SalasSV_11_04_ex_ans - ANSWERS TO ODD-NUMBERED EXERCISES...

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Unformatted text preview: 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 A-78 ANSWERS TO ODD-NUMBERED EXERCISES 25. converges absolutely; ratio test 33. 0. 1104 35. 0. 001 37. 10 11 23. diverges; ak → 0 / 31. diverges; ak → 0 / 45. No. 27. diverges; ak = 1 for all k k 29. converges absolutely; comparison 43. (a) 4 (b) 6 1/k 2 39. N = 39, 998 41. n = 999 For instance, set a2k = 2/k and a2k +1 = 1/k . |ak | converges, |ak |2 = a2 converges (Exercise 49, Section 11.2). k 47. (a) Since (b) 1/k 2 is convergent, (−1)k is not absolutely convergent. k 49. See the proof of Theorem 11.7.2. 2n 51. (a) k =1 ( − 1)k +1 = S2n − Sn k (b) Let P = {n, n + 1, n + 2, . . . , 2n} be the regular partition of the interval [n, 2n] into n subintervals; form the Riemann sum by evaluating f (x) = 1/x at the right endpoints (= the lower sum). 2n (c) n (1/x) dx = ln (2n) − ln (n) = ln 2. SECTION 11.5 1. −1 + x + 1 x2 − 2 9. P0 (x) = 1, m 2k 14 x 24 3. − 1 x2 − 2 14 x 12 5. 1 − x + x2 − x3 + x4 − x5 P3 (x) = 1 − x + 3x2 + 5x3 k 7. x + 1 x3 + 3 n 25 x 15 P1 (x) = 1 − x, P2 (x) = 1 − x + 3x2 , n 11. k =0 ( − 1)k xk k! 23. |x| < 1. 513 13. k =0 n x where m = and n is even (2k !) 2 (79/48 ∼ 1. 646) = |c| < |x| 35. 27. 5/6 −4 sin 2c 5 x, 15 41. 15. k =0 rk x k! 17. 0.00002 29. 13/24 4 19. n = 9 (13/24 ∼ 0. 542) = 2 21. n = 6 31. 0. 17 25. 79/48 33. 39. 4e 5 x, 15 2c (5/6 ∼ 0. 833) = |c| < |x| 37. 3 sec c − 2 sec c 3 x, 3 43. |c| < |x| |c| < |x| 3c 2 − 1 3 x, 3(1 + c2 )3 (b) 2 |c| < |x| (c) 999 ( − 1)n+1 e−c n+1 x, (n + 1) ! |c| < |x| 1 xn+1 , (1 − c)n+2 45. (a) 4 47. (a) 1. 649 (b) 0. 368 ∞ 49. For 0 ≤ k ≤ n, P (k ) (0) = k !ak ; for k > n, P (k ) (0) = 0. Thus P (x) = k =0 P (k ) (0) xk . k! 51. d 2k ( sinh x) dx2k x =0 = sinh (0) = 0 : d 2k +1 ( sinh x) dx2k +1 ∞ k =0 x =0 = cosh (0) = 1 Therefore sinh x = x + ∞ x3 x5 + + ··· = 3! 5! ∞ 1 x 2 k +1 (2k + 1)! ∞ 53. k =0 ak k x, k! (−∞, ∞) 1+ 1− x −n e1/x2 1 3 1 3 55. k =0 ( − 1)ka2k 2k x, (2k )! 1 3 3 (−∞, ∞) 5 57. ln a + k =1 ( − 1)k −1 k x, kak (−a, a] 59. In 2 = In ∼2 1 + 1 = 3 3 + 1 5 1 3 = 842 1215 842 ∼ = 0. 693 1215 61. routine; use u = (x − t )k and dv = f (k +1) (t ) dt cxk e1/x2 63. (b) f (x) = and lim f (x) has the form ∞ x→0 ∞. Successive applications of L opital’s rule will finally produce a quotient of the form ’H ˆ x→0 , where k is a nonnegative integer and c is a constant. It follows that lim f (x) = 0. (c) f (0) = lim e −1/x2 −0 x→0 x x→0 = 0 by part (b). Assume that f (k ) (0) = 0. Then f (k +1) (0) = lim f (k ) ( x ) − 0 f (k ) ( x ) = lim . x→0 x x ...
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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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