K k 1 k k0 k0 1 2k 1 1k converges radius

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Unformatted text preview: + 1/n)(2 + 1/n) = 1/3 lim an = lim n→+∞ n→+∞ 6 38. (a) 1, 39. Let an = 0, bn = sin2 n 1 , cn = ; then an ≤ bn ≤ cn , lim an = lim cn = 0, so lim bn = 0. n→+∞ n→+∞ n→+∞ n n n n 1+n 3 , cn = ; then (for n ≥ 2), an ≤ bn ≤ 2n 4 lim an = lim cn = 0, so lim bn = 0. 40. Let an = 0, bn = n→+∞ n→+∞ lim an = lim e2 n→+∞ n = cn , n→+∞ 41. (a) a1 = (0.5)2 , a2 = a2 = (0.5)4 , . . . , an = (0.5)2 1 (c) n/2 + n 2n n ln(0.5) n→+∞ n = 0, since ln(0.5) < 0. (d) Replace 0.5 in part (a) with a0 ; then the sequence converges for −1 ≤ a0 ≤ 1, because if a0 = ±1, then an = 1 for n ≥ 1; if a0 = 0 then an = 0 for n ≥ 1; and if 0 < |a0 | < 1 then n−1 ln a1 = 0 since 0 < a1 < 1. a1 = a2 > 0 and lim an = lim e2 0 n→+∞ n→+∞ 42. f (0.2) = 0.4, f (0.4) = 0.8, f (0.8) = 0.6, f (0.6) = 0.2 and then the cycle repeats, so the sequence does not converge. 30 43. (a) 0 5 0 ln(2x + 3x ) 2x ln 2 + 3x ln 3 = lim x→+∞ x→+∞ x 2x + 3x (b) Let y = (2x + 3x )1/x , lim ln y = lim x→+∞ (2/3)x ln 2 + ln 3 = ln 3, so lim (2n + 3n )1/n = eln 3 = 3 x→+∞ n→+∞ (2/3)x + 1 = lim Exercise Set 11.2 364 44. Let f (x) = 1/(1 + x), 0 ≤ x ≤ 1. Take ∆xk = 1/n and x∗ = k/n then k n an = k=1 1 (1/n) = 1 + (k/n) n k=1 1 ∆xk so lim an = n→+∞ 1 + x∗ k 1 0 1 dx = ln(1 + x) 1+x 1 = ln 2 0 n ln n 1 ln n 1 dx = , lim an = lim = lim = 0, n→+∞ n→+∞ n − 1 n→+∞ n n−1 1x ln n apply L’Hˆpital’s Rule to o , converges n−1 45. an = 1 n−1 46. (a) If n ≥ 1, then an+2 = an+1 + an , so an+2 an =1+ . an+1 an+1 (c) With L = lim (an+2 /an+1 ) = lim (an+1 /an ), L = 1 + 1/L, L2 − L − 1 = 0, n→+∞ n→+∞ √ √ L = (1 ± 5)/2, so L = (1 + 5)/2 because the limit cannot be negative. 47. 1 1 −0 = < n n if n > 1/ (a) 1/ = 1/0.5 = 2, N = 3 (b) 1/ = 1/0.1 = 10, N = 11 (c) 1/ = 1/0.001 = 1000, N = 1001 48. 1 n −1 = < n+1 n+1 if n + 1 > 1/ , n > 1/ − 1 (a) 1/ − 1 = 1/0.25 − 1 = 3, N = 4 (b) 1/ − 1 = 1/0.1 − 1 = 9, N = 10 (c) 1/ − 1 = 1/0.001 − 1 = 999, N = 1000 49. (a) (b) 1 1 −0 = < n n if n > 1/ , choose any N > 1/ . 1 n −1 = < n+1 n+1 50. If |r| < 1 then if n > 1/ − 1, choose any N > 1/ − 1. lim rn = 0; if r > 1 then n→+∞ lim rn = +∞, if r < −1 then rn oscillates between n→+∞ positive and negative values that grow in magnitude so lim rn does not exist for |r| > 1; if r = 1 n→+∞ then lim 1n = 1; if r = −1 then (−1)n oscillates between −1 and 1 so lim (−1)n does not exist. n→+∞ n→+∞ EXERCISE SET 11.2 1. an+1 − an = 1 1 1 − =− < 0 for n ≥ 1, so strictly decreasing. n+1 n n(n + 1) 2. an+1 − an = (1 − 3. an+1 − an = 1 1 1 ) − (1 − ) = > 0 for n ≥ 1, so strictly increasing. n+1 n n(n + 1) n 1 n+1 − = > 0 for n ≥ 1, so strictly increasing. 