# Ak 40 ak 3 3 34 05 error a11 12 000074

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Unformatted text preview: (r/n)P = P (1 + r/n), at the end of 2 intervals it is P (1 + r/n) + (r/n)P (1 + r/n) = P (1 + r/n)2 , and continuing in this fashion the value at the end of nt intervals is P (1 + r/n)nt . (b) Let x = r/n, then n = r/x and lim P (1 + r/n)nt = lim+ P (1 + x)rt/x = lim+ P [(1 + x)1/x ]rt = P ert . n→+∞ x→0 x→0 (c) The rate of increase is dA/dt = rP ert = rA. 27. (a) A = 1000e(0.08)(5) = 1000e0.4 ≈ \$1, 491.82 (b) P e(0.08)(10) = 10, 000, P e0.8 = 10, 000, P = 10, 000e−0.8 ≈ \$4, 493.29 (c) From (11) with k = r = 0.08, T = (ln 2)/0.08 ≈ 8.7 years. CHAPTER 11 Inﬁnite Series EXERCISE SET 11.1 1. (a) 1 (b) 3n−1 (−1)n−1 3n−1 (c) 2. (a) (−r)n−1 ; (−r)n 2n − 1 2n (d) n2 π 1/(n+1) (b) (−1)n+1 rn ; (−1)n rn+1 (b) 1, −1, 1, −1 3. (a) 2, 0, 2, 0 (c) 2(1 + (−1)n ); 2 + 2 cos nπ (b) (2n − 1)! 4. (a) (2n)! 5. 1/3, 2/4, 3/5, 4/6, 5/7, . . .; lim n→+∞ n = 1, converges n+2 n2 = +∞, diverges n→+∞ 2n + 1 6. 1/3, 4/5, 9/7, 16/9, 25/11, . . .; lim 7. 2, 2, 2, 2, 2, . . .; lim 2 = 2, converges n→+∞ 1 1 1 1 8. ln 1, ln , ln , ln , ln , . . .; lim ln(1/n) = −∞, diverges n→+∞ 2 3 4 5 9. ln 1 ln 2 ln 3 ln 4 ln 5 , , , , , . . .; 1 2 3 4 5 ln x ln n 1 = lim = 0 apply L’Hˆpital’s Rule to o , converges lim n→+∞ n n→+∞ n x 10. sin π , 2 sin(π/2), 3 sin(π/3), 4 sin(π/4), 5 sin(π/5), . . .; sin(π/n) (−π/n2 ) cos(π/n) = lim = π , converges n→+∞ n→+∞ 1/n −1/n2 lim n sin(π/n) = lim n→+∞ 11. 0, 2, 0, 2, 0, . . .; diverges (−1)n+1 = 0, converges n→+∞ n2 12. 1, −1/4, 1/9, −1/16, 1/25, . . .; lim 13. −1, 16/9, −54/28, 128/65, −250/126, . . .; diverges because odd-numbered terms approach −2, even-numbered terms approach 2. 14. 1/2, 2/4, 3/8, 4/16, 5/32, . . .; lim n→+∞ n 1 = 0, converges = lim n n n→+∞ 2 ln 2 2 15. 6/2, 12/8, 20/18, 30/32, 42/50, . . .; lim n→+∞ 1 (1 + 1/n)(1 + 2/n) = 1/2, converges 2 16. π/4, π 2 /42 , π 3 /43 , π 4 /44 , π 5 /45 , . . .; lim (π/4)n = 0, converges n→+∞ 17. cos(3), cos(3/2), cos(1), cos(3/4), cos(3/5), . . .; lim cos(3/n) = 1, converges n→+∞ 18. 0, −1, 0, 1, 0, . . .; diverges 361 Exercise Set 11.1 362 x2 = 0, so lim n2 e−n = 0, converges x→+∞ ex n→+∞ 19. e−1 , 4e−2 , 9e−3 , 16e−4 , 25e−5 , . . .; lim x2 e−x = lim x→+∞ √ 10 − 2, √ 18 − 3, √ 28 − 4, √ 40 − 5, . . .; 3n lim ( n2 + 3n − n) = lim √ = lim 2 + 3n + n n→+∞ n→+∞ n→+∞ n 20. 1, 21. 