# EVEN08 - CHAPTER Infinite Series Section 8.1 Section 8.2...

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Unformatted text preview: CHAPTER Infinite Series Section 8.1 Section 8.2 Section 8.3 Section 8.4 Section 8.5 Section 8.6 Section 8.7 Section 8.8 Section 8.9 8 Sequences . . . . . . . . . . . . . . . . . . . . . 369 Series and Convergence . . . . . . . . . . . . . . 373 The Integral Test and p-Series Comparisons of Series . . . . . . . . . . 378 . . . . . . . . . . . . . . 381 Alternating Series . . . . . . . . . . . . . . . . . 385 The Ratio and Root Tests . . . . . . . . . . . . . 389 Taylor Polynomials and Approximations . . . . . 393 Power Series . . . . . . . . . . . . . . . . . . . . 398 Representation of Functions by Power Series . . 403 Section 8.10 Taylor and Maclaurin Series . . . . . . . . . . . 408 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 414 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . 421 CHAPTER Infinite Series Section 8.1 8 Sequences Solutions to Even-Numbered Exercises 2. an a1 a2 a3 a4 a5 2n n 2 4 4 5 6 6 8 7 10 8 5 4 2 n 1 3 1 2 4. an a1 a2 a3 a4 a5 4 9 2 3 2 3 n 6. an a1 a2 cos cos cos cos n 2 2 0 1 3 2 0 1 0 8 27 16 81 32 243 2 n 2 1 2 3 1 2 2 5 6 n2 6 3 2 2 3 3 8 6 25 18 25 2 34 3 87 8 266 25 a3 a4 a5 cos 2 cos 5 2 8. an a1 a2 a3 a4 a5 2 5 2 3 2 1 1 n 1 10. an a1 a2 10 10 10 10 10 10 12. an a1 a2 a3 a4 a5 n 3n! 1! 3 6 9 12 15 3n 2 2 2 1 31 32 33 34 35 a3 a4 a5 2 4 1 2 14. a1 a2 a3 a4 a5 4, ak 1 2 2 2 3 2 4 2 k 1 1 2 ak 16. a1 a2 a3 a4 a5 6, ak 12 a 31 12 a 32 12 a 33 12 a 34 1 12 a 3k 12 6 3 12 12 3 12 48 3 1 768 3 2 1 1 1 1 a1 a2 a3 a4 4 6 12 30 12 48 768 196,608 369 370 Chapter 8 Infinite Series 18. Because the sequence tends to 8 as n tends to infinity, it matches (a). 22. 24. 20. This sequence increases for a few terms, then decreases a2 126 8. Matches (b). 26. 4 4 10 −1 12 −1 −3 −1 12 −1 −1 12 an 2 n 2 5 2 6 2 2! n! 4 ,n n 6 6 6 11 2 6 1, . . . , 10 an 8 0.75 n 1, n 1, 2, . . . , 10 an 25! 23! 3n2 ,n n2 1 23! 24 25 23! 24 25 1, . . . , 10 28. an a5 a6 30. an 1 2an, a1 2 40 2 80 5 80 160 32. a5 a6 600 34. n n! n n 1 n2 1n n! 1n 2 2 36. 2n 2 ! 2n ! 2n ! 2n 2n 1 2n 2n ! 2 2 1 2n 38. lim 5 n→ 5 0 5 40. lim n→ 5n n2 4 n→ lim 5 5 1 4 n2 42. lim cos n→ 2 n 1 5 1 44. 2 46. 4 −1 12 −1 −1 12 −1 The graph seems to indicate that the sequence converges to 0. Analytically, n→ The graph seems to indicate that the sequence converges to 3. Analytically, n→ lim an n→ lim 1 n3 2 x→ lim 1 x3 2 0. lim an n→ lim 3 1 2n 3 0 3. 3 48. lim 1 n→ 1 n 50. lim n→ n 1 3 n 1, converges does not exist, (alternates between 0 and 2), diverges. 1 n2 1 n 52. lim n→ 0, converges 54. lim n→ ln n n lim 1 2n n→ lim 1 2 ln n n n→ 0, converges (L’Hôpital’s Rule) n 56. lim 0.5 n→ 0, converges 58. lim n→ n n! 2! n→ lim 1 nn 1 0, converges Section 8.1 n2 2n 1 n 1 x sin . x 1 x x→ Sequences 371 60. lim n→ n2 1 2n 1 n→ lim 2n2 4n2 1 1 , converges 2 62. an n sin Let f x x→ lim x sin lim sin 1 x 1x sin y y x→ lim 1 x 2 cos 1 x 1 x2 1 n x→ lim cos 1 x cos 0 1 (L’Hôpital’s Rule) or, x→ lim sin 1 x 1x y→0 lim 1. Therefore lim n sin n→ 1. 64. lim 21 n→ n 20 1, converges 66. lim n→ cos n n2 0, converges 68. an 4n 1 70. an 1n n2 2n 1 72. an n 3n 2 1 74. an 1 n 3n 2n 2 1 76. an 1 2n 1 1 2n 1 78. an 1 n! 80. an xn 1 n 1! 2n 3x x 2 6 x 2 82. Let f x . Then f x 2. 84. an a1 a2 a3 ne n2 0.6065 0.7358 0.6694 Thus, f is increasing which implies an is increasing. an < 3, bounded Not monotonic; an ≤ 0.7358, bounded 86. an a1 a2 a3 4 9 8 27 2 3 2n 3 3n 2 3n 2 n→ 1 88. an < an 1 Monotonic; lim an , not bounded 2 Not monotonic; an ≤ 3 , bounded 90. an a1 a2 a3 a4 cos n n 0.5403 0.2081 0.3230 0.1634 Not monotonic; an ≤ 1, bounded 372 Chapter 8 3 n Infinite Series 92. (a) an 4 an 4 (b) 8 3 < 4 ⇒ bounded n 4 3 <3 n 4 n 1 an 1 ⇒ monotonic −1 12 0 Therefore, an converges. 1 2n an bounded n→ lim 4 3 n 4 94. (a) an 4 an 4 (b) 6 1 ≤ 4.5 ⇒ 2n 4 an 1 −1 12 −1 1 >4 2n ⇒ 1 2n 1 an monotonic n→ lim 4 1 2n 4 Therefore, an converges. 96. An 100 101 1.01 \$101.00 \$203.01 \$306.04 \$410.10 \$515.20 \$621.35 102. Impossible. An unbounded sequence diverges. 106. (a) an 12,000 n 1 (b) A60 (c) A240 \$8248.64 \$99,914.79 (a) A1 A2 A3 A4 A5 A6 98. The first sequence because every other point is below the x-axis. 100. Impossible. The sequence converges by Theorem 8.5. 104. Pn P1 P2 P3 P4 P5 16,000 1.045 \$16,720.00 \$17,472.40 \$18,258.66 \$19,080.30 \$19,938.91 n 410.9212n 6003.8545 0 0 10 (b) For 2004, n 14 and an \$11,757,000,000. 110. Since n→ 11,757, or 108. an a1 a2 a3 a4 a5 a6 n→ 1 1 n n 2.0000 2.2500 2.3704 2.4414 2.4883 2.5216 1 n n lim sn L > 0, there exists for each > 0, an integer N such that sn L < for every n > N. Let L > 0 and we have, sn L < L, L < sn L < L, or 0 < sn < 2L for each n > N. lim 1 e Section 8.2 Series and Convergence 373 112. If an is bounded, monotonic and nonincreasing, then a1 ≥ a2 ≥ a3 ≥ . . . ≥ an ≥ . . . . Then a1 ≤ a2 ≤ a3 ≤ . . . ≤ an ≤ . . . an converges, is a bounded, monotonic, nondecreasing sequence which converges by the first half of the theorem. Since then so does an . 114. True 1 x 2n 1 ,n xn 1 1.414214 1.414214 1.414114 1.414214 1.414214 2. In fact, this sequence is Newton’s Method applied to f x x2 116. True 118. x0 x1 x2 x3 x4 x5 1, xn 1.5 1.41667 1 1, 2, . . . x6 x7 x8 x9 1.414216 1.414214 1.414214 x10 The limit of the sequence appears to be 2. Section 8.2 2. S1 S2 S3 S4 S5 1 6 1 6 1 6 1 6 1 6 Series and Convergence 4. S1 S2 S3 S4 S5 1 1 1 1 1 n 0.1667 1 6 1 6 1 6 1 6 0.3333 3 20 3 20 3 20 1 3 1 3 1 3 1 3 1.3333 1 5 1 5 1 5 0.4833 2 15 2 15 1.5333 1 9 1 9 0.6167 5 42 1.6444 1 11 0.7357 1.7354 6. S1 S2 S3 S4 S5 1 1 1 1 1 1 2 1 2 1 2 1 2 8. n 0 4 3 Geometric series 0.5 1 6 1 6 1 6 0.6667 1 24 1 24 r 4 >1 3 0.6250 1 120 Diverges by Theorem 8.6 0.6333 10. n 0 2 r 1.03 n Geometric series 12. n n 1 2n 3 n 3 1 2 0 1.03 > 1 Diverges by Theorem 8.6 n→ lim 2n Diverges by Theorem 8.9 n n2 n n2 n! 2n 1 n! 2n 14. n 1 1 1 n→ 16. n n→ lim lim 1 1 1 n2 1 0 n→ lim Diverges by Theorem 8.9 Diverges by Theorem 8.9 374 Chapter 8 2 3 n Infinite Series 2 3 4 9 17 03 8 9 n 18. n 0 1 ... 20. n 17 1 3 0.63, S3 8 9 64 81 ... S0 1, S1 5 ,S 32 2.11, . . . . S0 17 ,S 31 5.1, . . . Matches graph (b). Analytically, the series is geometric: 2 3 n Matches (d). Analytically, the series is geometric: 17 03 1 8 1 4 3 4 8 9 n n 0 1 1 23 1 13 3 n 1 17 3 89 17 3 17 9 3 22. n 1 1nn 2 1 2 1 2 n n 1 1 2n 1 2n n→ 2 1 2 1 4 1 2 2n 1 6 1 1 1 4 2n 1 8 1 2 1 6 1 2 1 10 1 12 1 10 1 14 ... n 1 nn n→ lim Sn lim 24. n 0 2 26. n 0 0.6 n Geometric series with r Converges by Theorem 8.6 4 n 1 1 2 < 1. Geometric series with r Converges by Theorem 8.6 0.6 < 1. 28. (a) nn 1 1 1 5 1 2 4 n 1 1 n 1 6 1 n 1 3 25 12 4 1 7 1 4 1 8 1 5 1 9 1 6 1 10 ... 1 2 1 3 5 1 4 2.0833 20 1.9051 50 2.0071 100 2.0443 (b) n Sn 10 1.7607 1.5377 (c) 4 0 0 11 (d) The terms of the series decrease in magnitude slowly. Thus, the sequence of partial sums approaches the sum slowly. 3 0.85 10 16.0625 30. (a) n 3 0.85 1 n 1 1 5 20 20 (Geometric series) 50 19.9941 100 19.999998 (b) n Sn 11.1259 19.2248 (c) 24 0 0 11 Section 8.2 1 3 n 1 Series and Convergence 375 32. (a) n 1 5 5 1 13 15 4 3.75 (c) 7 (b) n Sn 5 3.7654 10 3.7499 20 3.7500 50 3.7500 100 0 3.7500 11 0 (d) The terms of the series decrease in magnitude rapidly. Thus, the sequence of partial sums approaches the sum rapidly. 4 n 1nn 34. 2 2 n 1 1 n 1 n 2 2 1 1 3 1 2 1 4 1 3 1 5 ... 