73 let r a sin n the proof for r a cos n is similar

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Unformatted text preview: m = 1/e < 1, converges k→+∞ k→+∞ e(k + 1) e ln k 35. Ratio Test, ρ = lim k+1 1 = lim = 0, converges 2k+1 2k+1 k→+∞ e k→+∞ 2e 36. Ratio Test, ρ = lim 37. Ratio Test, ρ = lim k→+∞ k+5 = 1/4, converges 4(k + 1) k3 = 1, +1 k→+∞ k 3 lim √ k→+∞ k = 1, k2 + k 383 Chapter 11 38. Root Test, ρ = lim ( k→+∞ 39. diverges because lim kk 1 ) = lim = 1/e, converges k→+∞ (1 + 1/k )k k+1 k→+∞ ∞ 40. k=1 √ k ln k = 3+1 k +∞ 2 41. 42. 1 = 1/4 = 0 4 + 2−k √ k ln k ln k k ln k k ln k because ln 1 = 0, 3 < = 2, 3+1 3 k k +1 k k √ ∞ k=2 ln x 1 ln x − dx = lim − →+∞ x2 x x tan−1 k π/2 < 2, k2 k ∞ k=1 2 π/2 converges so k2 2 5k 5k + 5k 5k + k < = , k! + 3 k! k! ∞ 1 = (ln 2 + 1) so 2 ∞ k=1 k=1 k=2 ln k converges and so does k2 ∞ k=1 √ k ln k . k3 + 1 tan−1 k converges k2 ∞ 5k k! 2 ∞ converges (Ratio Test) so k=1 5k + k converges k! + 3 (k + 1)2 = 1/4, converges k→+∞ (2k + 2)(2k + 1) 43. Ratio Test, ρ = lim 2(k + 1)2 = 1/2, converges k→+∞ (2k + 4)(2k + 3) 44. Ratio Test, ρ = lim 45. uk = k! k+1 , by the Ratio Test ρ = lim = 1/2; converges k→+∞ 2k + 1 1 · 3 · 5 · · · (2k − 1) 46. uk = 1 · 3 · 5 · · · (2k − 1) 1 , by the Ratio Test ρ = lim = 0; converges k→+∞ 2k (2k − 1)! 47. Root Test: ρ = lim k→+∞ 1 (ln k )1/k = 1/3, converges 3 k+1 π (k + 1) = π , diverges = lim π 1+1/k k→+∞ k k→+∞ k 48. Root Test: ρ = lim sin(π/k ) = 1 and 49. (b) ρ = lim k→+∞ π/k 50. (a) cos x ≈ 1 − x2 /2, 1 − cos √ ∞ π/k diverges k=1 1 k ≈ 1 2k 2 (b) ρ = lim k→+∞ 1 − cos(1/k ) = 2, converges 1/k 2 1 d 1 g (x) = √ − = 0 when x = 4. Since lim g (x) = lim g (x) = +∞ x→+∞ x→0+ dx 2x x √ √ it follows that g (x) has its minimum at x = 4, g (4) = 4 − ln 4 > 0, and thus x − ln x > 0 for x > 0. √ ∞ ∞ ln k 1 k 1 ln k (a) < 2 = 3/2 , converges so converges. k2 k k2 k k 3/2 51. Set g (x) = x − ln x; k=1 (b) 1 1 >, 2 (ln k ) k ∞ k=2 1 diverges so k k=1 ∞ k=2 1 diverges. (ln k )2 Exercise Set 11.7 384 52. By the Root Test, ρ = lim k→+∞ α α = α = α, the series converges if α < 1 and diverges 1 (k 1/k )α ∞ if α > 1. If α = 1 then the series is 1/k which diverges. k=1 bk . Then a1 + a2 + · · · + an ≤ b1 + b2 + · · · + bn ≤ M ; 53. (a) If bk converges, then set M = apply Theorem 11.4.6 to get convergence of ak . (b) Assume the contrary, that bk converges; then use part (a) of the Theorem to show that ak converges, a contradiction. 54. (a) If lim (ak /bk ) = 0 then for k ≥ K , ak /bk < 1, ak < bk so k→+∞ ak converges by the Comparison Test. (b) If lim (ak /bk ) = +∞ then for k ≥ K , ak /bk > 1, ak > bk so k→+∞ Comparison Test. EXERCISE SET 11.7 1. ak+1 < ak , lim ak = 0, ak > 0 k→+∞ 2. 1 k ak+1 < for k > 0, so {ak } is decreasing and tends to zero. = ak 3(k + 1) 3 3. diverges because lim ak = lim k→+∞ k→+∞ k+1 = 1/3 = 0 3k + 1 k+1 4. diverges because lim ak = lim √ = +∞ = 0 k→+∞ k→+∞ k+1 5. {e−k } is decreasing and lim e−k = 0, converges k→+∞ 6. ln k k ln k = 0, converges k→+∞ k is decreasing and lim (3/5)k+1 = 3/5, converges absolutely k→+∞ (3/5)k 7. ρ = lim 8. ρ = lim k→+∞ 2 = 0, converges absolutely k+1 3k 2 = 3, diverges k→+∞ (k + 1)2 9. ρ = lim 10. ρ = lim k→+∞ k+1 = 1/5, converges absolutely 5k (k + 1)3 = 1/e, converges absolutely k→+∞ ek 3 11. ρ = lim (k + 1)k+1 k ! = lim (1 + 1/k )k = e, diverges k→+∞ (k + 1)!k k k→+∞ 12. ρ = lim ak diverges by the 385 Chapter 11 ∞ (−1)k+1 converges by the Alternating Series Test but 3k 13. conditionally convergent, k=1 ∞ 1 14. absolutely convergent, k=1 k 4/3 ∞ k=1 1 diverges 3k converges 15. divergent, lim ak = 0 k→+∞ 16. absolutely convergent, Ratio Test for absolute convergence ∞ 17. k=1 cos kπ = k ∞ k=1 (−1)k is conditionally convergent, k ∞ k=1 (−1)k converges by the Alternating Series k ∞ Test but 1/k diverges. k=1 ∞ 18. conditionally convergent, k=3 (−1)k ln k converges by the Alternating Series Test but k diverges (Limit Comparison Test with ∞ (−1)k+1 19. conditionally convergent, k=1 ∞ Alternating Series Test but k=1 ∞ 20. conditionally convergent, ∞ k=1 k=1 ∞ k=3 ln k k 1/k ). k+2 converges by the k (k + 3) k+2 diverges (Limit Comparison Test with k (k + 3) 1/k ) (−1)k+1 k 2 converges by the Alternating Series Test but k3 + 1 k2 diverges (Limit Comparison Test with 3+1 k (1/k )) ∞ sin(kπ/2) = 1 + 0 − 1 + 0 + 1 + 0 − 1 + 0 + · · ·, divergent ( lim sin(kπ/2) does not exist) 21. k→+∞ k=1 ∞ 22. absolutely convergent, k=1 | sin k | converges (compare with k3 ∞ 23. conditionally convergent, k=2 1/k 3 ) (−1)k converges by the Alternating Series Test but k ln k ∞ k=2 1 diverges k ln k (Integral Test) ∞ 24. conditionally convergent, ∞ k=1 k=1 1 k (k + 1) (−1)k k (k + 1) converges by the Alternating Series Test but dive...
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