# The amount will grow arbitrarily large over time b p

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Unformatted text preview: llion \$2 billion 6.60n2 151.7n 387 1500 A2 \$1.6 billion A3 \$1.28 billion A4 \$1.024 billion (c) lim 0.8 n→ 113. an (c) A sequence is monotonic if its terms are nondecreasing, or nonincreasing. 9378.13 A5 n 9335.34 A4 105. an 9041.25 0.055 12 (b) A sequence converges if it has a limit. See the definition. n 2.5 5 500 0 (b) In 2008, n 13 18 and a18 979 endangered species. 10n n! (a) a9 a10 109 9! 1,000,000,000 362,880 1,562,500 567 (b) Decreasing (c) Factorials increase more rapidly than exponentials. S ection 9.1 n 115. an 11 a1 a2 1 n1 n Sequences 441 n 1 2 1.4142 a3 3 3 1.4422 a4 4 4 1.4142 a5 5 5 1.3797 a6 6 6 1.3480 Let y ln y lim n1 n. n→ 1 ln n n lim n→ Since ln y lim n→ ln n n e0 0, we have y lim n→ 1n n 0 1. Therefore, lim n→ 117. True 121. an 2 n n 1. 119. True an an 1 (a) a1 1 a7 8 a2 1 a8 13 8 a3 1 1 2 a9 21 13 a4 2 1 3 a10 34 21 3 2 5 a11 55 34 5 3 8 a12 89 55 144 1 an an 1 an 1 an 89 a6 1 55 a5 1 bn 34 (b) bn 5 13 (c) 1 an 1 ,n ≥ 1 an b1 1 1 b2 2 1 b3 1 21 an an an n→ n→ lim bn n→ 13 8 b7 21 13 3 2 b8 34 21 b4 5 3 b9 55 34 b5 8 5 b10 an 1 an , then lim 1 (d) If lim bn Since lim bn b6 1 1 n→ 1, bn 1 bn . 1 we have 89 55 1 2 1 . 1 2 0 2 1± 1 1 2 4 1± 5 2 1 Since an, and thus bn, is positive, 123. (a) a1 2 1.4142 a2 2 2 1.8478 a3 2 2 2 1.9616 a4 2 2 2 2 1.9904 a5 2 2 2 2 2 —CONTINUED— 1 (b) an 1.9976 2 an 1, n ≥ 2, a1 5 2 2 1.6180. 442 Chapter 9 Infinite Series 123. —CONTINUED— (c) We first use mathematical induction to show that an ≤ 2; clearly a1 ≤ 2. So assume ak ≤ 2. Then ak 2≤4 ak 2≤2 ak 1 ≤ 2. Now we show that an is an increasing sequence. Since an ≥ 0 and an ≤ 2, an 1 ≤0 2 an 2 an 2≤0 an an2 ≤ an an ≤ 2 an an ≤ an 2 1. Since an is a bounding increasing sequence, it converges to some number L, by Theorem 9.5. L⇒ lim an n→ 2 L⇒2 L ⇒L 2L L2 ⇒ L2 L 0⇒L 1 2 L L 1 1 2 4k (b) Since an is bounded and increasing, it has a limit L. (c) lim an . 1 k≤ 4k 2 ≤ 1 1 2 4k 21 4k 1 1 2 4k 1 1 2 4k 1 1 2 4k 4k So assume an ≤ an k≤ an k≤ an 1 ≤ 21 4k 4 1 1 2 4k 1 1 1 2 4k 4k 4k 4k 4k 2 . . Then k 1 2 1 2 1 2 4k k . an is increasing because an 1 1 2 4k an 1 2 1 an2 L L k Since L > 0, L 4k 1 2 1 1 4k 1 k 1 1 21 k k L2 ⇒L . 