21 1 2 3 36. n 1 2n n 1 1 2n 3 1 2n 1 1 1 1 2n 3 2n 1 2 1 3 1 5 1 5 1 7 n 1 7 1 9 ... 11 23 1 6 38. n 0 6 4 5 3 4 1 n 6 45 8 34 30 (Geometric) 40. n 0 2 2 3 1 2 2 1 23 4 1 12 6 5 8 3 n 42. n 0 8 1 32 44. n 0 4 46. n 1 0.7 n 0.9 n n 0 7 10 n n 0 9 10 n 2 1 1 7 10 1 1 9 10 2 10 3 10 2 34 3 48. 0.8181 n 81 1 0 100 100 n 50. 0.21515 81 100 1 5 n 3 1 0 200 100 3 200 n Geometric series with a S a 1 n 2n 1 n 2n r 1 1 1 1 1 2 0 1 81 100 1 100 and r 81 99 9 11 1 100 Geometric series with a S 1 5 a 1 r 1 5 and r 71 330 1 100 3 200 99 100 52. n n→ lim Diverges by Theorem 8.9 1 n 1 54. nn 3 1 3n 1 3 1 1 3 1 1 n 1 4 1 2 1 n 3 1 2 1 3 1 5 1 3 1 6 1 4 1 7 1 5 1 8 1 6 1 9 ... 1 11 , converges 18 56. n 3n 3 1n 3n n3 n→ n→ lim lim ln 2 3n 3n2 n→ lim ln 2 2 3n 6n n→ lim ln n 3 3n 6 (by L’Hôpital’s Rule) Diverges by Theorem 8.9 376 Chapter 8 1 n 04 Infinite Series 2n 1 100 2 k n n 58. n 60. n 62. n 1 1 lim 1 Geometric series with r Converges by Theorem 8.6 1 4 Geometric series with r Diverges by Theorem 8.6 n→ k n n ek 0 Diverges by Theorem 8.9 64. lim an n→ 5 means that the limit of the sequence an is 5. a1 a2 ... 5 means that the limit of the 66. If lim an n→ 0, then n 1 an diverges. an n 1 partial sums is 5. x 2 x2 4 x3 8 x 2 x n 68. (a) (c) y1 y2 x 2 is the common ratio. 2 2 1 x x 2 −5 5 5 (b) 1 1 1 x2 n 0 2 2 Geometric series: −5 ,x <2 a 1, r x , 2 x <1⇒ x <2 2 70. f x 2 1 1 0.8x 0.8 10 −2 12 y = 10 72. 1 < 0.0001 2n 10,000 < 2n This inequality is true when n 14. Horizontal asymptote: y 2 n 0 4 5 n 16 −2 0.01 n < 0.0001 S 1 2 45 10 10,000 < 10n This inequality is true when n at a faster rate. 5. This series converges The horizontal asymptote is the sum of the series. f n is the nth partial sum. n 1 74. V t V5 225,000 1 0.7 5 0.3 n 0.7 n 225,000 76. i 0 100 0.60 i 100 1 0.6n 1 0.6 250 1 0.6n million dollars 225,000 \$37,815.75 Sum 12 33 12 33 n n 250 million dollars 78. The ball in Exercise 77 takes the following times for each fall. s1 s2 s3 sn 16t 2 16t 2 16t 2 80. P n P2 12 03 3 16 16 0.81 16 0.81 2 s1 s2 s3 1 0 if t 0 if t 0 if t 1 0.9 0.9 2 2 4 27 13 23 1 n 1 16t 2 16 0.81 n sn 0 if t 0.9 n 1 Beginning with s2, the ball takes the same amount of time to bounce up as it takes to fall. The total elapsed time before the ball comes to rest is t 1 1 2 n 1 0.9 2 1 n 1 2 n 0 0.9 n 0.9 19 seconds. S ection 8.2 82. (a) 64 (b) n 0 Series and Convergence 377 32 64 1 2 n 16 1 8 4 2 126 in.2 64 12 128 in.2 16 in. Note: This is one-half of the area of the original square! 16 in. 84. Surface area 4 1 2 94 1 3 2 92 4 1 9 2 ... 4 ... 12t 1 86. n 0 P1 r 12 n P1 1 P 12 r 1 1 1 1 r 12 r 12 12t 88. P 75, r 75 0.05, t 12 0.05 1 25 0.05 12 1 1 12 25 (a) A 1 r 12 1 P ert er 12 1 1 92. T 12t 1 \$44,663.23 (b) A 12 P r 12t n 1 r 12 12t 75 e0.05 25 e0.05 12 \$44,732.85 P er 0 12 n P1 1 50 1 1 1 er 12 12t er 12 90. P 20, r 20 0.06, t 12 0.06 40,000 39 40,000 1.04 n ... 40,000 1.04 39 (a) A (b) A 0.06 12 12 50 1 \$75,743.82 n 0 40,000 1.04 40,000 1 1 20 e0.06 50 e0.06 12 \$76,151.45 1.0440 1.04 \$3,801,020 94. x 0.a1a2a3 . . . aka1a2a3 . . . ak 0.a1a2a3 . . . ak 1 0.a1a2a3 . . . ak n 0 1 10k 1 10k n 1 10k 2 1 10k 3 ... 0.a1a2a3 . . . ak 1 1 1 10k a rational number 96. Let Sn be the sequence of partial sums for the convergent series n n→ 1 an L. Then lim Sn L and since Rn k n 1 ak L Sn, we have n→ lim Rn n→ lim L Sn n→ lim L n→ lim Sn L L 0. 98. If an bn converged, then an bn diverges. an bn an bn would converge, which is a contradiction. Thus, 378 Chapter 8 Infinite Series n 1000 n 1 1000 100. True 102. True; n→ lim 1 0 104. 1 r 1 r2 1 r3 ... n 11 0r r n 1 1r 1r 1 r 1 since 1 <1 r This is a geometric series which converges if 1 < 1 ⇔ r > 1. r Section 8.3 2. n The Integral Test and p-Series 4. n 1 2 1 3n 5 2 3x 5 . ne n2 Let f x xe x 2. Let f x f is positive, continuous, and decreasing for x ≥ 3. Since f x 2 ln 3x 3 2x < 0 for x ≥ 3. 2e x 2 x2 f is positive, continuous, and decreasing for x ≥ 1. 2 1 3x 5 dx 5 1 3 xe dx 2x 2e x2 3 10e 32 Diverges by Theorem 8.10 Converges by Theorem 8.10 1 n 1 6. 2n 1 1 2x 1 . 8. n 1 n n2 3 x x2 3 . Let f x Let f x f is positive, continuous, and decreasing for x ≥ 1. 1 1 f x is positive, continuous, and decreasing for x ≥ 2 since fx x 1 2x 1 dx ln 2x 1 1 Diverges by Theorem 8.10 x2 3 3 x2 dx x2 < 0 for x ≥ 2. 3 ln x2 3 1 Diverges by Theorem 8.10 1 n1 3 1 1 . x1 3 10. n 1 nk e Let f x n 12. n xk . ex Let f x f is positive, continuous, and decreasing for x > k since fx xk 1 f is positive, continuous, and decreasing for x ≥ 1. 1 dx x1 3 32 x 2 3 1 k ex x <0 1 for x > k. We use integration by parts. xk e 1 x Diverges by Theorem 8.10 xk 1 dx 1 e xk e k e x 1 k 1 e x dx kk e 1 ... k! e Converges by Theorem 8.10 Section 8.3 3 n 1n 53 The Integral Test and p-Series 379 14. 16. n 1 1 n2 2>1 Convergent p-series with p 5 >1 3 Convergent p-series with p 18. n 1 1 n2 3 2 3 20. n 1 1 n >1 Divergent p-series with p 2 n 2 3 3.67 2 1 2 2 2 3 <1 Convergent p-series with p 2 n2 2 2.5 2.722 2 22 2 32 22. n 1 ... 24. n 1 2 ... S1 S2 S3 S1 S2 S3 Matches (d) Diverges—harmonic series 26. (a) n Sn 5 3.7488 10 3.75 20 3.75 50 3.75 100 3.75 0 0 Matches (c) Converges—p-series with p 2 > 1. 8 11 The partial sums approach the sum 3.75 very rapidly. (b) 5 n Sn 5 1.4636 10 1.5498 20 1.5962 50 1.6251 100 1.635 0 0 11 2 The partial sums approach the sum 6 1.6449 slower than the series in part (a). 28. x n 1 n x n 1 1 nx Converges for x > 1 by Theorem 8.11. ln n p 2n 1, then the series diverges by the Integral Test. If p ln x dx xp p x 2 p 30. n If p 1, p 1 ln x 2 ln x dx 2 xp1 p1 2 1 . (Use Integration by Parts.) Converges for 1 < 0 or p > 1. 1 is a p-series, p > 0. np 1 . n 1 32. A series of the form n 1 34. The harmonic series n The p-series converges if p > 1 and diverges if 0 < p ≤ 1. 380 Chapter 8 Infinite Series 1 25 1 dx x5 1 35 1 45 1 4x4 36. From Exercise 35, we have: 0≤S SN ≤ N 38. S4 R4 ≤ 1 1.0363 0.0010 4 f x dx f x dx N N 4 SN ≤ S ≤ SN N n 1 1.0363 ≤ n 1 1 ≤ 1.0363 n5 0.0010 1.0373 an ≤ S ≤ n an 1 N f x dx 40. S10 R10 ≤ 1 2 ln 2 3 1 3 ln 3 1 1 ln x n 1 e3 3 1 4 ln 4 1 3 3 ... 1 11 ln 11 1 1 2 3 1.9821 1 2 ln 11 2.0691 0.0870 10 x dx 2 ln x ≤ 1.9821 10 3 1.9821 ≤ n 1 1 1 ln n 1 e4 e n x 4 1 3 0.0870 42. S4 R4 ≤ 1 e 1 e2 e 4 x 0.5713 0.0183 0.0183 0.5896 44. 0 ≤ RN ≤ N 1 dx x3 2 2 x1 2 N 2 < 0.001 N dx e N 12 < 0.0005 N > 2000 N ≥ 4,000,000 0.5713 ≤ n 0 ≤ 0.5713 46. RN ≤ N e x2 dx 2e x2 N 2 < 0.001 eN 2 48. Rn ≤ N 2 x2 5 dx 2 1 arctan 5 x 5 N 2 eN eN 2 2 < 0.001 > 2000 2 52 2 arctan N 5 < 0.001118 arctan N 5 < 0.001 N > ln 2000 2 N > 2 ln 2000 N ≥ 16 15.2 1.56968 < arctan N 5 N > tan 1.56968 5 N ≥ 2004 50. (a) 10 1 dx xp 1 xp x p p 1 1 10 p 1 1 10 p 1 .p > 1 (b) f x R10 p n 1 ≤ Area under the graph of f over the interval 10, np 11 0. As n increases, the error decreases. n ln 4 2 ln 6 1 (n n2 ln 2 ln 8 1 n 2 (c) The horizontal asymptote is y 1 n2 n2 n2 ln 1 ln 7 1 52. n 2 ln 1 ln n 2 ln n 2 ln n ln 5 2 ln 7 1 ln 3 ln n 1 2 ln n ln 6 2 ln 8 ln 4 ... 2 ln 5 ln 2 ln 3 2 ln 2 ln 5 2 ln 3 ln 6 2 ln 4 ln 7 ln 9 S ection 8.4 1 n n2 1 n0.95 0.95 Comparisons of Series 381 54. n 2 1 1 x x2 1 . 56. 3 n 1 Let f x p-series with p Diverges by Theorem 8.11 f is positive, continuous, and decreasing for x ≥ 2. 1 x x2 1 dx arcsec x 2 2 2 3 Converges by Theorem 8.10 1 n2 1 n3 1 n2 1 n3 58. n 0 1.075 n 60. n 1 62. n n→ 2 ln n lim ln n n 1 n 1 Geometric series with r Diverges by Theorem 8.6 1.075 Since these are both convergent p-series, the difference is convergent. Diverges by Theorem 8.9 64. n ln n 3 2n ln x . x3 1 3 ln x < 0 for x ≥ 2. x4 ln 2 8 1 (Use Integration by Parts.) 