The singular point was discussed in part a there is a

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online