2 k k L ⇒ L2 ⇒ L 6, then Before proceeding to the induction step, note that 2 L implies that n→ [Note that if k 2, then an ≤ 3, and if k an ≤ 3.] Clearly, a1 0 1 125. (a) We use mathematical induction to show that an ≤ 2 4k an ≤0 k≤0 an2 ≤ an an ≤ an ≤ an k an 1. k 1 1± 1 2 1 2 4k 0 4k . . S ection 9.1 127. (a) f x sin x, an fx cos x, f 0 lim an (b) f 0 n→ f0 lim h h h→0 lim 1 n f1n 1n h→0 sin 1 n 1n lim n→ 1 f0 f0 fh h lim 1 lim n sin n→ 1 n n sin Sequences n→ lim n f n→ 1 n lim an n→ 129. (a) y (b) y y = ln x 2.0 y = ln x 2.0 1.5 1.5 1.0 1.0 0.5 0.5 2 3 x ... 4 n 2 n ln x dx < ln 2 n ... ln 3 ln n 4 ... x n+1 1 ln x dx > ln 2 1 ln 3 1 ln 1 (c) 3 ln x dx 3. . .n 2 x ln x x n 1 ln n! C n ln n ln n! n ln x dx ln nn n 1 1 From part (a): ln nn 1 < ln n! n n eln n n 1 n en n < n! n < n! 1 1 ln x dx n 1 ln n 1 n 1 1 ln n 1 n 1 n 1 From part (b): ln n n 1 1 1 n 1 en n nn (d) n e 1 n e1 < 1n < 1n lim n→ lim n→ n n 1 e1 n n < n! < 1 en n > ln n! n n 1 eln n 1 n > n! > n! 1 20 1 e n! n < n 11 ne 100: 0.3679 1n 1n 1 e1 1n 11 ne 50! 50 50: n 1n e lim n n→ 1 1 e 100 1n n n→ 1 e 1 e n By the Squeeze Theorem, lim 0.3897 50 1 e 1n 0.4152 20: n n n! < 20! 20 (e) n n! n 1 . e 1n 100! 100 0.3799 ... ln n 443 444 Chapter 9 Infinite Series an 133. If an is bounded, monotonic and nonincreasing, then a1 ≥ a2 ≥ a3 ≥ . . . ≥ an ≥ . . . . Then a1 ≤ a2 ≤ a3 ≤ . . . ≤ an ≤ . . . is a bounded, monotonic, nondecreasing sequence which converges by the first half of the theorem. Since an converges, then so does an . > 0, we must find M > 0 such that 131. For a given 1 < n3 L whenever n > M. That is, n3 > 1 or n > 1 13 . So, let > 0 be given. Let M be an integer satisfying M > 1 1 3. For n > M, we have 1 n> n3 > > 13 1 1 1 ⇒3 n3 n 1 n3 Thus, lim n→ 0. 2n n! 135. Tn 0<. We use mathematical induction to verify the formula. T0 1 1 2 T1 1 2 3 T2 2 4 6 Assume Tk Tk 2k. Then k! 1 k 1 k 4 Tk 5 k! k 4k 2k 5kk k2 5k 1! 2k 4k 1 4k k 1 Tk 1k 4k 4 1 4k 4k 1! 1k 1 1! 8 1 8 Tk 2k 1 4k 2k 4k 1k 2 4k 2! k 2k 2! 54 2 8k 1 1 2k 4k 2 1. By mathematical induction, the formula is valid for all n. Section 9.2 Series and Convergence 3. S1 1 S2 1 1 4 1 1 9 S4 1 1 9 1 16 1 1 4 1 9 1 16 1 25 5. S1 3 S2 3 3 2 4.5 S3 3 3 2 3 4 S4 3 3 4 3 8 5.625 S5 3 3 2 3 4 3 8 3 16 27 4 5.25 S4 3 9 2 27 4 81 8 S5 3 9 2 27 4 81 8 5.250 3 2 3 9 2 1.4236 S5 9 2 1.3611 1 4 3 1.2500 1 4 3 S2 1. S1 S3 S3 1.4636 7. 3 n 0 3 2 1.5 n Geometric series 5.8125 r 3 >1 2 Diverges by Theorem 9.6 4.875 243 16 10.3125 2 S ection 9.2 n 1000 1.055 9. n n 11. 0 n Geometric series r 1n 1.055 > 1 n 1 n 1 2n 1 n1 12 15. n n→ n→ 17. n 2n 1 2n 1 lim li...
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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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