Ln n for k k n n 1 1 1 1 10x2 n 0 1 x 1 x 2 1 x 0x2

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Unformatted text preview: 1 0.75n million dollars 400 million dollars 101. P n 0.81 16 0.75 n 1 0.75 400 1 Sum 16 100 1 0 0.9n , n > 0 80,000 1 99. D1 i 5. This series converges 1 S ection 9.2 1 2 103. (a) n 1 n n 0 449 n 11 22 1 12 1 21 Series and Convergence 1 (b) No, the series is not geometric. (c) n n 1 2 n 1 105. (a) 64 2 32 1 2 64 (b) n 0 16 8 n 4 64 12 1 126 in.2 2 128 in.2 16 in. Note: This is one-half of the area of the original square! 16 in. 20 107. 50,000 n 1 1 1.06 50,000 19 1 1.06 i 0 1.06 n 50,000 1 1.06 1 i 1.06 1.06 20 1 , n 20, r 1.06 1 573,496.06 n 1 0.01 2 109. w i i 0 0.01 1 2n 12 0.01 2n 1 111. P 50, r 29: w (b) When n 30: w 50 (b) A $5,368,709.11 50 e0.03 20 e0.03 12 $10,737,418.23 113. P 31: w 100, r $21,474,836.47 0.04, t 12 0.04 20 0.03 12 1 1 1 12 20 1 $16,415.10 $16,421.83 40 0.04 12 (a) A 100 (b) A 1 100 e0.04 40 e0.04 12 1 115. (a) an 12 0.03 (a) A (a) When n (c) When n 0.03, t 1 12 40 1 $118,196.13 $118,393.43 3484.1363 1.0502 n 3484.1363e0.04897n (b) Adding the ten values a3, a4, . . . , a12 you obtain 12 6500 an n 50,809, or $50,809,000,000. 3 12 3484.1363e0.04897n (c) 2 3500 117. False. lim n→ n 13 50,803, or $50,803,000,000 3 (Answers will vary.) 1 n 0, but n 1 1 diverges. n a a 1r The formula requires that the geometric series begins with n 0. arn 119. False; n 1 450 Chapter 9 Infinite Series 121. True 123. By letting S0 0.74999 . . . 9 103 0.74 9 103 n 0.74 1 and n k 129. (a) 1 1an an 2 1. 1 n 1 3 1 can c 0 an an 1 1an 1 3 2 2an 3 1 k ak 1ak n k 0 1 r 1 r2 2 3 1 a2a3 1 a3a4 ... an 1 an 2an 3 1 1 131. 1 2ak ak 1 a2a3 1 a1a2 n 3 1 1ak ak 1 a1a2 0 an 3 1 0 an 1 1an 1 r3 2an 3 lim Sn 3 lim n→ ... n 1 n→ 11 0r r 1 an n 1 since 1 <1 r This is a geometric series which converges if 1 < 1 ⇔ r > 1. r 1 2an 1r 1r 1 r Sn 1 Sn 1 1 c c c Sn . can converges. Because an converges. This is a contradiction since Hence, can diverges. an 3 an an 1an 2an an (b) Sn Thus, 1 127. Suppose, on the contrary, that c 0, 0 an n 1. 1 n 0 1 1 2an Sn 1 Sn n Sn c 0 an Sn Sn 1 1 1 10 n n ak an n bn 1 1 0.75 0 bn k 1 n Both are divergent series. an 1 ak 10 9 1 100 0.74 n an n 1 9 103 0.74 an 0 n ... 1 10 9 103 0.74 125. Let 9 104 0, we have 1 3 1 1an 2 an 1 2an 3 can diverged. S ection 9.3 The Integral Test and p-Series 133. Let H represent the half-life of the drug. If a patient receives n equal doses of P units each of this drug, administered at equal time interval of length t, the total amount of the drug in the patient’s system at the time the last dose is administered is given by Tn P Pekt Pe2kt . . . Pe n 1 kt where k ln 2 H. One time interval after the last dose is administered is given by Tn 1 Pekt Pe2kt Pe3kt . . . Penkt. Two time intervals after the last dose is administered is given by Tn 2 Pe2kt Pe3kt Pe4kt . . . Pe n and so on. Since k < 0, Tn s→ 0 as s → , where s is an integer. 135. f 1 0, f 2 1, f 3 2, f 4 451 1 kt 4, . . . 2 n 4, n even n2 1 4, n odd. In general: f n (See below for a proof of this.) x y and x fx y y are either both odd or both even. If both even, then fx y fx y x y 2 x 4 y 2 xy. 4 If both odd, fx y y2 4 x 1 x y2 4 1 xy. Proof by induction that the formula for f n is correct. It is true for n k2 4 and is valid for k. If k is even, then f k fk 1 k 2 fk k2 2k k2 4 k 2 k 12 4 1. Assume that the formula 4 1 . The argument is similar if k is odd. Section 9.3 The Integral Test and p-Series 1 1. 1n n 3. 1 1 Let f x x 1 x 1 dx e x. f is positive, continuous, and decreasing for x ≥ 1. e 1 n 1 Let f x . f is positive, continuous, and decreasing for x ≥ 1. 1 e n ln x 1 x dx 1 e x e 1 1 1 Converges by Theorem 9.10 Diverges by Theorem 9.10 1 5. n 1 n2 7. 1 n 1 Let f x x2 1 . 1 1 x 1 dx ln n n arctan x 1 4 1 1 ln x x Let f x f is positive, continuous, and decreasing for x ≥ 1. 2 1 1 . 1 f is positive, continuous, and decreasing for x ≥ 2 since fx Converges by Theorem 9.10 1 1 ln x x ln x x1 1 dx 1 1 2 < 0 for x ≥ 2. ln x Diverges by Theorem 9.10 1 2 2 1 452 Chapter 9 9. n 1 n n 1 Infinite Series Let f x n 1 x x 1 1 ,f x 2x3 2 2x x1 2 < 0. x 1 x 1 dx 2 ln x 1 ln x ,f x x2 1 ln x dx x2 ln x x 2 ln x . x3 1 2 n 1 1 ,f x 2 1 x2 1 2 2x x2 1 ln x 2 dx 1 1 , diverges 1 arctan x ,f x x2 1 1 2x arctan x < 0 for x ≥ 1. x2 1 2 arctan x 2 2 32 , converges 32 1 1 c xk xk Let f x < 0 for x > 1. 1 c . f is positive, continuous, and decreasing for x > k c k 1 since f is positive, continuous, and decreasing for x > 1. 1 < 0. arctan n n2 1 nk k 1n n 2x 32 Hence, the series converges by Theorem 9.10. 19. x2 2x arctan x dx x2 1 1 1 Let f x 1 1, converges x2 1 1 2x f is positive, continuous, and decreasing for x ≥ 1. 1.6. 1 2n n2 dx 1 Let f x Hence, the series converges by Theorem 9.10. 17. 1 ,f x Hence, the series diverges by Theorem 9.10. n f is positive, continuous, and decreasing for x > e1 1 x 1 15. Let f x x 1 , diverges 1 ln n 2 1n n 1 f is positive, continuous, and decreasing for x ≥ 1. Hence, the series diverges by Theorem 9.10. 13. 1 Let f x f is positive, continuous, and decreasing for x...
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