2, (5/3)2 , (6/4)3 , (7/5)4 , (8/6)5 , . . .; let y = ln lim ln y = lim x→+∞ x→+∞ x+3 x+1 3 1 + 3/n + 1 = 3 , converges 2 x , converges because x+3 2x2 n+3 x + 1 = lim = 2, so lim x→+∞ (x + 1)(x + 3) n→+∞ n + 1 1/x n = e2 22. −1, 0, (1/3)3 , (2/4)4 , (3/5)5 , . . .; let y = (1 − 2/x)x , converges because lim ln y = lim x→+∞ 23. 24. x→+∞ 2n − 1 2n n−1 n2 ln(1 − 2/x) −2 = lim = −2, lim (1 − 2/n)n = lim y = e−2 x→+∞ 1 − 2/x n→+∞ x→+∞ 1/x +∞ ; lim n→+∞ n=1 +∞ 2n − 1 = 1, converges 2n n−1 = 0, converges n→+∞ n2 ; lim n=1 25. 1 3n +∞ ; lim n=1 n→+∞ 1 = 0, converges 3n +∞ 26. {(−1)n n}n=1 ; diverges because odd-numbered terms tend toward −∞, even-numbered terms tend toward +∞. 27. 28. 29. 1 1 − n n+1 3/2n−1 +∞ 1 1 − n n+1 ; lim n=1 n→+∞ +∞ ; lim 3/2n−1 n=1 n→+∞ = 0, converges = 0, converges √ +∞ n + 1 − n + 2 n=1 ; converges because √ √ (n + 1) − (n + 2) −1 √ √ = lim √ =0 lim ( n + 1 − n + 2) = lim √ n→+∞ n→+∞ n + 1 + n + 2 n→+∞ n + 1 + n + 2 √ 30. (−1)n+1 /3n+4 32. lim n→+∞ √ n +∞ ; lim (−1)n+1 /3n+4 n=1 n→+∞ n = 1, so lim n→+∞ √ n = 0, converges n3 = 13 = 1 n, 33. (a) 1, 2, 1, 4, 1, 6 (b) an = n odd n 1/2 , n even 1/n, (c) an = n odd 1/(n + 1), n even (d) In part (a) the sequence diverges, since the even terms diverge to +∞ and the odd terms equal 1; in part (b) the sequence diverges, since the odd terms diverge to +∞ and the even terms tend to zero; in part (c) lim an = 0. n→+∞ 34. The even terms are zero, so the odd terms must converge to zero, and this is true if and only if lim bn = 0, or −1 &lt; b &lt; 1. n→+∞ 363 35. Chapter 11 1 1 √ (yn + p/yn ), L = (L + p/L), L2 = p, L = ± p; lim yn+1 = lim n→+∞ 2 2 √ √ √ L = − p (reject, because the terms in the sequence are positive) or L = p; lim yn = p. n→+∞ n→+∞ 36. (a) an+1 = (b) √ 6 + an lim an+1 = lim n→+∞ √ n→+∞ √ 6 + an , L = 6 + L, L2 − L − 6 = 0, (L − 3)(L + 2) = 0, L = −2 (reject, because the terms in the sequence are positive) or L = 3; lim an = 3. n→+∞ 1 21 2 31 +, ++, + 4 4 9 9 9 16 1 (c) an = 2 (1 + 2 + · · · + n) = n 37. (a) 1, 2 3 4 325 + + = 1, , , 16 16 16 438 1n+1 11 n(n + 1) = , lim an = 1/2 n2 2 2 n n→+∞ 1 41 4 91 4 9 16 5 14 15 +, + +, + + + = 1, , , 8 8 27 27 27 64 64 64 64 8 27 32 1 (n + 1)(2n + 1) 1 11 , (c) an = 3 (12 + 22 + · · · + n2 ) = 3 n(n + 1)(2n + 1) = n n6 6 n2 1 (1...
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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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