16 Let f x f is positive, continuous, and decreasing for x ≥ 2 since f x ln x dx x3 ln x 2x 2 1 2 1 dx x3 ln 2 8 1 4x 2 2 2 2 2 Converges by Theorem 8.10. See Exercise 14. Section 8.4 2. (a) n 1 Comparisons of Series 2 2 2 2 0.5 4 1.5 ... S 1 2 2 0.5 4 2.5 2 4 an 2 n 2 n 4 n an = 2 n − 0.5 an = 4 n + 0.5 an = 2 n n 1 0.5 0.5 ... S 1 ...S 1 1 2 4 3 2 3.3 < 1. n 1 1 n 2 4 6 8 10 (b) The first series is a p-series. It diverges p (c) The magnitude of the terms of the other two series are greater than the corresponding terms of the divergent p-series. Hence, the other two series diverge. (d) The larger the magnitude of the terms, the larger the magnitude of the terms of the sequence of partial sums. Sn 20 16 12 8 4 Σ 2 n − 0.5 Σ 4 n + 0.5 Σ 2 4 6 8 2 n n 10 382 Chapter 8 1 3n2 2 1 3n2 Infinite Series 1 > n1 Therefore, 1 for n ≥ 2. n 4. < 6. Therefore, 1 n 2 1 3n 2 n 2 1 n1 converges by comparison with the convergent p-series 1 3n 3n 4 n diverges by comparison with the divergent p-series 1 . n 1 n3 2 1 1 . n2 3 4 n n 2 8. 5 < 10. 1 n3 1 < Therefore, 3n n n 04 Therefore, 5 1 n3 1 n 1 converges by comparison with the convergent geometric series 3n . 4 1 4 4 converges by comparison with the convergent p-series 1 . n3 2 1 n n 0 12. 4 3 1 n 1 > n 14. 4n 3n 1 > 4n 3n Therefore, 14 3 Therefore, 1 n 4n 1 n 1 n 3n 1 diverges by comparison with the divergent p-series 1 4n 1 1 4 diverges by comparison with the divergent geometric series 4n . 3 n2 1n 4 3n n2 4 n . n 1 16. lim n→ 2 3n 5 1 3n n→ lim 2 3n 3n 5 2 18. lim n→ 3 n→ lim 3 Therefore, 2 n n 13 Therefore, 3 5 n 3 n 2 4 converges by a limit comparison with the convergent geometric series 1n . 3 5n 20. lim n→ diverges by a limit comparison with the divergent harmonic series 1 . 3n n 1 n 3 5 n→ n2 2n 1n lim 5n2 n2 2n 3n 5 5 1 n n2 1 22. lim n→ 1 n3 Therefore, n→ lim n3 n3 n 1 Therefore, 5n n 1 3 2n 5 n 1 1 n n2 1 converges by a limit comparison with the convergent p-series 1 . n3 n2 diverges by a limit comparison with the divergent p-series 1 . n n 1 n 1 Section 8.4 n n 1 2n 1 2n 1 1 n→ Comparisons of Series 5n n2 5 2 383 24. lim n→ lim n n 1 1 26. lim n→ 5n n2 1n 4 n→ lim n 4 Therefore, n n 1 2n 1 Therefore, n 5 n2 4 n 1 n 1 converges by a limit comparison with the convergent geometric series 1 2 n 1 diverges by a limit comparison with the divergent harmonic series 1 . 1n . n n 1 28. lim n→ tan 1 n 1n n→ lim 1 n2 sec2 1 n 1 n2 n→ lim sec2 1 n 1 Therefore, tan n 1 1 n diverges by a limit comparison with the divergent p-series 1 . n 1 5 n n 1 30. n 0 5 32. n 4 3n2 1 2n 15 1 n2 1 Converges Geometric series with r 1 5 Converges Limit comparison with n 34. n 1 1 n 1 n 1 2 1 2 1 3 1 3 1 4 1 4 1 5 ... 1 2 Converges; telescoping series 3 n 1nn 36. 3 Converges; telescoping series 1 n 1 n 3 1, then k Pn Qn 1 nk j j > 1. The p-series with p L > 0, 1, then k j ≤ 1 which implies that k j converges and since n 1 38. If j < k n→ lim the series n Pn converges by the limit comparison test. Similarly, if j ≥ k Qn 1 n Pn Qn 1 diverges by the limit comparison test. 1 3 1 8 1 15 1 24 1 35 1 n 2 2n 40. ... 1 , 42. n 1 n2 n 3 1 which converges since the degree of the numerator is two less than the degree of the denominator. diverges since the degree of the numerator is only one less than the degree of the denominator. 384 Chapter 8 n ln n Infinite Series 1 1n 44. lim n→ n→ lim n→ lim n 0 Therefore, 1 diverges. ln n 48. This is not correct. The beginning terms do not affect the convergence or divergence of a series. 1 In fact, 1 1000 and 1 1 4 1 208 1 9 1 227 ... n n 2 46. See Theorem 8.13, page 585. One example is 1 1n1 diverges because lim n→ n1 1n n 2 and 1 diverges (p-series). n 1 1001 ... n 1 1000 n diverges (harmonic) n 2 1 1n 2 converges (p-series). 50. 1 200 1 210 1 220 ... n 0 200 1 10n , 52. 1 201 1 264 ... n 1 200 1 n3 , diverges 1 n 1 converges 1 4n 54. (a) 2n 1 2 n 1 4n2 1 converges since the degree of the numerator is two less than the degree of the denominator. (See Exercise 38.) (b) n Sn (c) n 3 5 1.1839 1 2n 1 2n 1 2 10 1.02087 2 20 1.2212 0.1226 0.0277 50 1.2287 100 1.2312 8 2 2 S2 S9 (d) n 10 1 8 56. True 58. False. Let an 1 n, bn 1 n, cn 1 n2. Then, an ≤ bn cn, but n 1 cn converges. 60. Since n 1 an converges, then n 1 an an n 1 an2 62. 1 converge, and hence so does n2 1 n2 2 1 . n4 converges by Exercise 59. 1 and n3 an bn n→ 64. (a) an lim bn lim 1 n 1 . Since n2 n→ (b) 1 n2 an lim an bn 1 and n n→ bn lim 1 . Since n n and 1 n n→ lim 1 n3 1 n2 0 and n→ lim 1n 1n 1 . n n→ converges, so does 1 . n3 diverges, so does Section 8.5 Alternating Series 385 Section 8.5 2. n 1 Alternating Series 6 1 6 4 6 9 ... 4. n 1 1n n2 6, S2 16 1 n 1 10 n2n 5, S2 3.75 10 2 10 8 ... S1 4.5 S1 Matches (d) 1 n 1 n 1 Matches (a) 1 e 2 0 6. (a) n n Sn 1! 1 1 0.3679 3 0.5 4 0.3333 5 0.375 6 0.3667 7 0.3681 8 0.3679 9 0.3679 10 0.3679 (b) 2 0 0 11 (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. (d) The distance in part (c) is always less than the magnitude of the next series. 1 n 1 n 1 8. (a) 2n n Sn 1! 1 1 sin 1 2 0.8333 0.8415 3 0.8417 4 0.8415 5 0.8415 6 0.8415 7 0.8415 8 0.8415 9 0.8415 10 0.8415 (b) 2 (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. 11 0 0 (d) The distance in part (c) is always less than the magnitude of the next series. 10. n 1 1 2n 2n n 1 n 1 2 1 1 12. n 1 1 ln n n 1 1 < n→ lim n an 1 ln n 1 1 ln n 1 ln n 1 2 an Diverges by the nth Term Test. n→ lim 0 Converges by Theorem 8.14 1 n 1 n 1 14. an n 1 1 2 n2 1 1 n n n 1 < n n2 1 an 16. n 1 1 n2 n2 n 1n2 5 5 1 n→ lim n2 n→ lim n2 1 0 Diverges by nth Term Test Converges by Theorem 8.14 386 Chapter 8 1 n 1 n Infinite Series 1 ln n 1 < 1 n n→ 18. an ln n n1 1 1 1 20. n 1 2n 1 sin n 2 1 n 1 1n n 1 1 ln n n ln n n 1 1 1 for n ≥ 2 1 0 Converges; (see Exercise 9) n→ lim lim 1n 1 1 Converges by Theorem 8.14 1 cos n 1n 1 n n 22. n 24. n 0 1 2n 1 n n 1 1! 1 2n 1 3! 1! < Converges; (see Exercise 9) an 1 2n 1! an n→ lim 2n 0 Converges by Theorem 8.14 1 n 1 n 3 1 26. lim n n n→ 28. n 6 2 1n n e 1e 1 n n 1 1n e2n 1 2en 1 n1 2 n1 3 n→ lim n1 Let f x fx 2ex e2x 1 2e2x 1 e2x . Then e2x < 0 for x > 0. 12 Diverges by the nth Term Test Thus, f x is decreasing for x > 0 which implies an 1 < an. n→ lim 2en e2n 1 n→ lim 2en 2e2n n→ lim 1 en 0 The series converges by Theorem 8.14. 6 30. S6 n 4 1n 1 1 1 ln n S S6 ≤ a7 2.7067 4 ln 8 1.9236; 0.7831 ≤ S ≤ 4.6303 R6 6 32. S6 n 1 1n 2n 1n 0.1875 7 27 0.05469; 0.1328 ≤ S ≤ 0.2422 R6 S S6 ≤ a7 34. n 0 1n 2 n! n 36. n 0 1n 2n ! (a) By Theorem 8.15, Rn ≤ aN 1 (a) By Theorem 8.15, 2N 1 1 N 1! 4. < 0.001. RN ≤ aN 1 1 2N 2! < 0.001. 3. This inequality is valid when N This inequality is valid when N (b) We may approximate the series by 1 348 3 (b) We may approximate the series by 4 n 0 1 n 2n n! 1 1 2 1 8 1 48 0.607. n 0 1n 2n ! 1 1 2 1 24 1 720 0.540. 0.) (5 terms. Note that the sum begins with n 0.) (4 terms. Note that the sum begins with n Section 8.5 1n 4n n 1 Alternating Series 387 38. n 1 (a) By Theorem 8.15, RN ≤ aN 1 (b) We may approximate the series by 4N 1 1 N 3 1 < 0.001. n 1 1n 4n n 1 1 4 1 32 1 192 0.224. This inequality is valid when N 1n n4 1 3. (3 terms) 1 n 1 n 1 40. n 1 42. 1 1 n 1 By Theorem 8.15, RN ≤ aN This inequality is valid when N N 5. 1 4 < 0.001. The given series converges by the Alternating Series Test, but does not converge absolutely since the series 1 n 1 n 1 diverges by the Integral Test. Therefore, the series converge conditionally. 