2n + 3 2n + 1 (2n + 1)(2n + 3) 365 Chapter 11 4. an+1 − an = n 1 n+1 − =− < 0 for n ≥ 1, so strictly decreasing. 4n + 3 4n − 1 (4n − 1)(4n + 3) 5. an+1 − an = (n + 1 − 2n+1 ) − (n − 2n ) = 1 − 2n < 0 for n ≥ 1, so strictly decreasing. 6. an+1 − an = [(n + 1) − (n + 1)2 ] − (n − n2 ) = −2n < 0 for n ≥ 1, so strictly decreasing. 7. (n + 1)(2n + 1) 2n2 + 3n + 1 (n + 1)/(2n + 3) an+1 = = > 1 for n ≥ 1, so strictly increasing. = an n/(2n + 1) n(2n + 3) 2n2 + 3n 8. an+1 2n+1 1 + 2n 2 + 2n+1 1 = · = =1+ > 1 for n ≥ 1, so strictly increasing. n+1 n an 1+2 2 1 + 2n+1 1 + 2n+1 9. an+1 (n + 1)e−(n+1) = = (1 + 1/n)e−1 < 1 for n ≥ 1, so strictly decreasing. an ne−n 10. an+1 (2n)! 10n+1 10 · < 1 for n ≥ 1, so strictly decreasing. = = an (2n + 2)! 10n (2n + 2)(2n + 1) 11. an+1 (n + 1)n+1 n! (n + 1)n · n= = = (1 + 1/n)n > 1 for n ≥ 1, so strictly increasing. an (n + 1)! n nn 12. an+1 5n+1 2n 5 = (n+1)2 · n = 2n+1 < 1 for n ≥ 1, so strictly decreasing. an 5 2 2 2 13. f (x) = x/(2x + 1), f (x) = 1/(2x + 1)2 > 0 for x ≥ 1, so strictly increasing. 14. f (x) = 3 − 1/x, f (x) = 1/x2 > 0 for x ≥ 1, so strictly increasing. 15. f (x) = 1/(x + ln x), f (x) = − 1 + 1/x < 0 for x 1, so strictly decreasing. (x + ln x)2 16. f (x) = xe−2x , f (x) = (1 − 2x)e−2x < 0 for x ≥ 1, so strictly decreasing. 17. f (x) = 1 − ln(x + 2) ln(x + 2) , f (x) = < 0 for x ≥ 1, so strictly decreasing. x+2 (x + 2)2 18. f (x) = tan−1 x, f (x) = 1/(1 + x2 ) > 0 for x ≥ 1, so strictly increasing. 19. f (x) = 2x2 − 7x, f (x) = 4x − 7 > 0 for x ≥ 2, so eventually strictly increasing. 20. f (x) = x3 − 4x2 , f (x) = 3x2 − 8x = x(3x − 8) > 0 for x ≥ 3, so eventually strictly increasing. 21. f (x) = 10 − x2 x , f (x) = 2 < 0 for x ≥ 4, so eventually strictly decreasing. x2 + 10 (x + 10)2 22. f (x) = x + 23. 17 x2 − 17 , f (x) = > 0 for x ≥ 5, so eventually strictly increasing. x x2 an+1 n+1 (n + 1)! 3n = > 1 for n ≥ 3, so eventually strictly increasing. = · an 3n+1 n! 3 24. f (x) = x5 e−x , f (x) = x4 (5 − x)e−x < 0 for x ≥ 6, so eventually str...
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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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