Converges absolutely ratio test 21 diverges ak 0 15

Unformatted text preview: onally; 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 =...
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