ODDREV08 - R eview Exercises for Chapter 8 167 Review...

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Unformatted text preview: R eview Exercises for Chapter 8 167 Review Exercises for Chapter 8 1. an 1 n! 3. an 2 : 6, 5, 4.67, . . . n Matches (a) 4 5. an 10 0.3 n 1: 10, 3, . . . Matches (d) n3 n 2 7. an 8 5n n 2 9. lim n→ n n 2 1 0 11. lim n→ 1 Converges 0 0 12 The sequence seems to converge to 5. n→ lim an n→ lim 5n n 2 2 n n→ lim 5 5 13. lim n→ n 1 n n→ lim n 1 n n n 1 1 n n n→ lim n 1 1 n 0.05 4 0 Converges 15. lim n→ sin n n Converges n 0 17. An n 5000 1 1, 2, 3 5000 1.0125 n (a) A1 A2 A3 A4 (b) A40 19. (a) 5062.50 5125.78 5189.85 5254.73 8218.10 (b) A5 A6 A7 5320.41 5386.92 5454.25 A8 5522.43 k Sk 5 13.2 10 113.3 15 873.8 20 6448.5 3 2 25 50,500.3 120 (c) The series diverges geometric r >1 0 − 10 12 21. (a) k Sk 5 0.4597 10 0.4597 15 0.4597 20 0.4597 25 0.4597 (b) 1 (c) The series converges by the Alternating Series Test. 0 − 0.1 12 23. Converges. Geometric series, r 0.82, r < 1. 25. Diverges. nth Term Test. lim an n→ 0. 168 Chapter 8 2 3 n Infinite Series 1 2n 1 3n 1n 2 1 12 1 3 n 27. n 0 29. n 0 n 0 n 0 Geometric series with a S a 1 r 1 1 and r 1 13 2 3. 1 1 1 13 2 3 2 1 2 1 23 3 31. 0.09 0.09 0.0009 0.000009 ... 0.09 1 0.01 0.0001 ... n 0 0.09 0.01 n 1 0.09 0.01 1 11 33. D1 D2 D 8 0.7 8 0.7 8 16 0.7 2 35. See Exercise 86 in Section 8.2. A ... 8 16 0.7 16 0.7 b n P e rt er 12 1 1 8 8 16 0.7 16 0.7 n ... 200 e 0.06 2 1 e0.06 12 1 $5087.14 16 0.7 n 0 1 451 meters 3 37. 1 x 4 ln x dx b→ lim 1 9 ln x 3x3 1 9 1 9x3 39. 1 n 1 1 n2 1 n n 1 n2 1 n 1 n 1 0 By the Integral Test, the series converges. 1 n 1 Since the second series is a divergent p-series while the first series is a convergent p-series, the difference diverges. 41. lim 1 43. 2n 2n 2 n→ n 1 1 2 1 3 n3 5 . . . 2n 1 4 6 . . . 2n 3 5 . . . 2n 1 4 6 . . . 2n n→ n3 1 n3 lim n3 n3 2 2n 1 an 2 3 2 By a limit comparison test with the convergent p-series 1 , the series converges. n3 2 1 n 5 . . . 2n 1 1 1 > 4 2n 2 2n 2n 1 1 1 Since diverges (harmonic series), 2n 2 n 1 n n1 so does the original series. 45. Converges by the Alternating Series Test (Conditional convergence) n n2 1e an 1 an n→ 47. Diverges by the nth Term Test 49. n 51. n 2n 3 1n an 1 an n→ n→ lim lim lim lim 01 n1 2 en 1 2 en 2 en 2 en n 1 1n n→ lim lim lim 2n 1 n1 2n3 n1 3 n3 2n 2 n 2n n→ n→ 3 n→ 1 e2n n 1 1 n Therefore, by the Ratio Test, the series diverges. 0<1 By the Ratio Test, the series converges. Review Exercises for Chapter 8 an 1 an n n n 1 35 n3 5n 1 3 5 n 1 169 53. (a) Ratio Test: lim n→ n→ lim lim n→ 3 <1 5 Converges (b) x Sn (c) 4 5 2.8752 10 3.6366 15 3.7377 20 3.7488 25 3.7499 (d) The sum is approximately 3.75. 0 −1 12 55. (a) N 1 dx x2 1 x N 1 N N N n 5 1 n2 1 1 dx x2 1.4636 10 1.5498 20 1.5962 30 1.6122 40 1.6202 0.2000 0.1000 0.0500 0.0333 0.0250 N (b) N 1 dx x5 1 4x 4 N 1 4N4 N N n 1 5 1 n5 1 dx x5 1.0367 10 1.0369 20 1.0369 30 1.0369 40 1.0369 0.0004 0.0000 0.0000 0.0000 0.0000 N The series in part (b) converges more rapidly. The integral values represent the remainders of the partial sums. 57. f x fx fx fx P3 x e x2 f0 x2 1 1 2 1 4 1 8 x2 2! f 0 x3 3! 