f is positive, continuous, and decreasing on 1, n 1≤ 1 Sn 1015. Or, n > e40. Then ln e40 Sn ≥ 1 1 Thus, ln n (b) Since ln n ln n 1 1 2 ln n. Similarly, 1 dx x ln n 1 1. 1 ≤ Sn ≤ 1 Sn ln n 1 Sn 1 ln n 1 1 1 dx x 1 n 1 and the sequence is decreasing. (d) Since the sequence is bounded and monotonic, it converges to a limit, . (e) a100 S100 ln 100 3 ... n −1 x n n +1 1 ln n ≤ Sn ln n ≤ 1. Also, since ln x is an increasing function, ln n ≤ 1 and the sequence an is bounded. n Thus, an ≥ an 2 ln n. 1 ≤ Sn ≤ 1 ln n, we have ln n ln n > 0 for n ≥ 1. Thus, 0 ≤ Sn 1 55. 1 n an e4 y 1 ≤ ln n...
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