1n nn 1 n 1 1 44. n 1 46. n 1 1 n 1 n 2n 3 n 10 2n 10 3 1 3 1n 2 nn n Therefore, the given series converges absolutely. n which is a convergent p-series. n→ lim 2 Therefore, the series diverges by the nth Term Test. 1n n1.5 1 48. n 0 1 2 en 1 2 en 50. n 1 n 0 n 1 1 is a convergent p-series. n1.5 converges by a comparison to the convergent geometric series 1n . e Therefore, the given series converge absolutely. n 0 Therefore, the given series converges absolutely. 1 n 0 n 52. n 4 54. n 1 1 n 1 arctan n 0 Therefore, the series diverges by The given series converges by the Alternating Series Test, but 1 n4 n→ lim arctan n 2 n 0 the nth Term Test. diverges by a limit comparison to the divergent p-series 1 . n n 1 Therefore, the given series converges conditionally. 388 Chapter 8 sin 2n n 1 Infinite Series 2 n 1 56. 1 n 1n n 1 58. S Sn Rn ≤ an 1 (Theorem 8.15) The given series converges by the Alternating Series Test, but sin 2n n 1 1 n 2 n 1 1 n is a divergent p-series. Therefore, the series converges conditionally. 1n n n 1 60. n 1 1 1 2 1 3 1 4 . . . (Alternating Harmonic Series) 62. n 1 1 np 64. n 1 1n n 1 converges, but n 1 1 diverges n If p n→ 0, then lim 1 np 1 and the series diverges. If p > 0, then n→ lim 1 np 0 and 1 n 1 p < 1 . np Therefore, the series converge by the Alternating Series Test. xn 1n 66. (a) n (b) When x 1, we have the convergent alternating series 1n . n 1, we have the divergent harmonic series converges absolutely (by comparison) for n 1 1 < x < 1, since xn < xn and n xn 1 < x < 1. When x 1 . n Therefore, xn 1n is a convergent geometric series for n converges conditionally for x 3 n 1n 2 1. 68. True, equivalent to Theorem 8.16 70. 5 converges by limit comparison to convergent p-series 1 . n2 3 2 72. Converges by limit comparison to convergent geometric 1 . series 2n 76. Converges (conditionally) by Alternating Series Test. 74. Diverges by nth Term Test. lim an n→ 78. Diverges by comparison to Divergent Harmonic Series: ln n 1 > for n ≥ 3. n n Section 8.6 The Ratio and Root Tests 389 Section 8.6 2. 2k 2! 2k ! The Ratio and Root Tests 2k 2k 2k 2! 1 2k 1 2k 2k 2! 1 1 1 233! 3 5 6! 4. Use the Principle of Mathematical Induction. When k 1 5 . . . 2n 2n n! 2n 5 3 2n 2n ! 1 3, the formula is valid since 1. Assume that 1 3 and show that 1 3 1 5 . . . 2n 2n 5 2n 3 1 n 1 ! 2n 1 2n 2n 2 ! 1 . To do this, note that: 1 3 1 5 . . . 2n 5 2n 3 1 2n 3 1 5 . . . 2n 3 2n 2n ! 2n 2n 1 5 1 2n 1 2n 1 2n 2n 1 3 2 2 1 1 3 n! 2n 2n n! 2n 1 2n ! 2n 2 n 1 n! 2n 1 2n 2n ! 2n 1 2n 2 2n The formula is valid for all n ≥ 3. 3 4 n 1 n 1 ! 2n 1 2n 2n 2 ! 6. n 1 1 n! 3 4 1.03 91 16 2 ... 8. n 1 1 n 14 (2n ! 2 4 2 4 24 ... S1 3 ,S 42 S1 Matches (b) Matches (c) 4 e 10. n 0 4e S1 4 n 4 ... Matches (e) n a 12. (a) Ratio Test: lim n 1 n→ an (b) n→ 1 n n2 2 1 n→ lim 1! 1 n! lim n2 n2 2n 1 2 n 1 1 0 < 1. Converges n Sn 5 7.0917 10 7.1548 15 7.1548 20 7.1548 25 7.1548 (c) 10 (d) The sum is approximately 7.15485 (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. 0 0 11 390 Chapter 8 3n 0 n! an 1 an n→ Infinite Series 3 2 n 14. n 16. n 1 n lim n→ lim lim lim 3 n 3 n n 1 1! 1 0 n! 3n n→ an 1 an n→ lim n 1 3n 2n 1 1 2n n3n n→ n→ lim 3n 1 2n 3 2 Therefore, by the Ratio Test, the series converges. n3 n 12 an an 1 n→ Therefore, by the Ratio Test, the series diverges. 1n 1n 2 nn 1 an 1 18. n 20. n 1 n→ lim lim n n 1 n3 2n 1 2n3 3 3 2n 1 n 2 1 n3 1n 0 2 ≤ n nn 2 1 an n→ lim 1 2 n→ lim n nn Therefore, by the Ratio Test, the series converges. Therefore, by Theorem 8.14, the series converges. Note: The Ratio Test is inconclusive since lim n→ an 1 an 1. The series converges conditionally. 1 n 1 n 1 22. lim 32 n n2 an 1 an n→ 24. n 1 2n ! n5 an 1 an n→ n→ lim lim n2 3 2n 2n 3n2 2n 1 1 1 n2 32 n n→ lim lim lim 2n n 2n 2! 15 2 2n n1 n5 2n ! 1 n5 5 n→ 2 n2 3 >1 2 n→ Therefore, by the Ratio Test, the series diverges. nn 1 n! an 1 an n→ Therefore, by the Ratio Test, the series diverges. n! 2 3n ! an 1 an n→ 26. n 28. n 0 n→ lim lim lim n n n n n n 1 n 1 1! n! nn n→ lim lim n 3n 3n 1!2 3! n 3 3n 3n ! n! 2 12 2 3n 1 0 n→ 1 n 1 n n! 1 n!nn 1 n n→ lim n→ lim e>1 Therefore, by the Ratio Test, the series converges. Therefore, by the Ratio Test, the series diverges. 1 n 24n 2n 1 ! an 1 an n→ 30. n 0 n→ lim lim 24n 4 2n 3 ! 2n 24n 1! n→ lim 2n 24 3 2n 2 0 Therefore, by the Ratio Test, the series converges. Section 8.6 1 n 2 4 6 . . . 2n 5 8 . . . 3n 1 an 1 an n→ The Ratio and Root Tests 391 32. n 1 2 n→ lim lim 2 4 . . . 2n 2n 2 2 5 . . . 3n 1 3n 2 2 5 . . . 3n 1 2 4 . . . 2n n→ lim 2n 3n 2 2 2 3 Therefore, by the Ratio Test, the series converges. Note: The first few terms of this series are 2 2 2 2 4 5 2 2 4 5 6 8 ... 34. (a) n 1 1 n4 an 1 an n→ n→ lim lim 1 n 1 4 n4 1 n→ lim n n 1 4 1 (b) n 1 np 1 an 1 an n n→ n→ lim lim 1 n 1 p np 1 n→ lim n n 1 p 1 36. n 1 2n n n 1 an n→ 38. n 1 3n 2n n 3n 1 an n→ n→ lim lim n 2n n 1 2 n n→ lim lim lim n 3n 2n 3 3n 1 3 2 3 n→ lim 2n n 1 n→ 3n 2n 1 27 8 Therefore, by the Root Test, the series diverges. Therefore, by the Root Test, the series diverges. 40. n 0 e lim n n n→ an n→ lim n 1 en 1 e Therefore, by the Root Test, the series converges. n n 0 42. lim 1 3n n 44. n 1 5 n 5 n 1 1 n n→ an y ln y n→ lim n→ n n 3 x x n 1 1 x 1 1 1 n n→ lim n 3 1 This is the divergent harmonic series. Let lim x n→ lim ln lim n→ 1 ln x x ln x x e0 1 . 3 n→ lim 1 x 1 0. Since ln y n→ 0, y n1 3 1, so lim Therefore, by the Root Test, the series converges. 392 Chapter 8 n Infinite Series n n 1 46. n 1 4 4 < 1,this is convergent geometric series. 48. lim 2n2 1 n→ Since n→ n 2n2 1 1n lim n2 2n2 1 1 >0 2 This series diverges by limit comparison to the divergent harmonic series 1 . n n 1 50. n 10 n3 13 10 1 n3 3n3 2 2 52. n 2n 4n2 1 1 lim 2n 4n2 1 n→ n→ lim 10 3 n→ lim ln 2 2n 8n n→ lim ln 2 22n 8 Therefore, the series converges by a limit comparison test with the p-series 1 . n3 2 Therefore, the series diverges by the nth Term Test for Divergence. n 1 54. n 1n 2 n ln n 1 56. n ln n 2 1n an n 1 n ln n 1 1 ln n 0 1 ≤ 1 n ln n an 1 ln n ≤ 32 n2 n Therefore, the series converges by comparison with the p-series 1 . n3 2 n→ lim Therefore, by the Alternating Series Test, the series converges. 1 n 3n n2n an 1 an n→ n 1 58. n 1 n→ lim lim n 3n 1 1 2n 1 n2n 3n n→ lim 3n 2n 1 3 2 Therefore, by the Ratio Test, the series diverges. 60. n 1 3 5 7 . . . 2n 1 18n 2n 1 n! an 1 an n→ n→ lim lim 3 5 7 . . . 2n 1 2n 3 18n 1 2n 1 2n 1 n! 3 18n 2n 1 n! 5 7 . . . 2n 1 n→ lim 2n 3 2n 1 18 2n 1 2n 1 2 18 1 9 Therefore, by the Ratio Test, the series converge. 62. (b) and (c) n n 0 64. (a) and (b) are the same. 3 4 n 1 n n 1 3 4 3 4 n 1 n 2 n 1n 1 2n 1n n2n 1 1 1 2 1 2 1 2 22 22 1 3 23 3 1 23 ... 1 2 3 3 4 2 4 3 4 3 ... n 1 1 2 ... Section 8.7 Taylor Polynomials and Approximations 3k 5 . . . 2k 3 k 2k k! 2k ! 2k 1 6 k k! 2k 1 ! 393 66. Replace n with n 2n n 2 2. 2n 2 0 n! 68. k 01 3 1 k 0 n 2! n k 0 0.40967 (See Exercise 3 and use 10 terms, k 1 . n 9.) 70. See Theorem 8.18. 