# The singular point was discussed in part a there is a

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Unformatted text preview: rges (Limit Comparison Test with 1/k ) ∞ 25. absolutely convergent, (1/ ln k )k converges by the Root Test k=2 Exercise Set 11.7 386 ∞ (−1)k+1 √ converges by the Alternating Series Test but k+1+ k k=1 √ diverges (Limit Comparison Test with 1/ k ) √ 26. conditionally convergent, ∞ √ k=1 1 √ k+1+ k 27. conditionally convergent, let f (x) = +∞ {ak }k=2 = k2 + 1 k3 + 2 +∞ is nonincreasing, lim ak = 0; the series converges by the k→+∞ k=2 ∞ Alternating Series Test but k=2 ∞ 28. k=1 k cos kπ = k2 + 1 ∞ k=1 x2 + 1 x(4 − 3x − x3 ) then f (x) = ≤ 0 for x ≥ 2 so 3+2 x (x3 + 2)2 2 k +1 diverges (Limit Comparison Test with k3 + 2 (−1)k k is conditionally convergent, k2 + 1 ∞ Alternating Series Test but k=1 ∞ k=1 1/k ) (−1)k k converges by the k2 + 1 k diverges 2+1 k k+1 =0 k→+∞ (2k + 1)(2k ) 29. absolutely convergent by the Ratio Test, ρ = lim 30. divergent, lim ak = +∞ 31. |error| < a8 = 1/8 = 0.125 32. |error| < a6 = 1/6! < 0.0014 √ 33. |error| < a100 = 1/ 100 = 0.1 k→+∞ 34. |error| < a4 = 1/(5 ln 5) < 0.125 35. |error| < 0.0001 if an+1 ≤ 0.0001, 1/(n + 1) ≤ 0.0001, n + 1 ≥ 10, 000, n ≥ 9, 999, n = 9, 999 36. |error| < 0.00001 if an+1 ≤ 0.00001, 1/(n + 1)! ≤ 0.00001, (n + 1)! ≥ 100, 000. But 8! = 40, 320, 9! = 362, 880 so (n + 1)! ≥ 100, 000 if n + 1 ≥ 9, n ≥ 8, n = 8 √ √ 37. |error| < 0.005 if an+1 ≤ 0.005, 1/ n + 1 ≤ 0.005, n + 1 ≥ 200, n + 1 ≥ 40, 000, n ≥ 39, 999, n = 39, 999 38. |error| < 0.05 if an+1 ≤ 0.05, 1/[(n + 2) ln(n + 2)] ≤ 0.05, (n + 2) ln(n + 2) ≥ 20. But 9 ln 9 ≈ 19.8 and 10 ln 10 ≈ 23.0 so (n + 2) ln(n + 2) ≥ 20 if n + 2 ≥ 10, n ≥ 8, n = 8 39. ak = 40. ak = 3 3 3/4 = 0.5 , |error| < a11 = 12 < 0.00074; s10 ≈ 0.4995; S = 2k+1 2 1 − (−1/2) 2 3 k−1 , |error| < a11 = 2 3 10 < 0.01735; s10 ≈ 0.5896; S = 1 = 0.6 1 − (−2/3) 1 1 , an+1 = ≤ 0.005, (2n + 1)! ≥ 200, 2n + 1 ≥ 6, n ≥ 2.5; n = 3, (2k − 1)! (2n + 1)! s3 = 1 − 1/6 + 1/120 ≈ 0.84 41. ak = 42. ak = 1 1 , an+1 = ≤ 0.005, (2n)! ≥ 200, 2n ≥ 6, n ≥ 3; n = 3, s3 ≈ 0.54 (2k − 2)! (2n)! 387 Chapter 11 43. ak = 44. ak = 1 1 , an+1 = ≤ 0.005, (n + 1)2n+1 ≥ 200, n + 1 ≥ 6, n ≥ 5; n = 5, s5 ≈ 0.41 k 2k (n + 1)2n+1 (2k − 1)5 1 1 , an+1 = ≤ 0.005, 5 + 4(2n + 1) + 4(2k − 1) (2n + 1) (2n + 1)5 + 4(2n + 1) ≥ 200, 2n + 1 ≥ 3, n ≥ 1; n = 1, s1 = 0.20 45. (c) ak = 1 1 , an+1 = ≤ 10−2 , 2n + 1 ≥ 100, n ≥ 49.5; n = 50 2k − 1 2n + 1 ∞ (1/k p ) converges if p > 1 and diverges if p ≤ 1, so 46. (−1)k k=1 1 converges absolutely if p > 1, kp and converges conditionally if 0 < p ≤ 1 since it satisﬁes the Alternating Series Test; it diverges for p ≤ 0 since lim ak = 0. k→+∞ 47. 