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

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

This is the end of the preview. Sign up to access the rest of the document.

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...
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