SalasSV_10_03_ex_ans - A-72 ANSWERS TO ODD-NUMBERED...

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Unformatted text preview: A-72 ANSWERS TO ODD-NUMBERED EXERCISES (−1)n−1 , 2n − 1 n2 + 1 , n SECTION 10.2 1. an = 2 + 3(n − 1), 7. an = n 1/n n = 1, 2, 3, . . . 3. an = n = 1, 2, 3, . . . 5. an = n = 1, 2, 3, . . . if n = 2k − 1, where k = 1, 2, 3, . . . if n = 2k , 3 2 9. decreasing; bounded below by 0 and above by 2 13. decreasing; bounded below by 0 and above by 0.9 17. increasing; bounded below by 4 5 11. not monotonic; bounded below by 0 and above by 15. increasing; bounded below by 19. increasing; bounded below by 1 2 2 51 but not bounded above but not bounded above √ 5 and above by 2 21. increasing; bounded below by 0 and above by ln 2 √ 3 and above by 2 1 2 23. decreasing; bounded below by 1 and above by 4 25. increasing; bounded below by 27. decreasing; bounded above by −1 but not bounded below 31. decreasing; bounded below by 0 and above by 1 35. decreasing; bounded below by 0 and above by 39. increasing; bounded below by 3 4 1 2 29. increasing; bounded below by and above by 1 5 6 1 3 33. decreasing: bounded below by 0 and above by 37. decreasing; bounded below by 0 and above by 41. for n ≥ 5, ln 3 but not bounded above a n +1 5n+1 q n! 5 = = <1 an (n + 1)! 5n n+1 43. boundedness : 0 < (c + d ) < (2d ) = 2 d ≤ 2d . monotonicity : an+1 = cn+1 + d n+1 = ccn + dd n < (cn + d n )1/n cn + (cn + d n )1/n d n n +1 = (cn + d n )1+(1/n) = (cn + d n )(n+1)/n = an+1 . n n n 1/n n 1/n 1/n and 25 2 an+1 < an . Sequence is not nonincreasing : a1 = 5 < = a2 . Taking the n + 1-th root of each side we have an+1 < an . The sequence is monotonic decreasing. 45. a1 = 1, a2 = 1 , a3 = 1 , a4 = 2 6 1 , a5 24 = 1 , a6 120 = 1 ; 720 an = 1/n! 47. a1 = a2 = a3 = a4 = a5 = a6 = 1; an = 1 49. a1 = 1, a2 = 3, a3 = 5, a4 = 7, a5 = 9, a6 = 11; an = 2n − 1 51. a1 = 1, a2 = 4, a3 = 9, a4 = 16, a5 = 25, a6 = 36; an = n2 53. a1 = 1, a2 = 1, a3 = 2, a4 = 4, a5 = 8, a6 = 16; an = 2n−2 for n ≥ 3 55. a1 = 1, a2 = 3, a3 = 5, a4 = 7, a5 = 9, a6 = 11; an = 2n − 1 57. First a1 = 21 − 1 = 1. Next suppose ak = 2k − 1 for some k ≥ 1. Then ak +1 = 2ak + 1 = 2(2k − 1) + 1 = 2k +1 − 1. 1 k k +1 k +1 k k +1 = 1. Next suppose ak = k −1 for some k ≥ 1. Then ak +1 = ak = = . 20 2 2k 2k 2k −1 2k √ n−1 1 − rn 53 3 2 n−1 61. (a) n (b) 63. (a) 150 3 (b) 65. {an } is increasing; limit 1 4 2 1−r 2 4 √ √ √ √ 67. (a) a2 = 1 + a1 = 2 > 1 = a1 . Assume ak = 1 + ak −1 > ak −1 . Then ak +1 = 1 + ak > 1 + ak −1 = ak . 59. First a1 = Thus {an } is an increasing sequence. √ √ (b) an = 1 + an−1 < 1 + an , since an−1 < an . an − √ √ √ 1+ 5 3+ 5 √ √ √ an − 1 < 0, or ( an )2 − an − 1 < 0, which implies (solve the inequality) that an < , hence an < for all n. 2 2 (c) lub {an } ∼ 2. 6180. = n→∞ 69. (b), (c) It appears that lim an = 1; the sequence is bounded. SECTION 10.3 1. diverges 3. converges to 0 15. converges to 1 17. converges to 5. converges to 1 4 9 7. converges to 0 9. converges to 0 11. diverges 13. converges to 0 √ 1 19. converges to 2 2 21. diverges 23. converges to 1 25. converges to 0 ANSWERS TO ODD-NUMBERED EXERCISES 27. converges to 35. b < 1 2 A-73 29. converges to e2 31. diverges 33. (a) 0 (b) π 4 (c) 1 2 √ √ √ nn a + bn = b n (a/b)n + 1 < b n 2. Since 21/n → 1 as n → ∞, it follows that n an + bn → b by the pinching theorem. 