1 e 2 1x e 4 1 e 8 f0 1 1 f0 f0 f0 f0 2 x2 f 0x 1 x 2 1 x 2 95 180 1 x2 4 2! 12 x 8 1 x3 8 3! 13 x 48 95 180 95 3 18033! 0.75 3 3 59. sin 95 sin 95 5 18055! 0.75 4 4 95 7 18077! 0.75 5 5 95 9 18099! 0.75 6 6 0.996 61. ln 1.75 0.75 0.75 2 2 ... 0.75 15 15 0.560 170 63. f x Chapter 8 cos x, c f n 1 Infinite Series 0 1 Rn x f n 1 n z zn x 1! ≤ 1 ⇒ Rn x ≤ xn 1 n 1! (b) Rn x ≤ 4. 1 n n 1 (a) Rn x ≤ 0.5 n 1 < 0.001 n 1! 1! < 0.001 6. This inequality is true for n (c) Rn x ≤ 0.5 n 1 < 0.0001 n 1! This inequality is true for n (d) Rn x ≤ 2n 1 < 0.0001 n 1! This inequality is true for n x 10 n 5. This inequality is true for n 10. 65. n 0 Geometric series which converges only if x 10 < 1 or 1nx 2 n 12 un 1 un n 10 < x < 10. 67. n 0 69. n 0 n! x lim 2 n n→ lim n→ lim x 2 1 n 1 x 2 2 2 n 1 n n 12 1nx 2 n n→ un 1 un n→ lim n 1!x 2 n! x 2 n n 1 R1 Center: 2 Since the series converges when x 1 and when x the interval of convergence is 1 ≤ x ≤ 3. x2n n! 3, which implies that the series converges only at the center x 2. 71. y n 0 1 1 n 1 n 4n n 2 1 n 0 y y n 0 4n 1 1 n 2n x2n n! 2 1 1n 4n 1 x2n 1 1 2n n 2 x2n 1!2 1 2n 1 4n n 1 2 2n n 1! 2 2 n 0 n 1 1 x2y xy x2y n 0 2n 2 2n 1 x2n 4n 1 n 1 ! 2 1 1n 4n 2n 1! n 1 1 2n n 2 x2n 1!2 1 4n n! n 2 2 1 n 0 n x2n 1 4n n! 2 1 n 0 n 2n 4n 1 2 2n 1 n 1!2 1 2 1 4n 1 1 4n n! x 2n 2 2 2 n 1 x 2n 0 x 2n 2 1 n 0 n 2n 2 2n 4n 1 n 1 ! 14 1 1 4n n! 2 2 x 2n 2 1 n 0 n 1 n 1! 1 2 2 4n n 1 n n 2 n 0 1 n 11 4n n! 2 a 1 1 4n n! 2 73. 2 3 x 2x 33 23 1 x3 n n 0 1 2x n 1 r 75. Derivative: n 2nx n 1 n1 13 n 0 3n Review Exercises for Chapter 8 2 x 3 fx fx fx fx sin x n 0 171 77. 1 42 x 9 sin x cos x sin x 83 x 27 ... n 0 2x 3 n 1 1 2x 3 3 3 2x , 3 3 <x< 2 2 79. cos x , . . . f 2 2 n xx n! 2 x 2 3 3 4 4 n 2 2 x 2! 3 4 2 ... 2 2n 1 0 nn 12 x n! 3 4 n 81. 3x 3x eln 3 x ex ln 3 and since ex n xn , we have 0 n! 83. fx fx 1 x 1 x2 2 x3 6 ... , x4 f n 0 n n 0 x ln 3 n! x ln 3 n 1 x2 ln2 3 2! x3 ln3 3 3! x 4 ln4 3 4! . . .. fx f x 1 x 1x n! n! x 1 n! n 1n x n 0 1 n n 0 85. 1 1 x x k 1 1 1 1 1 kx x 5 1 x 5 x 5 x 5 kk 2! 15 1 1 x2 kk 1k 3! 15 2 x3 45 3! ... ... 9 5 x3 ... 15 4 5 x2 2! 1 n 14 4x2 52 2! 1 4 9x3 533! 9 n 2 14 . . . 5n 5nn! ... 6 xn 22 x 25 63 x 125 87. ln ln x n 1 1 1 n 1 n 1 x n 1 n , 1 n 0<x≤2 89. ex n xn , n! 0 1 2nn! <x< 1.6487 5 4 n 1 54 n 1 4nn e1 2 n 0 1 n 1 n 1 0.2231 172 Chapter 8 Infinite Series x2n , 2n ! 22n 32n 2n ! 91. cos x n 0 1 1 n 0 n <x< 0.7859 93. The series for Exercise 41 converges very slowly because the terms approach 0 at a slow rate. cos 2 3 n 95. (a) f x fx fx fx e2x 1 1 (c) e2x ex 1 e2x 2e2x 4e2x 8e2x 2x 2x ex x f0 f0 f0 f0 4x2 2! 2x2 1 x x x2 1 2 4 8 8x3 3! 43 x 3 x2 2 x2 2 ... ... x3 6 x2 2 ... x3 6 x2 2 x3 2 x3 6 x3 2 (b) ex n xn 0 n! 2x n n! 2x 1 2x2 2x 43 x 3 4x2 2! ... 