1 + 1 1 1 1 1 1 1 + 2 + ···= 1 + 2 + 2 + ··· − 2 + 2 + 2 + ··· 2 3 5 2 3 2 4 6 = 48. 1 + 1 1 π2 π2 1 1 π2 π2 − 2 1 + 2 + 2 + ··· = − = 6 2 2 3 6 46 8 1 1 1 1 1 1 1 + 4 + ···= 1 + 4 + 4 + ··· − 4 + 4 + 4 + ··· 34 5 2 3 2 4 6 = 1 1 π4 π4 1 1 π4 π4 − 4 1 + 4 + 4 + ··· = − = 90 2 2 3 90 16 90 96 49. Every positive integer can be written in exactly one of the three forms 2k − 1 or 4k − 2 or 4k , so a rearrangement is 1− = 50. (a) 11 − 24 11 − 24 + + 111 −− 368 11 − 68 + + 1 1 1 − − 5 10 12 1 1 − 10 12 + ··· + 1.5 + ··· + 1 1 1 − − 2k − 1 4k − 2 4k 1 1 − 4k − 2 4k (b) + ··· = + ··· 1 ln 2 2 Yes; since f (x) is decreasing for x ≥ 1 and lim f (x) = 0, so the series x→+∞ satisﬁes the Alternating Series Test. 0 10 0 51. (a) The distance d from the starting point is 180 180 180 11 1 + − ··· − = 180 1 − + − · · · − . 2 3 1000 23 1000 1 11 diﬀers from ln 2 by less than 1/1001 so From Theorem 11.7.2, 1 − + − · · · − 23 1000 180(ln 2 − 1/1001) < d < 180 ln 2, 124.58 < d < 124.77. d = 180 − Exercise Set 11.8 388 180 180 180 + + ··· + , and from inequality (2) in (b) The total distance traveled is s = 180 + 2 3 1000 Section 11.4, 1001 1 180 dx < s < 180 + x 1000 1 180 dx x 180 ln 1001 < s < 180(1 + ln 1000) 1243 < s < 1424 52. (a) Suppose Σ|ak | converges, then lim |ak | = 0 so |ak | < 1 for k ≥ K and thus |ak |2 < |ak |, k→+∞ a2 < |ak | hence Σa2 converges by the Comparison Test. k k (b) Let ak = 1 , then k a2 converges but k ak diverges. EXERCISE SET 11.8 uk+1 = |x|, so the interval of convergence is −1 < x < 1, converges uk 1. geometric series, ρ = lim k→+∞ there to 1 (the series diverges for x = ±1) 1+x uk+1 = |x|2 , so the interval of convergence is −1 < x < 1, converges uk 2. geometric series, ρ = lim k→+∞ there to 1 (the series diverges for x = ±1) 1 − x2 3. geometric series, ρ = lim k→+∞ there to uk+1 = |x − 2|, so the interval of convergence is 1 < x < 3, converges uk 1 1 = (the series diverges for x = 1, 3) 1 − (x − 2) 3−x 4. geometric series, ρ = converges there to lim k→+∞ uk+1 = |x + 3...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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