39. 1+ 1 n n +1 37. Use |(an + bn ) − (L + M )| ≤ |an − L| + |bn − M |. = 1+ 1 n n 1+ 1 1 . Note that 1 + n n n → e and 1+ 1 n → 1. 41. Imitate the proof given for the nondecreasing case in Theorem 10. 3. 6. 43. Let > 0. Choose k so that, for n ≥ k , L − < cn < L + and an ≤ bn ≤ cn . L − < an < L + , For such n, L − < bn < L + . 45. Let > 0. Since an → L, there exists a positive integer N such that L − < an < L + for all n ≥ N . Now an ≤ M for all n, so L − < M , or L < M + . Since is arbitrary, L ≤ M . 47. Assume an → 0. Let > 0. There exists a positive integer N such that |an − 0| < Now assume |an | → 0. Since −|an | ≤ an ≤ |an |, an → 0 by the pinching theorem. 49. By the continuity of f , f (L) = f ( lim an ) = lim f (an ) = lim an+1 = L. n→∞ n→∞ n→∞ for all n ≥ N . Since ||an | − 0| ≤ |an − 0|, it follows that |an | → 0. 53. converges to 0 51. Use Theorem 10.3.12 with f (x) = x1/p . 63. L = 0, n = 7 55. converges to 0 57. diverges √ 3+ 5 67. (a) (b) 3 2 69. (a) a2 0.540302 a3 0.857553 59. L = 0, n = 32 61. L = 0, n = 4 65. L = 0, n = 65 a4 0.654290 a5 0.793480 a6 0.701369 a7 0.763960 a8 0.722102 a9 0.750418 a10 0.731404 (b) 0. 739085; it is the fixed point of f (x) = cos x. SECTION 10.4 1. converges to 1 15. converges to 1 27. converges to e−1 3. converges to 0 17. converges to π 29. converges to 0 5. converges to 0 19. converges to 1 7. converges to 0 9. converges to 1 23. diverges 11. converges to 0 25. converges to 0 37. (a) 2 13. converges to 1 21. converges to 0 33. converges to e x 31. converges to 0 35. converges to 0 (b) 0 (c) 1 √ √ √ √ √ n+1− n √ 1 39. n + 1 − n = √ √ ( n + 1 + n) = √ √ →0 n+1+ n n+1+ n 41. (b) 2π r . As n → ∞, the perimeter of the polygon tends to the circumference of the circle. 47. (a) mn+1 − mn = 43. 1 2 45. 1 8 1 1 (a1 + · · · + an + an+1 ) − (a1 + · · · + an ) n+1 n 1 = nan+1 − (a1 + · · · + an ) > 0 since {an } is increasing. n(n + 1) |a1 + · · · + aj | |a1 + · · · + aj | n−j + → 0, and therefore for n sufficiently large (b) We begin with the hint mn < . Since j is fixed, n 2 n n |a1 + · · · + aj | n−j < . Since < , we see that, for n sufficiently large, |mn | < . This shows that mn → 0. n 2 2 n 2 √ √ √ √ √ 49. (a) Let S be the set of positive integers n(n ≥ 2) for which the inequalities hold. Since ( b)2 − 2 ab + ( a)2 = ( b − a)2 > 0, it follows that √ √ a1 + b 1 a1 + b1 a+b > ab and a1 > b1 . Now a2 = < a1 and b2 = a1 b1 > b1 . Also, by the argument above, a2 = > a1 b1 = b2 , and so 2 2 2 ak + bk ak + a k ak + bk a1 > a2 > b2 > b1 . Thus 2 ∈ S . Assume that k ∈ S . Then ak +1 = < = ak , bk +1 = ak bk > b2 = bk , and ak +1 = > k 2 2 2 ak bk = bk +1 . Thus k + 1 ∈ S . Therefore the inequalities hold for all n ≥ 2. (b) {an } is a decreasing sequence which is bounded below. {bn } is an increasing sequence which is bounded above. Let La = lim an , Lb = lim bn . Then an = n→∞ n→∞ an−1 + bn−1 La + L b implies La = and La = Lb . 2 2 ...
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This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston - Downtown.

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