SalasSV_11_02_ex_ans - A-76 ANSWERS TO ODD-NUMBERED...

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Unformatted text preview: A-76 ANSWERS TO ODD-NUMBERED EXERCISES CHAPTER 11 SECTION 11.1 1. 12 10 3. 15 ( − 1)k −1 2 5. − 15 7 11 7. k =1 (2k − 1) 35 9. k =1 k (k + 1) n 10 11. k =1 Mk xk 13. k =3 1 , 2k 7 i =0 1 2 i +3 25. 1 2 15. k =3 10 3 k , k +1 i =0 ( − 1)i i+3 i+4 ∞ 17. let k = n + 3 7 7 = 10k 9 ∞ 19. let k = n − 3 24 8 = 100k 33 21. 140 23. 680 27. 11 18 29. 31. − 3 2 33. 24 35. k =1 37. k =1 39. 62 1 ∞ 45 687 + = 100 100 k =1 100k 1100 41. Let x = a1a2 . . . an a1a2 . . . an . . . . Then x= ∞ k =1 ∞ a1 a2 · · · an 1 = a1 a2 · · · an n (10n )k k =1 10 k = a1 a2 · · · an ∞ 1 − 1 = a1 a2 · · · an . 1 1 − 10n 10n − 1 x k +1 ∞ 43. ∞ ∞ 1 1 = = ( − x )k = (−1)k xk 1+x 1 − ( − x ) k =0 k =0 45. k =0 47. k =0 (−1)k x2k +1 ∞ 49. k =0 3 2 ∞ k ; geometric series with r = r 100 −k 3 >1 2 51. lim k →∞ k +1 k k =e=0 53. (a) 87. 9935; diverges (b) 8.17837; diverges (c) 1.64443; converges, S ∼ 1. 645 = (d) 1.71828; converges, the sum is e − 1 55. 18 57. k =1 nk 1 + 59. $9 61. 32 63. lim Sn = L = n→∞ ∞ ∞ k =0 ak . Thus lim Rn = lim n→∞ ∞ k =1 n→∞ Sn − ∞ k =0 ak = lim Sn − L = 0. n→∞ 65. 1 + ( − 1)n , 2 n = 0, 1, 2, . . . 67. Sn = ln k =1 n k =1 k +1 k = [ln (k + 1) − ln k ] = ln(n + 1) → ∞ 69. (a) Sn = ∞ (dk − dk +1 ) = d1 − dn+1 → d1 = ∞ k =1 (b) (i) k =1 √ √ k +1− k k (k + 1) 1 1 √ −√ k +1 k 75. N = =1 ∞ (ii) k =1 ∞1 2k + 1 = 2 (k + 1)2 2k k =1 2 1 1 − k2 (k + 1)2 = 1 2 71. N = 5 73. N = 9999 ln( [1 − x]) , where [[ ln|x| ] denotes the greatest integer function. SECTION 11.2 1. converges; comparison 1/k 2 1/k 3. converges; comparison 9. converges; integral test / 17. diverges; ak → 0 2k 5k 1/k 2 5. diverges; comparison 11. diverges; p-series with p = 1/k 2 1/(k + 1) 2 3 7. diverges; limit comparison 15. diverges; comparison 1/k ≤1 13. diverges; ak → 0 / 19. converges; limit comparison 1/k 21. diverges; integral test 1/k 3/2 23. converges; limit comparison with 29. converges; integral test 25. diverges; limit comparison 3/k 2 27. converges; limit comparison 2/k 2 31. converges; comparison 33. converges; comparison 35. (a) converges; S ∼ 3. 18459 = (b) converges; S ∼ 1. 1752 = (c) converges; S = 2 37. p > 1 ∞ 39. (a) The improper integral 0 ∞ e−αx dx = 1 converges. α ∞ (b) The improper integral 0 x e−αx dx = 1 converges. α2 (c) The improper integral 0 xn e−αx dx = n! converges. α n +1 ANSWERS TO ODD-NUMBERED EXERCISES 41. (a) 1.1777 1 < 1. 209 43. (a) 1/101 < R100 < 1/100 k3 47. (a) If ak /bk → 0, then ak /bk < 1 for all k ≥ K for some K . But then ak < bk for all k ≥ K and, since comparison test, 11.2.5] (b) 0. 02 < R4 < 0. 0313 (c) 1. 