72. One example is n 1 100 74. Assume that n→ lim an 1 an L > 1 or that lim an n→ 1 1 an 0⇒ . Then there exists N > 0 such that an an 1 an > 1 for all n > N. Therefore, an diverges > an , n > N ⇒ lim an n→ 76. The differentiation test states that if Un n 1 is an infinite series with real terms and f x is a real function such that f 1 n exists at x 0, then Un n 1 Un for all positive integers n and d 2 f dx 2 converges absolutely if f 0 Convergent Series 1 ,f x n3 1 cos x3 1 ,f x n 1 f0 0 and diverges otherwise. Below are some examples. Divergent Series 1 ,f x n x sin x cos x 1 sin , f x n Section 8.7 2. y 14 8x 12 2x Taylor Polynomials and Approximations 1 4. y e 121 3 x 1 3 x 1 1 y-axis symmetry Three relative extrema Matches (c) 4 3 Cubic Matches (b) 6. f x fx P1 x x 4 x 3 4x 43 13 f8 f8 8 8 2 1 12 8 (8, 2) −4 14 f8 2 f8x 1 x 12 −4 P1 x 1 x 12 8 3 394 Chapter 8 Infinite Series 3 8. f x fx P1 tan x sec2 x f 1 f 2x 1 f f 4 4 1 − 2 2 2 −3 4 4 4 2 x 4 P1 x 2x 10. f x fx fx P2 x P2 x x fx P2 x sec x sec x tan x sec3 x f sec x tan2 x f x f f f f 4 3 2 4 4 4 4 2 2x 4 4 2 2 −3 4 3 0 32 2 4 2 2.15 4 2x x 2 4 4 0.585 0.685 1.2913 1.2936 0.885 1.5791 1.5761 0.985 1.8088 1.7810 1.785 4.7043 4.9475 1.8270 1.1995 15.5414 1.2160 1.4142 1.4142 12. f x (a) x 2ex, f 0 fx fx fx f 4 0 2x ex 4x 6x 8x x2 6x3 3! x3 x2 12x 4 4! f0 1 1 1 1 1 1 f03 x 3! f 4 x2 x2 x2 x2 2x 2 2! x2 x2 f0 2 ex f0 f0 f 4 0 2 6 12 (b) P 4 f −3 2 P2 6 ex 12 ex 3 x 0 P3 −1 P2 x P3 x P4 x (c) x3 x2 x3 x4 2 f0 f0 f 4 2 6 12 P2 0 P3 0 P4 4 0 0 0 (d) f n Pn n 0 14. fx fx fx f f f 4 5 e x e e x x f0 f0 x x x e e x x f0 f 4 5 0 0 e f0 x f f 0x P5 x f0 2 x 2! 04 x 4! f 5 05 x 5! 1 x x2 2 x3 6 x4 24 x5 120 Section 8.7 16. fx fx fx fx f 4 Taylor Polynomials and Approximations 395 e3x 3e3x 9e3x f0 f0 f0 1 3 9 27 81 27 3 x 3! f0 81 4 x 4! 0 1 3x 92 x 2 93 x 2 20. 27 4 x 8 fx fx 0 3 3 3 27e3x f 0 81e3x f 1 3x 4 x 0 92 x 2! P4 x 18. fx fx fx fx P3 x sin cos 2 3 x x x x 02 x 2! x 2e 2xe 2e x x x f0 x 2e x 4xe x x x x 0 0 2 6 12 f0 f0 f0 3! x3 f0 x2e x x 2e x 2e x x sin cos x fx fx x x3 f 4 f0 f0 f 4 6e 12e 0 x2 x 6xe 8xe 0 x 0 6 P4 x 0x x3 22 x 2! 14 x 2 63 x 3! 12 4 x 4! 22. fx fx fx fx f 4 x x x 2x 6x 24 x 0 1x 1 1 1 x x 2 1 1 1 1 x 1 1 f0 f0 0 1 2 6 24 x4 1 4 3 f0 f0 5 x 1 f 63 x 6 24 4 x 24 x x2 4 0 P4 x 22 x 2 x3 24. f x fx fx fx P3 x tan x sec2 x 2 4 0 sec2 sec2 x tan x x tan2 x 0 2 sec4 x x f0 f0 f0 f0 13 x 3 0 1 0 2 1x 23 x 6 26. fx fx fx fx f 4 2x 2 f2 3 1 2 1 2 3 4 3 2 15 4 2 2 4x 12x 4 f2 f2 5 48x 240x 1 2 6 fx f 2 4 x x 3 x 8 P4 x 1 x 2 1 x 4 2 3 5 x 32 2 4 396 Chapter 8 x1 1 x 3 3 Infinite Series f8 2 1 12 1 144 10 27 1 x 288 1 28 8 2 28. f x fx fx fx 30. f x fx fx P2 x 5 3456 5 x 20,736 8 3 x 2 cos x cos x 2 cos x 2 f x 2 sin x 4x sin x 2x x2 f cos x f 2 2 23 f8 f8 f8 2 2 x 2 2 2 x 9 10 x 27 2 53 2 2 83 P3 x 1 x 12 1 8 32. fx fx fx f f 4 x2 x2 1 2x 1 2 −3 2 P4 f 3 Q4 P2 −2 2 3x 2 1 x2 1 3 24x 1 x 2 x2 1 4 24 5x 4 10x 2 x2 1 5 2, c 1 4, c 1 2 ln x x x x ln x P1 x P4 x 1 1 1.00 0.0000 0.0000 0.0000 1 2 x x 1 (b) n 4, c 1 0 0x 22 x 2! 03 x 3! 24 4 x 4! 1 x2 x4 (a) n P2 x (c) n Q4 x 0 0x 1 1 x 2 1 12 x 2! 1 2 22 x 2! 1 x2 P4 x 0 x 3! 1 3 3 x 4! 1 4 1 2 1 x 2 1 1 x 4 1 2 1 x 8 1 4 34. fx P1 x P4 x (a) x 1 2 1 3 x 1.50 1 3 1 4 x 1 4 1.25 0.2231 0.2500 0.2230 1.75 0.5596 0.7500 0.5303 2.00 0.6931 1.0000 (b) −1 2 P1 f 5 0.4055 0.5000 0.4010 P4 0.5833 −2 (c) As the distance increases, the accuracy decreases. 36. (a) fx P3 x (b) x fx P3 x arctan x x x3 3 0.75 0.6435 0.6094 0.50 0.4636 0.4583 0.25 0.2450 0.2448 0 0 0 0.25 0.2450 0.2448 0.50 0.4636 0.4583 0.75 0.6435 0.6094 f −1 −π 4 −π 2 (c) π 2 π 4 y P3 x 1 2 1 Section 8.7 38. f x P7 2 Taylor Polynomials and Approximations 4xe y 4 2 397 arctan x y 40. f x x2 4 P5 P1 2 y = 4xe(− x /4) 1 f (x) = arctan x −3 −2 x 1 3 −4 x 4 −2 P9 P13 P11 P3 42. f fx 1 5 x 2e x x2 x3 14 x 2 0.0328 2 44. f fx 7 8 x 2 cos x 6.7954 2 2x 2 2 x 2 46. f x ex; f 6 x 6 ex ⇒ Max on 0, 1 is e1. 0.00378 3.78 10 3 R5 x ≤ e1 1 6! 48. f x arctan x; f 4 x 24x x 2 1 1 x2 4 4 50. f n fx 1 ex ex 1.8221. 1 ⇒Max on 0, 0.4 is f 22.3672 0.4 R3 x ≤ 4! 4 0.4 22.3672. x Max on 0, 0.6 is e0.6 0.0239 Rn ≤ 1.8221 0.6 n 1! n < 0.001 By trial and error, n 5. x3 3! 52. fx gx fx cos cos x g 1 1 x2 x2 1 x2 2! x4 4! x6 6! ... 54. fx R3 x sin x x x4 sin z 4 x≤ < 0.001 4! 4! x 4 < 0.024 x2 2! 2x 4 2 x2 4! 4x8 4 x2 6! 6x12 6 ... x < 0.3936 0.3936 < x < 0.3936 ... 6 2! 2 4! 4 6! 4! 0.6 8 f 0.6 1 2! 0.6 4 6! 0.6 12 ... Since this is an alternating series, Rn ≤ an 2n 1 2n ! 0.6 4n < 0.0001. 0.4257. By trial and error, n 4. Using 4 terms f 0.6 398 56. f c Chapter 8 P2 c , f c Infinite Series P2 c , and f c P2 c 58. See Theorem 8.19, page 611. x3 3! x2 2! x5 for f x 5! x4 4! 60. P2 10 8 6 4 2 −20 y 62. (a) P5 x f P1 x 1 sin x P5 x x P3 −2 −4 10 20 This is the Maclaurin polynomial of degree 4 for gx cos x. x6 x2 x 4 (b) Q6 x for cos x 1 2 4! 6! Q6 x (c) R x Rx 1 1 x x5 x3 3! 5! x3 x2 x 2! 3! x3 x2 x 2! 3! P5 x x4 4! The first four terms are the same! 64. Let f be an odd function and Pn be the nth Maclaurin polynomial for f. Since f is odd, f is even: f x h→0 lim f x h h f x h→0 4 lim fx h h fx h→0 lim fx h h fx f x. f0 ... 0. Similarly, f is odd, f is even, etc. Therefore, f, f , f Hence, in the formula Pn x f0 f 0x f 0 x2 2! , etc. are all odd functions, which implies that f 0 . . . all the coefficients of the even power of x are zero. 66. Let Pn x Pn c a0 a0 a1 x fc c a2 x c 2 ... an x c n where ai f i c i! . For 1 ≤ k ≤ n, Pn k c an k! f k c k! k! f k c. Section 8.8 2. Centered at 0 Power Series 4. Centered at 1 n xn 2n n→ 6. n 0 2x L n 8. n 0 n→ lim un un 1 1 2 n→ lim 2x 2x n n 1 2x L lim un 1 un n→ lim 1 n 1xn 2n 1 1 2n 1 n xn 2x <1 ⇒ R 1 x 2 1 x < 1⇒R 2 2 Section 8.8 2n !x2n n! n→ Power Series 399 10. n 0 12. n 0 x k n L lim un un 1 n→ lim lim 2n 2n 2 n 2n !x2n n! 2 2n n1 0. R 0. 1 x2 !x2n 2 1! Since the series is geometric, it converges only if x k < 1 or k < x < k. n→ The series only converges at x 14. n 0 1 lim un n 1 n 1 xn lim n n 1 n 2 n 16. n 0 3x n 2n ! un 1 un n→ 1 un n→ n→ 1 n n 2x 1 xn x n 1 n→ lim lim lim 3x 2n 2n n 1 1! 3x 2 2n 2n ! 3x n 1 0 <x< . n→ lim 2x 1 n→ Interval: When x When x 1<x<1 1, the series n 0 Therefore, the interval of convergence is 1 n 1 n 1 diverges. 1, the series n 0 n 1 diverges. 1 < x < 1. Therefore, the interval of convergence is 1 n xn 1n 2 n→ 18. n 0 n n→ lim un 1 un lim 1 n n 1xn 1 2n 3 n 1n 2 1 n xn 1n 1n 2 n→ lim n n 1x 3 x Interval: When x When x 1<x<1 1, the alternating series n 0 n 2 converges. 1 . n2 1, the series n 0 n 1 1n converges by limit comparison to n 1 Therefore, the interval of convergence is 1 n n! x 3n un 1 un 0 4 4 n 1 ≤ x ≤ 1. 20. n 0 n→ lim n→ lim 1 n 1 n 3n 1!x 1 4 n 1 3n 1 n! x n 4 n n→ lim n 1x 3 4 R Center: x Therefore, the series converges only for x 4. 400 Chapter 8 x n 2n 1 4n un 1 un 4 2 4<x 1 1 Infinite Series 22. n 0 n→ lim n→ lim x n 2n 2 4n 2 2 n x 1 4n 2n 1 1 n→ lim x 2n 1 4n 2 1 x 4 2 R Center: x Interval: When x When x 2 < 4 or 2<x<6 1 n 0 n 1 2, the alternating series 6, the series n 0 n 1 converges. 1 n 1 diverges. 2 ≤ x < 6. Therefore, the interval of convergence is 1 n 1 n 1 24. lim x c n ncn un 1 un c c c<x c < c or 0 < x < 2c 1 diverges. n 1n n 1 n→ n→ lim 1 n 2 n x cn 1 cn 1 1 1 n ncn 1x c n n→ lim nx cn c 1 1 x c c R Center: x Interval: When x When x 0, the p-series n 1 2c, the alternating series n 1 converges. Therefore, the interval of convergence is 0 < x ≤ 2c. 26. n 0 1 nx2n 2n 1 un 1 un 1 1 n→ lim n→ lim 1n 2n 1x2n 3 3 2n 1 1 nx2n 1 n→ lim 2n 2n 12 x 3 x2 R Interval: When x When x 1<x<1 1, n 0 1 2n n 1 converges. 1, n 1n 1 converges. 1 0 2n 1 ≤ x ≤ 1. n!xn 2n ! un 1 un n→ Therefore, the interval of convergence is 1 n x 2n n! un 1 un n→ 28. n 0 30. n 1 n→ lim lim lim 1 n x2 n n 1x 2n 2 1! 1 0 n! 1 n x 2n n→ lim lim lim n 1 !xn 2n 2 ! n 1 2n ! n!xn 0 <x< . n→ n→ 2n 1x 2 2n 1 Therefore, the interval of convergence is <x< . Therefore, the interval of convergence is Section 8.8 2 n 13 Power Series 401 32. 