8x3 3! ... e2x n 0 1 1 x x3 6 ... ... 43 x 3 ... 1 2x 2x2 97. sin t n 0 1 nt 2n 1 2n 1 ! 1 n t 2n 2n 1 ! 2n 2n 1 n t 2n 1 2n 1 nx 2n 1 2n 1 1 x 99. 1 1 ln 1 ln t t x t t 1 1 ntn n 0 sin t t x 0 1 1 n t dt n nt n 0 n 0 1 ntn n1 1 sin t dt t 1 n 0 n 0 1! 1! 0 1 1 nt n 1 n 12 x 0 n 0 ln t t 1 n 0 dt n 0 0 1 nxn 1 n 12 101. arctan x arctan x x x→0 x x 0 x3 3 x5 5 x5 2 3 x7 7 x9 2 5 x9 9 x13 7 2 ... x17 9 2 ... lim arctan x x 1 arctan x By L’Hôpital’s Rule, lim x→0 x x→0 lim 1 1 x2 x→0 lim 2x 1 x2 0. 2x Problem Solving for Chapter 8 1. (a) 1 1 3 2 1 9 4 1 27 ... n 12 03 3 n 1 13 23 1 12 (b) 0, , , 1, etc. 33 (c) lim Cn n→ 1 n 0 12 33 n 1 1 0 Problem Solving for Chapter 8 nn 2 1 173 3. If there are n rows, then an For one circle, . 1 3 2 r1 1 2 a1 1 and r1 1 3 3 2 3 6 1 23 For three circles, a2 r2 3 and 1 2 1 23 1 2 3r2 2r2 2r2 3 r2 r2 For six circles, a3 r3 6 and 1 1 23 4 1 2 3 r3 4r3 2r3 3 r3 r3 Continuing this pattern, rn Total Area rn2 an An lim An 1 4 1 1 1 1 R (b) 23 23 1 2n 1 2n 1 1 1 . 2 nn 2 1 nn 1 2 2 3 2n 2 n→ 2 8 2x x3 x3 2x 3x2 x6 x6 3x2 x3 ... ... 1 1 1 x3 1. a1x p ... ... 1 5. (a) an x n 2x4 2x 2x 3x5 x4 3x2 ... x7 ... 3 x2 x5 x8 ... 1 because each series in the second line has R an x n a0 a0 1 a0 a0 R 1 a1x xp a1 x a1 x ... ... ... ... ap 1x p 1 a0 x p xp ... xp 1 1 xp . ... ap 1x p ... 1 a1 x 1 ap ap 1x 1x p 1 1 xp ... 1 p 1 174 7. Chapter 8 ex xex xex dx Letting x 1 n 0 Infinite Series x x2 ex x2 2! x3 2! C n 0 1 x xex ... ... n xn 1 0 n! 2 xn n 1, 2 n! 0, C 1 n 1 1. Letting x 2 n! 1 2 1 . 2 1 n 1 n 2 n! . Thus, n 1 n 2 n! 9. Let a1 0 sin x dx, a2 x 2 sin x dx, a3 x 3 2 sin x dx, etc. x Then, sin x dx x a1 0 and an a2 1 a3 a4 . . .. 0 Since lim an n→ < an, this series converges. 11. (a) a1 a2 a3 a4 a5 a6 n→ 3.0 1.73205 2.17533 2.27493 2.29672 2.30146 1 2 13 [See part (b) for proof.] a a1 a a> a a1. lim an (b) Use mathematical induction to show the sequence is increasing. Clearly, a2 Now assume an > an an an an a > an a> 1 1 1. Then a an 1 a > an. Use mathematical induction to show that the sequence is bounded above by a. Clearly, a1 Now assume an < a. Then a > an and a aa a2 1 > an 1 a > an a2 > an a> an L⇒ 1 2 1 a a an 1. 1 a < a. 1 > 1 implies Hence, the sequence converges to some number L. To find L, assume an L L a 1± L2 . 4a . a L⇒ L2 L a 0 an L: 4a 1 2 Hence, L Problem Solving for Chapter 8 1 n n 12 1 n 175 13. (a) 1 21 1 20 1 9 8 11 8 45 32 1 n 1 1 22 1 1 8 1 4 1 32 1 16 9 8 1 1 23 1 1 24 1 1 25 1 ... S1 S1 S3 S4 S5 (b) an 1 an 2n 2n 1 11 8 45 32 47 32 2 1 1 n n 1 1 n 1 21 This sequence is 1, 2, 1, 2, . . . which diverges. 8 8 (c) n 1 2n 1 n 2n 2 n 1 2 1 2 1n 1 n 1 n → 1 < 1 converges because 2 2 1 n 1 2, 2, 1, 2, . . . and 2 n 1 2 → 1 and n 2 → 1. 15. S6 S7 S8 S9 S10 130 240 440 810 1490 70 130 240 440 810 40 70 130 240 240 440 810 1490 2740 440 ...
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This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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