1977 < k =1 ∞ A-77 (b) 1.082 (b) 10, 001 bk converges, 45. (a) 15 ak converges. [The basic (b) Similar to (a) except that this time we appeal to part (ii) of Theorem 11.2.5. (c) ak = ak = (d) bk = bk = 49. (a) Since 1 converges, k2 1 converges, k2 1 √ diverges, k 1 √ diverges, k k →∞ bk = bk = ak = ak = 1 converges, k 3/2 1 √ diverges, k 1 converges, k2 1 diverges, k 1/k 2 1 = √ →0 1/k 3/2 k 1 1/k 2 √ = 3/2 → 0 k 1/ k 1 1/k 2 √ = 3/2 → 0 k 1/ k 1/k 1 √ = √ →0 1/ k k ak converges, lim ak = 0. Therefore there exists a positive integer N such that 0 < ak < 1 for k ≥ N . a2 converges by the comparison test. k 1/k 2 converges and ∞ Thus, for k ≥ N , a2 < ak and so k (b) 1/k 4 converges and 51. 0 < L − n k =1 ak may either converge or diverge. 1/k 2 converges; ∞ k =n+1 1/k diverges. f ( k ) = L − Sn = f (k ) < n f (x) dx 53. N = 3. 55. (a) Set f (x) = x1/4 − ln x. Then f (x) = 1 −3/4 1 1 1/4 x −= (x − 4). Since f (e12 ) = e3 − 12 > 0 and f (x) > 0 for x > e12 , we know that k 1/4 > 4 x 4x 1 ln k 1 ln k ln k and therefore 5/4 > 3/2 for sufficiently large k . Since is a convergent p-series, converges by the basic comparison test. k k k 5/4 k 3/2 ln x 1 (b) By L opital’s rule lim ’H ˆ =0 x→∞ x3/2 x5/4 SECTION 11.3 1. converges; ratio test 3. converges; root test 5. diverges; ak → 0 / 7. diverges; limit comparison √ 11. diverges; limit comparison 1/ k 13. diverges; ratio test 15. converges; comparison 1/k 3/2 1/k 2 19. diverges; integral test 1/k 3/2 21. diverges; ak → e−100 = 0 29. converges; ratio test 1/k 9. converges; root test 17. converges; comparison 25. converges; ratio test 33. converges; ratio test 23. diverges; limit comparison 4 27 1/k 27. converges; comparison 35. converges; root test 43. 10 81 31. converges; ratio test: ak +1 /ak → 37. converges; root test 39. converges; ratio test 47. p ≥ 2 41. (a) converges; ak +1 /ak → 0 (b) diverges; ak +1 /ak → 2 45. The series k! k! converges. Therefore lim k = 0 by Theorem 11.1.5. k →∞ k kk 49. Set bk = ak r k . If (ak )1/k → ρ and ρ < 1/r , then (bk )1/k = (ak r k )1/k = (ak )1/k r → ρ r < 1 and thus, by the root test, bk = ak r k converges. SECTION 11.4 / 1. diverges; ak → 0 3. diverges; ak → 0 / 1/k 5. (a) does not converge absolutely; integral test (b) converges conditionally; Theorem 11.4.3 7. diverges; limit comparison 11. diverges; ak → 0 / 9. (a) does not converge absolutely; limit comparison 1/k (b) converges conditionally; Theorem 11.4.3 √ 13. (a) does not converge absolutely; comparison 1 k + 1 (b) converges conditionally; Theorem 11.4.3 sin π 4k 2 1/k ≤ π π = 4k 2 4 1 k2 (| sin x| ≤ |x|) 17. converges absolutely; ratio test 21. diverges; ak → 0 / 15. converges absolutely (terms already positive); 19. (a) does not converge absolutely; limit comparison (b) converges conditionally; Theorem 11.4.3 ...
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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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