4 5 6 . . . 2n x 2n 7 . . . 2n 1 n→ 1 n→ lim un 1 un 1 lim 2 3 4 . . . 2n 2n 2 x 2n 3 5 7 . . . 2n 1 2n 3 3 5 . . . 2n 1 2 4 . . . 2n x 2n 1 n→ lim 2n 2n 2 x2 3 x2 R When x 1 2n ± 1, the series diverges by comparing it to 1 n 1 1 < x < 1. which diverges. Therefore, the interval of convergence is n! x c n 3 5 . . . 2n un 1 un 2 2<x c < 2 or c 2<x<c 2 n→ 34. n 1 1 1 3 n 1!x 5 . . . 2n cn 1 1 2n 1 1 3 5 . . . 2n n! x c 1 n→ n→ lim lim 1 lim n 1x c 2n 1 1 x 2 c R Interval: The series diverges at the endpoints. Therefore, the interval of convergence is c 1 n! c 2 c n 3 5 . . . 2n 1 1 n 1 n 1x n5n 1 2<x<c 2. 1 5 5 n 3 n! 22 . . . 2n 5 2 1 1 3 4 6 . . . 2n . . . 2n 1 > 1 5 1 n 1 n 1 36. (a) f x (b) f x n 1 , 0 < x ≤ 10 1 38. (a) f x (b) f x n 1 x 2 n n 1 1 n 1 ,1 < x ≤ 3 1, 1 1 n 2 n x 5n n n 1 n , 0 < x < 10 5 n n 2 x n n 1 2 n 1<x<3 2 n 1 n 2, n 1 (c) f x (d) f x dx n 1x 5n x 5 1 5n , 0 < x < 10 1 (c) f x n 2 n 1 1x x 1 2 1<x<3 1 1 nn 2 3 4 9 , 0 ≤ x ≤ 10 (d) f x dx n 1 1 nn ,1 ≤ x ≤ 3 40. g 2 n 0 2 3 n 1 ... 42. g 2 n 0 2n alternating. Matches (d) 3 S1 1, S2 1.67. Matches (a) 44. The set of all values of x for which the power series converges is the interval of convergence. If the power series converges for all x, then the radius of convergence is R . If the power series converges at only c, then R 0. Otherwise, according to Theorem 8.20, there exists a real number R > 0 (radius of convergence) such that the series converges absolutely for x c < R and diverges for x c > R. 46. You differentiate and integrate the power series term by term. The radius of convergence remains the same. However, the interval of convergence might change. xn , 0 n! nx n n! 1 1 n 1 48. (a) f x n <x< (See Exercise 11) (c) f x n xn 1 n! 1 x x2 2! x3 3! x4 4! ... (b) f x n xn 1 n 1! n xn 0 n! fx f0 (d) f x 1 ex 402 50. Chapter 8 y y n Infinite Series 1 n x 4n 7 11 . . . 4n 1 x2 n 2 1 n 1 22n n! 3 3 1 2n 1 2 n! 1 n 4nx 4n 1 7 11 . . . 4n y n 1 22n 1 n 4n 4n 1 x 4n 2 n! 3 7 11 . . . 4n 1 22n n! 3 1 7 n 22n n! 22n n! 22n n! 3 3 3 1 n 4nx 4n 2 7 11 . . . 4n 1 n 5 1 x2 1 1 0 y x 2y x2 n 2 4nx 4n 2 x 4n 2 11 . . . 4n 1 x4n 2 11 . . . 4n 1 k x 2k 1 k 1! 22k 2! 5 1 n 1 7 11 . . . 4n n 1 22n 2 n 1 n 14 n 1! 3 7 n 1 1 n 1 x 4n 2 7 11 . . . 4n 22 n 1 22 n 52. J1 x x k 2k 02 1k! 1 k x 2k k 1! 22k 3 k 2k 02 1 x 2k 1!k 1k! 3 u (a) lim k 1 k→ uk k→ lim 1 k k 1k! k 1! 1 k x 2k 1 . k→ lim 22 k 1 x2 2k 1 0 Therefore, the interval of convergence is (b) J1 x k <x< 1 2k 02 k x2k k 1 1k! 1! 1 x2k 1! 1 J1 x k 0 1 k 2k 22k 1k! k 1 J1 x k 1 k 2k 1 2k x 2k 22k 1k! k 1 ! x 2J1 xJ1 x2 1 J1 k 1 1 k 2k 1k! 22k 1 2k x 2k k 1! 1 kx 2k 3 1k! 1 k 0 1 k 2k 1 x 2k 22k 1k! k 1 ! 1k! 1 k 0 22k k 1! 1 k 0 22k 1 k x 2k 1 k 1! 1 k 2k 1 x 2k 22k 1k! k 1 ! 1k! 1 k 1 1 k 2k 1 2k x 2k 22k 1k! k 1 ! x 2 1 1 k kx 2k 2k 12 1 kx2k 1k! x 2 1! k 1 1 k 2k 02 k 1 kx2k 3 k 1! 2 2k 1 k 1 2k 1 2k 2k 22k 1k! k 1 ! k 1! 1 k 2k 02 1 1 kx2k 1 k 3 0 1 kx 2k 3 k! k 1 ! k 1 1 22k 1 kx2k 1k! 14k k 1k! k 1! k 2k 12 1 kx 2k 1 k 1 !k! k 1x 2k 3 k 2k 02 1 1 k x2k 3 k! k 1 ! 1 kx 2k 3 k! k 1 ! 1 k 2k 02 1 k! k 3 1k! 1! k k 2k 02 1 k 0 1 22k kx 2k 1 1 1! 0 (d) J0 x k 0 (c) P7 x x 2 4 13 x 16 15 x 384 1 x7 18,432 1 22k k 2 12 k 1 k 1 x 2k 1 1!k 1! 1 1 k 0 k 1x 2k 1 22k k 1 1k! 1x 2k k 1 1! J1 x k 6 2k 02 1 k! k kx 2k 1 k 2k 02 1! k! k 1! −6 Note: J0 x J1 x −4 Section 8.9 x 2n 1 2n 1 ! Representation of Functions by Power Series x 2n 1 2n 1 403 54. f x n 0 1 n sin x 56. f x n 0 1 n arctan x, 1≤x≤1 (See Exercise 47.) 2 (See Exercise 38 in Section 8.7.) 2 − − 2.5 2.5 −2 − 2 58. n 0 x 2n 1 2n 1 ! n 1 x 2n 1 2n 1 ! 60. True; if an xn n 0 Replace n with n 1. converges for x on 2, 2 . 62. True 1 1 2, then we know that it must converge f x dx 0 0 n 0 an xn dx n an xn 1 1 0n 1 0 n 0n an 1 Section 8.9 4 5 x Representation of Functions by Power Series 45 x5 a 1 r 4. (a) 1 1 x 1 x n 0 2. (a) f x 1 x n n a 1 r 1 n xn 0 1 n n 4x 05 5 n 4xn n1 05 5, 5 . 4x3 625 ... 1 x)1 1 This series converges on 4 5 x)4 4 4 x 25 4 x 5 4 x 5 4 x 5 42 x 125 This series converges on x x x x x2 x3 (b) 1 1, 1 . ... (b) 5 x2 x2 x2 42 x 25 42 x 25 42 x 25 x3 x3 x3 x4 4x3 125 4x3 125 4x3 125 4x 4 625 404 Chapter 8 Infinite Series a 1 r 1 6. Writing f x in the form 4 5 x 7 4 x , we have 47 1 7x a 2 1 r . 8. Writing f x in the form a 1 3 2x 1 3 3 2x 2 r , we have 1 1 23 x a 2 1 r 2 Therefore, the power series for f x is given by 4 5 x n 0 which implies that a 1 and r 2 3 x 2. Therefore, the power series for f x is given by 3 2x 1 n 0 ar n n 41 x 07 7 n n 2 ar n n 0 2 x 3 2 n n 2 , n 4x 2 . 7n 1 0 n 0 2 n x 3n , x 2 < 7 or 5<x<9 x 2< 31 7 or < x < . 22 2 10. Writing f x in the form a 1 1 2x 5 5 1 2x 1 r , we have 15 2 5x a 1 r 12. Writing f x in the form a 1 4 3x 2 8 4 3x 2 r , we have 1 12 38 x a 2 2. 1 r which implies that a 1 5 and r the power series for f x is given by 1 2x 5 n 0 2 5 x. Therefore, 2 x 5 n n which implies that a 1 2 and r 38 x Therefore, the power series for f x is given by 4 3x 2 n 0 ar n n 0 1 5 2n xn n 1, 05 arn n 1 02 3 x 8 3 n n 2 x 8n 2 n 5 x < or 2 5 5 <x< . 2 2 x 2< 8 or 3 1 2n , 0 2 14 <x< . 3 3 14. 4x 2x2 3x 7 2 x 3 2 2x 2 1 2 3 x 1 2 2x 1 32 1 2x 2 1 2x Writing f x as a sum of two geometric series, we have 4x 2x 2 3x 7 2 n 0 3 2 1 x 2 n 2 2x n 0 n n 0 3 2n 1 1 n 2n 1 xn, x < 1 or 2 1 1 <x< . 2 2 16. First finding the power series for 4 4 1 1 1 4x 1 x 4 n n 0 x , we have 1 n xn 4n n 0 Now replace x with 4 4 x2 n x2 . 1 n x 2n . 4n 2 < x < 2 since 4n 1 n x 2n 1 2n 1 0 2 x2n n 0 0 The interval of convergence is x 2 < 4 or n→ lim un 1 un x n→ lim 1 n 1 2n x 2 4 n 1 x2 4 1 2n x2 . 4 18. h x 1 1 21 1 2n 1 2n x 1 0 n 1 21 1 xn 1 x2 x 1 nxn 0 xn 0 2x 1, 0x2 2x3 0x4 2x5 ... 2 x2n 0 1<x<1 Section 8.9 d2 1 dx 2 x 1 Representation of Functions by Power Series 405 20. By taking the second derivative, we have 2 x 1 3 2 x 1 3 . Therefore, d2 dx 2 d dx 1 n xn n 0 1 nnx n n 1 1 n 2 1 nn n 1 xn 2 n 0 1 n n 2n 1 x n, 1 < x < 1. 22. By integrating, we have 1 1 fx ln 1 ln 1 x2 x dx x2 ln 1 ln 1 1 1 x x x dx 1 C1 and ln 1 1 x dx x n dx n 0 1 1 x dx ln 1 x C2. x . Therefore, 1 n x n dx n 0 C1 n 0 1 n xn n1 C n 0 1 C2 n xn 1 1 0n C n 0 1 n n 1 1 xn 1 C n 0 2x 2n 2n 2 2 1 n x 2n 1 2 To solve for C, let x ln 1 x2 n 0 and conclude that C x2n 2 , 1 0n 1<x<1 0. Therefore, 24. 2x x2 1 2x n 0 1 n x 2n (See Exercise 23.) 1 n 2x 2n 1 n 0 Since d ln x 2 dx 1 1 2x x2 1 , we have 1 ln x 2 1 n 2x 2n n 0 dx C n 0 1 n x 2n n1 2 , 1 ≤ x ≤ 1. To solve for C, let x ln x 2 1 n 0 0 and conclude that C 1x n1 n 2n 2 0. Therefore, , 1 ≤ x ≤ 1. 26. Since 1 4x 2 1 dx 2 1 arctan 2x , we can use the result of Exercise 25 to obtain 2 1 4x 2 1 dx 2 n 0 arctan 2x 1 n 4n x 2n dx 0. Therefore, C 2 n 0 1 n 4n x 2n 2n 1 1 , 1 1 <x≤ . 2 2 To solve for C, let x arctan 2x 2 n 0 0 and conclude that C 1 2n n 4nx 2n 1 1 , 1 1 <x≤ . 2 2 406 Chapter 8 x2 2 x3 3 Infinite Series x4 ≤ ln x 4 ≤x x2 2 28. x 1 x3 3 x4 4 x5 5 5 S5 f −4 8 S4 −3 x x ln x x x2 2 x 2 2 0.0 x3 3 1 x3 3 x4 4 x5 5 x4 4 0.0 0.0 0.0 0.2 0.18227 0.18232 0.18233 1 n 0.4 0.33493 0.33647 0.33698 0.6 0.45960 0.47000 0.47515 0.8 0.54827 0.58779 0.61380 1.0 0.58333 0.69315 0.78333 In Exercise 35–38, arctan x n 0 x 2n 1 . 2n 1 x3 3 x5 5 x7 , 7 30. g x x x3 , cubic with 3 zeros. 3 32. g x x Matches (d) Matches (b) x2n 1 . 2n 1 34. The approximations of degree 3, 7, 11, . . . 