SalasSV_10_03_ex_ans

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

## 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.

Ask a homework question - tutors are online