4n 1, n 1, 2, . . . have relative extrema. In Exercises 36 and 38, arctan x n 0 1 n 36. arctan x 2 n 0 1 n x 4n 2 2n 1 4n x 4n 3 3 2n 1 C, C 0 arctan x 2 dx n 34 0 1n 1 n 0 n arctan x 2 dx 0 3 4 4n 3 4n 3 2n 1 4n 34n 3 2n 3 1 n 0 n 1 44n 3 27 192 2187 344,064 177,147 230,686,720 Since 177,147 230,686,720 < 0.001, we can approximate the series by its first two terms: 0.13427 38. x 2 arctan x n 0 1 1 n 0 n x 2n 3 2n 1 2n x 2n 4 4 2n 1 4 2n 1 1 22n 4 x 2 arctan x dx 12 n x 2 arctan x dx 0 n 0 1 n 2n 1 64 1 1152 12 ... Since 1 < 0.001, we can approximate the series by its first term: 1152 x 2 arctan x dx 0 0.015625. S ection 8.9 1 1 x n 0 Representation of Functions by Power Series 407 In Exercises 40 and 42, x n, x 1. 40. Replace n with n nx n n 1 1 n 1. n 0 42. (a) 1 xn (b) 1 3n 1 10 n n 1 2 3 n n 2 9n n n 1 2 3 n 1 2 91 n 1 1 23 2 2 1 9 10 9 100 n 9 100 n 1 9 10 1 1 9 10 2 9 44. Replace x with x2. 48. (a) From Exercise 47, we have arctan 120 119 arctan 1 239 arctan arctan (b) 2 arctan 4 arctan 4 arctan 1 5 1 5 1 5 arctan 1 5 1 5 arctan 1 5 120 119 1 arctan 1 239 1 239 1 239 2 46. Integrate the series and multiply by 1. 120 119 120 119 1 arctan 10 24 28,561 28,561 arctan 5 12 2 arctan 1 4 arctan 1 5 21 5 15 5 12 1 239 arctan 5 12 2 arctan arctan 2 arctan arctan arctan arctan arctan 1 2 5 12 5 12 arctan 120 119 1 239 120 119 arctan 4 (see part (a).) 50. (a) arctan (b) 1 2 arctan 1 2 1 3 arctan 1 3 5 12 1 13 12 13 arctan 56 56 4 4 arctan 4 1 2 arctan 3 12 3 12 5 12 7 3.14 7 4 1 3 13 3 3 13 5 5 13 7 7 4 0.4635 4 0.3217 52. From Exercise 51, we have 1 n 1 n 1 54. From Example 5, we have arctan x n 0 1 1 1 2n 4 2n n x 2n 1 . 2n 1 1 3n n 1 n 1 n 1 13 4 ln 3 n n n 0 1 0.2877. n 1 2n 1 n 0 1 arctan 1 n 1 0.7854 1 ln 3 56. From Exercise 54, we have 1 n 1 n 1 1 32n 1 1 2n 1 1 n 0 n 32n n 1 1 2n 1 1 1 n 0 13 2n 2n 1 arctan 1 3 0.3218. 408 Chapter 8 Infinite Series x 2n 1 . 2n 1 3 3 3 1 2n 1 58. From Example 5, we have arctan x n 0 1 1n 2n n n 1 n n 0 3 2n n 1 n 0 3 2n 1 3 n 0 1 1 2n 3 arctan 3 1 3 23 6 Section 8.10 2. For c fx f n Taylor and Maclaurin Series 0, we have e3x 3ne 3x ⇒ f 1 3x n x e 3x 0 3n 27x3 3! ... n 0 9x2 2! 3x n n! 4. For c fx fx fx fx f 4 4, we have: sin x cos x sin x cos x sin x f f f f f 4 4 4 4 4 4 2 2 2 2 2 2 2 2 2 2 x and so on. Therefore we have: sin x n 0 f n 4x n! x 1 4 nn 12 4 x n 2 1 2 2 2 6. For c fx f ex n 0 n 4 2! 4 1! n 2 x 3! 1 4 3 x 4! 4 4 ... x n 0 n 1 1, we have: ex x ex ⇒ f f n n 1 1 n e e1 x 1 x 2! 1 2 1x n! x 3! 1 3 x 4! 1 4 ... e n 0 x n! 1 n Section 8.10 Taylor and Maclaurin Series 409 8. For c 0, we have: fx fx fx ln x 2 2x x2 2 x2 1 2x 2 12 1 f0 f0 f0 f0 6x 2 14 1 5 15x 2 6 0 0 2 0 12 0 240 fx f f f 4 4x x 2 3 x2 1 3 12 x4 x2 f f 1 f x x x 4 0 0 0 5 48x x 4 10x 2 x2 1 5 240 5x6 x2 5 6 15x 4 1 6 and so on. Therefore, we have: ln x 2 1 n 0 f n 0 xn n! x6 3 0 0x ... n 2x 2 2! 0x3 3! 1 n x 2n n1 12x 4 4! 2 0x5 5! 240x6 6! ... x2 x4 2 0 10. For c 0, we have; fx fx fx tan x sec2 x 2 sec2 x tan x 2 sec4 x 2 sec2 x tan2 x sec2 x tan3 x 11 sec4 x tan2 x x 2x3 3! 16x5 5! 2 sec2 x tan4 x ... x x3 3 f f f0 f0 f0 f0 4 5 0 1 0 2 0 16 ... fx f f 4 5 x x 8 sec4 x tan x 8 2 sec6 x f n 0 n 0 0 tan x 0 xn n! 25 x 15 12. The Maclaurin Series for f x f n 1 e 2x is n 0 2x n . n! x 2 n 1e 2x. Hence, by Taylor’s Theorem, zn x 1! n→ 1 0 ≤ Rn x Since lim n→ f n 1 n 2 n 1xn n 1! 1 2 n n 1e 2z 1! xn 1 . lim 2x n 1 n 1! 2x 0, it follows that Rn x → 0 as n → 2x . Hence, the Maclaurin Series for e converges to e for all x. 410 Chapter 8 Infinite Series kk 2! 1 x 2 x 2 1 n 1 14. Since 1 1 x x k 1 12 kx 1 1 1 1 x2 1 k 2 x3 . . . , we have 3! 1 2 3 2 x2 1 2 3 2 5 2 x3 . . . 2! 3! kk 1 3 5 x3 233! 1 xn ... 1 3 x2 222! 3 5 . . . 2n 2n n! 16. Since 1 1 x x k 1 1 1 1 kx 1 x 3 x 3 x 3 kk 2! 1 x2 13 kk 1k 3! 13 5 8x4 344! 2 x3 23 3! ... 4 . . . , we have 5 3 x3 ... 13 2 3 x2 2! 5x3 333! 12 2x2 322! 1 n 2 2 n 2 5 8 . . . 3n 3nn! . 18. Since 1 we have 1 x 12 1 12 x 2 1 1 n 2 n 1 1 n 1 3 1 5 . . . 2n 2n n! 3 3 xn (Exercise 14) 3 x3n . x3 x3 2 x2 2! 1 n 2 5 . . . 2n 2n n! 20. e ex n 3x n xn n! 0 3x n! 1 n x x3 3! 1 n 3n xn n! x2 2! x4 4! x4 4! 1 x5 5! 3x ... 9x 2 2! 27x3 3! 81x 4 4! 243x5 5! ... 0 n 0 22. cos x n 0 1 nx2n 2n ! 1 4x 2n ! 16x2 2! n 2n 1 x6 6! ... 24. sin x n 0 1 nx2n 1 2n 1 ! 1 n x3 2n 2n 1 ! x9 3! 2x9 3! x15 5! 2x15 5! 1 cos 4x n 0 n 0 1 2n ! n42nx2n 2 sin x3 2 n 0 1 256x4 4! ... 2 x3 2x3 ... ... 26. e ex e ex x 1 1 2 ex x x 2x2 2! e x x2 2! x2 2! 2x4 4! x3 3! x3 3! ... ... ... x2n 2n ! x 2 cos h x 2 n 0 S ection 8.10 1 n1 n 1 Taylor and Maclaurin Series 1 xn 411 28. The formula for the binomial series gives 1 1 ln x x2 x2 12 x 3 12 1 1 x 2n 3 5 . . . 2n 2n n! , which implies that 1 n 1 1 n1 1 x2 1 5 . . . 2n 2n n! 1 x dx 1 n1 3 5 . . . 2n 2n 2n 1 n! 1 2n 1 n 1 x x3 2 3 1 3x5 245 1 3 5x7 2467 . . .. 30. x cos x x1 x x3 2! x2 2! x5 4! x4 4! ... ... 32. arcsin x x n 0 2n !x2n 2nn! 2 2n 1 1 1 x 0 n 1 0 2n !x2n ,x 2nn! 2 2n 1 n 0 1 nx2n 2n ! 2x 2 2! x2 2! 34. eix eix 2 e e ix 2 1 2x 4 4! x4 4! x6 6! 2x6 6! . . . (See Exercise 33.) ... n 0 ix 1 n x 2n 2n ! cos x 36. g x ex cos x 1 1 x x x2 2 x2 2 x 6 4 8 x 24 x3 6 4 ... x3 2 1 x2 2 x4 24 x 24 x4 4 4 ... x4 24 ... x3 3 x4 6 ... −6 6 g P4 x2 2 1 x − 40 38. f x ex ln 1 1 x x x x2 x2 2 3 x x2 2 x2 2 x3 3 P5 3 x3 6 x3 3 3x5 40 x4 24 x3 2 ... ... x3 2 x x2 2 x4 4 x3 3 x4 3 x4 4 x4 4 x5 5 x4 6 ... x5 5 x5 4 x5 6 x5 12 x5 24 ... f −3 −3 412 Chapter 8 ex 1 1 1 x1 1 x Infinite Series x2 2 6 40. f x . Divide the series for ex by 1 x3 3 x2 2 x2 2 x2 2 3x 4 8 x3 6 x3 6 x3 2 x3 3 x3 3 ... x4 24 x5 120 x. fx 1 x3 3 f 3x 4 8 ... x2 2 x x P4 ... −6 −6 6 0 x4 24 x4 3 3x 4 8 3x4 8 x5 120 3x5 8 42. y x x3 2! x5 4! x1 x2 2! x4 4! x cos x. 44. y x2 x3 x4 x2 1 x x2 x2 1 1 x . Matches (b) x x Matches (d) 5 . . . 2n 2n n! 46. 0 1 t3 dt 0 1 t x t4 8 x4 8 t3 2 1 n 2 n 11 3 3 t3n 1x 0 dt 1 n 2 n 11 3 5 . . . 2n 3n 1 2n n! 3 5 . . . 2n 3n 1 2n n! x5 5! 1 7! x7 7! ... 3 t3n 3 x3n 1 n 2 n 11 1 48. Since sin x n 0 1 n x 2n 1 2n 1 ! n x 1 5! x3 3! . . . , we have 0.8415. (4 terms) sin 1 n 0 1 2n xn 0 n! 1 e 1! 1 1 3! 50. Since ex n 1 1 x 1 x2 2! 1 2! x3 3! 1 3! x4 4! 1 4! x5 5! 1 5! . . . , we have e 1 7! ... n 1 1 1 1n n! 1 1 1 2! 1 3! 1 4! 1 5! ... and e e 1 0.6321. (6 terms) 52. Since sin x n 0 1 n x 2n 1 2n 1 ! x2 3! x4 5! x→0 n x x6 7! x3 3! ... x5 5! x7 7! ... sin x x we have lim x→0 1 n n x 2n 0 1 n x 2n 2n 1 ! sin x x lim 0 1 2n 1 ! 1. Section 8.10 12 Taylor and Maclaurin Series 12 0 413 54. 0 arctan x dx x 9229 12 1 0 x2 3 x4 5 x6 7 . . . dx x x3 32 x5 52 x7 72 ... Since 1 12 0 < 0.0001, we have 1 2 x→0 arctan x dx x 1 3223 1 5225 1. 1 7227 1 9229 0.4872. Note: We are using lim 1 1 arctan x x x 2! x2 4! 56. 0.5 cos Since x dx 0.5 1 0.55 x3 6! x4 8! . . . dx x x2 2 2! x3 3 4! x4 4 6! x5 5 8! ... 1 0.5 1 210, 600 1 1 x dx < 0.0001, we have 1 0.5 1 1 4 0.52 1 1 72 0.53 1 1 2880 0.54 1 1 201,600 0.5 5 cos 0.5 0.3243. 14 14 58. 0 x ln x 1 dx 0 x2 x3 3 x4 4 2 x3 2 5 x4 3 x5 3 x5 4 x6 6 . . . dx ... 4 14 0 Since 14 15 14 5 < 0.0001, x ln x 0 1 dx 14 3 3 14 8 4 0.00472. 60. From Exercise 19, we have 1 2 2 e 1 x2 2 dx 1 2 1 2 1 2 2 1n 0 1 n x 2n dx 2nn! 1 1 2 n 0 2n n! 1 n x 2n 1 2n 1 2 1 n 0 1 n 2n 2nn! 2n 2 7 1 3 1 1 22 27 31 2! 5 23 127 3! 7 24 511 4! 9 25 2047 5! 11 0.1359. 1 26 x ln 1 2 x3 4 8191 6! 13 32,767 7! 15 131,071 28 8! 17 524,287 29 9! 19 62. f x P5 x sin x2 2 x 7x 4 48 11x5 96 −4 4 f 8 P5 −4 The polynomial is a reasonable approximation on the interval 0.60, 0.73 . 64. fx P5 x 3 x arctan x, c 0.7618 x x 4! 1 4 1 1 0.3412 x 5! x 2! 1 5 3 0.7854 1.3025 1 2 0.0424 x 3! 1 3 f P5 −2 −1 4 5.5913 The polynomial is a reasonable approximation on the interval 0.48, 1.75 . 414 66. a2n Chapter 8 Infinite Series 68. Answers will vary. 1 0 (odd coefficients are zero) 1 ,g 16 2 70. y 60 , v0 3x 3x 3x 64, k 32 1 16 32 x3 3 64 3 1 2 3 1 16 2 32 x 4 4 64 4 1 2 4 2 32x 2 2 64 2 1 2 32 32 n ... 22x 2 2 64 2 23x3 3 64 316 n 2 24x 4 4 64 4 16 3x 32 ... xn 16 2 n 32 n 2n xn n 16 2 n 64 n n 2 72. (a) f x ln x 2 1 . x2 1.5 From Exercise 8, you obtain P P8 (c) F x 0 x (b) 0 1 x 2n 1 x 0 1 n x 2n n1 x4 3 2 n 1 n x 2n n1 0 − 0.5 2 x2 2 x6 4 x8 5 ln t2 1 dt t2 P8 t d t Gx 0 x Fx Gx 0.25 0.2475 0.2475 0.50 0.4810 0.4810 0.75 0.6920 0.6920 1.00 0.8776 0.8805 1.50 1.1798 5.3064 2.00 1.4096 652.21 (d) The curves are nearly identical for 0 < x < 1. Hence, the integrals nearly agree on that interval. 74. Assume e e Set a a 1 N! e N! 1 N 1! 1 N 1 N N p q is rational. Let N > q and form the following. 1 1 1 2! 1 ... 1 N! 1 N! 1 N 1! N 1 2! ... ... 1 2! , a positive integer. But, 1 N 1 1 N 1 1 N 1 1N 1 1 N 1 ...< 1 N 1 N 1 1 2 ... ... 2 1 N 1 1 1 1 2 ... 1 , a contradiction. N Review Exercises for Chapter 8 2. an n n2 1 4. an n : 3.5, 3, . . . 2 Matches (c) 4 6. an 2 3 Matches (b) 6 n 1 : 6, 4, . . . R eview Exercises for Chapter 8 n 2 1 0 n Converges n→ 415 8. an 2 sin 10. lim 0 12 −2 The sequence seems to diverge (oscillates). sin n : 1, 0, 2 1, 0, 1, 0, . . . 12. lim n→ n ln n n→ lim 1 1n 14. lim 1 n→ 1 2n n k→ lim 1 1 k k 12 e1 2 Diverges 16. Let y ln y bn ln bn n n→ Converges; k cn 1n 2n 1, 2, 3, 4, 5 \$20,168.40 18. (a) Vn (b) V5 1 120,000 0.70 n, n 120,000 0.70 5 cn n→ lim ln y lim bn cn bn ln b cn ln c Assume b ≥ c and note that the terms bn ln b bn cn ln c cn bn ln b bn cn cn ln c bn cn converge as n → 20. (a) . Hence an converges. (b) 1 k Sk 5 0.3917 10 0.3228 15 0.3627 20 0.3344 25 0.3564 (c) The series converges by the Alternating Series Test. 0 0 12 22. (a) k Sk 5 0.8333 10 0.9091 15 0.9375 20 0.9524 25 0.9615 1 . n2 (b) 1 (c) The series converges, by the limit comparison test with 0 0 12 24. Diverges. Geometric series, r 2n 2 n 03 2 3 n 1.82 > 1. 26. Diverges. nth Term Test, lim an n→ 2 . 3 28. n 4 n 0 43 12 See Exercise 27. 2 3 n 30. n 0 n 1 1n 2 n 0 2 3 n n 0 1 n 1 1 1 2 n 1 2 1 2 1 3 1 3 1 4 ... 3 1 2 1 1 23 416 Chapter 8 Infinite Series 32. 0.923076 0.923076 1 0.000001 n 0.000001 1 2 ... 923,076 999,999 12 76,923 13 76,923 12 13 0.923076 0.000001 n 0 0.923076 0.000001 39 34. S n 0 32,000 1.055 \$4,371,379.65 n 32,000 1 1.05540 1 1.055 36. See Exercise 86 in Section 8.2. A P 12 r 1 r 12 1 12t 1 0.065 12 120 100 12 0.065 1 \$16,840.32 1 n3 1 n3 4 3 4 38. n 1 4 40. n 1 n 1 1 n2 1 2n n 1 n2 1 n 1 2n 1 Divergent p-series, p <1 The first series is a convergent p-series and the second series is a convergent geometric series. Therefore, their difference converges. 1 n 13 n 42. n 1 n nn n 1 2 1 nn 1n 2 n→ 44. Since lim n n 1 2 1 converges, n 1 13 n 5 converges by the Limit Comparison Test. n→ lim By a limit comparison test with n 1 , the series diverges. 1n 46. n 1 1n n n1 1 48. Converges by the Alternating Series Test. an an 1 an n1 n ≤ n2 n1 n n 1 0 3 ln n 1 3 ln n < n1 n an, lim n→ 3 ln n n 0 n→ lim By the Alternating Series Test, the series converges. n! n 1e an 1 an n→ 50. n n→ lim lim lim n 1! en 1 n e 1 en n! n→ By the Ratio Test, the series diverges. 52. n 1 12 3 5 an 1 an 5 . . . 2n 8 . . . 3n n→ 1 1 3 . . . 2n 5 . . . 3n 1 2n 1 3n 1 2 2 1 5 . . . 3n 3 . . . 2n 1 1 n→ n→ lim lim 1 2 lim 2n 3n 1 2 2 3 By the Ratio Test, the series converges. Review Exercises for Chapter 8 54. (a) The series converges by the Alternating Series Test. (b) x Sn (c) 0.3 417 5 0.0871 10 0.0669 15 0.0734 20 0.0702 25 0.0721 (d) The sum is approximately 0.0714. 0 0 12 56. No. Let an 3937.5 , then a75 n2 0.7. The series n 1 3937.5 is a convergent p-series. n2 0.75 2! 2 58. f x fx fx fx P3 x tan x sec2 x 2 sec2 x tan x 4 sec2 x tan2 x 1 2x 2 sec4 x 2x f f f f 2 4 4 4 4 8 x 3 0.25 4! 4 1 2 4 16 3 60. cos 0.75 1 0.75 4! 4 0.75 6! 6 0.7317 4 0.25 2! 2 4 0.25 3! 3 4 62. e 0.25 1 0.25 0.779 64. fx P4 x P6 x P10 x cos x 1 1 1 x2 2! x2 2! x2 2! 2 n un 1 un n→ n 10 66. P4 n 10 2x 0 n x4 4! x4 4! x4 4! x6 6! x6 6! − 10 f P6 − 10 P10 Geometric series which converges only if 2x < 1 or 1 1 2 < x < 2. x8 8! x10 10! x n 0 68. n 1 3n x 70. lim 3 n 1 2 2n n n 0 x 2 2 n n→ lim x n 2 1 n 1 n 3n x 2 n Geometric series which converges only if x 2 2 < 1 or 0 < x < 4. 3x 1 3 Center: 2 R 2 Since the series converges at 3 and diverges at 7 , the 3 interval of convergence is 5 ≤ x < 7 . 3 3 5 418 Chapter 8 Infinite Series 3 nx 2n 2nn! 3 n 1 n 72. y n 0 y y n 0 2n x2n 2nn! 1 1 n 0 3 n 1 2n 1 x2n 1 x2n 1 2n n 2 x2n 1! 1 3 3 n 0 n 2n n 1 2 n 1 2 2n n 1! 2 2n n 1! 2x y 3xy 3y 2n n 1 1 1 n 0 n 1n 3 2 1 2n 2 2 1 n 2 n 1n n n 2 x2n 1! 2 n nn 0 1 2n 1 n3n 1x2n 2nn! 1n 3 n 0 2n 2nn! 2n 2n 2n n 0 1 3x 2nn! 1 n 2 2n n 2 0 13 x 2nn! n 0 1 n3n 1x2n 2nn! 13 x 2nn! 1 n 1n nn 1 2n 1 1 n 0 3 2x 2n 2nn! 2 1n 1 n 0 n n 0 3 2x 2n 2nn! 1n 3 1 2n x n 1 2nn! 1 n3n 1x2n 2nn! 32 x2 2n n 1 2n 0 1 n3n 1x 2n 1n 1! 2n 2n 2n 2n n 1 74. 3 2 x 3 02 1 x 2 n 32 x2 a 1 r 1 76. Integral: n 0 n 1 n3xn 1 2n 1 1 n n 0 1 n3xn 2n 1 1 x 8 x 4 32 3 1 x , 3 n 78. 8 2x 3 1 x 2 3 2 3 3 ... n 0 8 32 x 1 8 x 34 4 1<x<7 80. fx fx fx fx cos x cos x sin x cos x sin x f n 0 n 4x n! x 4 4 n 2 2 1 nn 1 2 2 x 2 x 1! n 4 4 n 2 1 2 x 2! 2 4 2 2 x 3! 3 4 2 2 x 4! 4 ... 4 2 1 2 82. fx fx fx fx f 4 n 1 csc x csc x cot x csc3 x csc x cot2 x csc x cot3 x csc x cot4 x 1 1 x 2! 2 5 csc3 x cot x 5 csc5 x f n 0 n x 15 csc3 x cot2 x 2x n! 2 n csc x 2 5 x 4! 4 ... 2 Review Exercises for Chapter 8 84. fx fx fx fx f 4 419 x1 1 x 2 2 12 1 2 1 2 1 2 f n 0 n 1 x 2 1 2 1 2 32 3 x 2 3 2 52 x x 2 2 5 x 2 4 n 7 2, ... 4x n! 4 22 4 22 x x x 4 252! 2 1 n 11 3x 4 283! 3 3 1 3 5x 2114! 4 4 n 4 ... 1 n 2 5 . . . 2n 23n 1n! 3x 86. hx hx hx hx h4 x h5 x 1 1 x 3 1 31 12 1 x 3 x x x x 5 4 60 1 360 1 2520 1 1 3x 6 7 x 12x2 2! x n 8 60x3 3! 1 n 360x4 4! 2520x5 5! ... n 0 1 n n 2 !x n 2n! xn , n! 0 23 n! n n 1 n 0 n n 2n 2 1x 88. ln ln x n 1 1 1 n 1 n 1 , 1 n 0<x≤2 90. ex n <x< 2n n 0 3 n! 1.9477 6 5 n 1 65 n 1 5nn e2 3 n 0 1 n 1 n 1 0.1823 92. sin x n 0 1 1 n 0 n x2n 1 , 2n 1 ! 32n 1 <x< 0.3272 sin 1 3 n 1 2n 1! 94. ex n xn 0 n! xn 1 0 n! 1 x xex x x2 x2 2! x3 2! 1 ... ... xex n 1 xex dx 0 1 0n ex 0 e xn 2 e 1 0 0 1 1 1 1 xn 1 dx 0 n! n 0 n 2 n! n 0 n 2 n! 420 Chapter 8 fx fx fx fx f f f f 4 5 6 7 Infinite Series f0 f0 f0 f0 f f f f 8x3 3! 4 5 6 7 96. (a) sin 2x 2 cos 2x 4 sin 2x 8 cos 2x 16 sin 2x 32 cos 2x 64 sin 2x 128 cos 2x 0 2x 0x2 2! 0 2 0 8 0 32 0 128 32 x5 5! 0x6 6! 128x7 7! ... 2x 43 x 3 45 x 15 87 x 315 ... x x x x 0 0 0 0 0x4 4! sin 2x (b) sin x n 0 1 nx2n 1 2n 1 ! 1 n 2x 2n 2n 1 ! 8x3 6 32x5 120 1 sin 2x n 0 2x 128x7 5040 2x 3 3! ... 2x 5 5! 2x 2x 7 7! 43 x 3 ... 45 x 15 87 x 315 ... 2x (c) sin 2x 2 sin x cos x 2x 2x 2x 2 x3 3 x3 6 x3 2 x5 120 x3 6 2 x5 15 x7 5040 x5 24 4 x7 315 ... x5 12 ... 1 x2 2 x4 24 x6 720 x7 720 43 x 3 45 x 15 ... x7 144 x7 240 x7 5040 ... ... x5 120 2x 87 x 315 98. cos t n 0 1 n t 2n 2n ! 1t 2 2n ! 2n nn 100. et ntn 1 x et n tn n! 0 tn n! 1 tn 1 1 n! tn n! 1n x 0 n cos x t 2 t dt n n 0 1 n cos 0 2 0 2 2n 1 2n ! n nn et t x 1 n 1 1 0 n 0 22n x3 2 x2 2 1x 2n ! n et t 1 1 1 3 5x7 2467 1 3 5x6 2467 dt n 0 xn n! 1n 102. arcsin x arcsin x x x→0 x 1 1 3 3 1 3x5 245 1 3x4 245 ... ... lim arcsin x x 1 arcsin x By L’Hôpital’s Rule, lim x→0 x x→0 lim 1 1 x2 1. ...
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