# 34 2 345 3 34 2 k k x x x 2 3 k k

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Unformatted text preview: · · = 5 + 0.37 = 5 + 37/99 = 532/99 1 − 0.01 18. 0.159159159 · · · = 0.159 + 0.000159 + 0.000000159 + · · · = 0.159 = 159/999 = 53/333 1 − 0.001 19. 0.782178217821 · · · = 0.7821 + 0.00007821 + 0.000000007821 + · · · = 0.7821 7821 869 = = 1 − 0.0001 9999 1111 20. 0.451141414 · · · = 0.451 + 0.00014 + 0.0000014 + 0.000000014 + · · · = 0.451 + ∞ 22. (a) geometric; 18/5 (b) geometric; diverges (c) k=1 23. d = 10 + 2 · = 10 + 20 1 2 0.00014 44663 = 1 − 0.01 99000 1 1 − 2k − 1 2k + 1 33 333 3 · 10 + 2 · · · 10 + 2 · · · · 10 + · · · 4 44 444 3 4 + 20 3 4 2 + 20 3 4 3 + · · · = 10 + 20(3/4) = 10 + 60 = 70 meters 1 − 3/4 = 1/2 369 Chapter 11 24. volume = 13 + = 1 2 3 + 1 4 3 3 1 2n + ··· + + ··· = 1 + 1 + 8 1 8 2 + ··· + 1 8 n + ··· 1 = 8/7 1 − (1/8) 2 3 n 1 + ln + ln + · · · + ln = ln 2 3 4 n+1 lim sn = −∞, series diverges. n 123 · · ··· 234 n+1 25. (a) sn = ln = ln 1 = − ln(n + 1), n+1 n→+∞ (b) ln(1 − 1/k 2 ) = ln n+1 sn = ln k=2 = ln = ln k+1 k−1 k k2 − 1 (k − 1)(k + 1) k−1 + ln = ln − ln , = ln = ln 2 2 k k k k k k+1 k k−1 − ln k k+1 2 1 − ln 2 3 + ln 2 3 − ln 3 4 + ln (−1)k xk = 1 − x + x2 − x3 + · · · = k=0 (x − 3)k = 1 + (x − 3) + (x − 3)2 + · · · = k=0 ∞ (−1)k x2k = 1 − x2 + x4 − x6 + · · · = (c) n n+1 − ln n+1 n+2 1 1 = if | − x| < 1, |x| < 1, −1 < x < 1. 1 − (−x) 1+x ∞ (b) + · · · + ln n+1 1 1 − ln , lim sn = ln = − ln 2 2 n + 2 n→+∞ 2 ∞ 26. (a) 3 4 − ln 4 5 k=0 1 1 = if |x − 3| < 1, 2 < x < 4. 1 − (x − 3) 4−x 1 1 = if |− x2 | < 1, |x| < 1, −1 < x < 1. 1 − (−x2 ) 1 + x2 27. (a) Geometric series, a = x, r = −x2 . Converges for | − x2 | < 1, |x| < 1; x x = . S= 2) 1 − (−x 1 + x2 (b) Geometric series, a = 1/x2 , r = 2/x. Converges for |2/x| < 1, |x| > 2; S= 1 1/x2 =2 . 1 − 2/x x − 2x (c) Geometric series, a = e−x , r = e−x . Converges for |e−x | < 1, e−x < 1, ex > 1, x > 0; 1 e−x . =x S= −x 1−e e −1 √ √ √ √ 1 1 k+1− k k+1− k √ , 28. =√ − = √√ k+1 k2 + k k k+1 k n sn = k=1 1 1 √ −√ k+1 k +··· + = 1 1 √ −√ n n+1 1 1 √ −√ 1 2 =1− √ + 1 1 √ −√ 2 3 + 1 1 √ −√ 3 4 1 ; lim sn = 1 n + 1 n→+∞ 29. sn = (1 − 1/3) + (1/2 − 1/4) + (1/3 − 1/5) + (1/4 − 1/6) + · · · + [1/n − 1/(n + 2)] = (1 + 1/2 + 1/3 + · · · + 1/n) − (1/3 + 1/4 + 1/5 + · · · + 1/(n + 2)) = 3/2 − 1/(n + 1) − 1/(n + 2), lim sn = 3/2 n→+∞ Exercise Set 11.3 n 30. sn = k=1 = 1 2 n 31. sn = k=1 = 1 2 370 n 1 = k (k + 2) n k=1 1 − k 1/2 1 1/2 − = k k+2 2 k=1 n+2 k=3 1 k = k=1 1 − 2k − 1 n+1 k=2 k=1 1 − k n k=1 1 1 1 1 1+ − − ; 2 2 n+1 n+2 1 = (2k − 1)(2k + 1) n n n 1/2 1 1/2 − = 2k − 1 2k + 1 2 k=1 1 1 1 = 1− ; 2k − 1 2 2n + 1 1 k+2 lim sn = n→+∞ n k=1 3 4 1 − 2k − 1 lim sn = n→+∞ n k=1 1 2k + 1 1 2 1 1 32. Geometric series, a = sin x, r = − sin x. Converges for | − sin x| < 1, | sin x| < 2, 2 2 sin x 2 sin x so converges for all values of x. S = = . 1 2 + sin x 1 + sin x 2 33. a2 = a5 = 1 1 1 1 1 1 1 1 1 1 1 1 1 a1 + , a3 = a2 + = 2 a1 + 2 + , a4 = a3 + = 3 a1 + 3 + 2 + , 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 a4 + = 4 a1 + 4 + 3 + 2 + , . . . , an = n−1 a1 + n−1 + n−2 + · · · + , 2 2 2 2 2 2 2 2 2 2 2 lim an = lim n→+∞ a1 n→+∞ 2n−1 ∞ + n=1 1 2 n =0+ 1/2 =1 1 − 1/2 34. 0.a1 a2 · · · an 9999 · · · = 0.a1 a2 · · · an + 0.9 (10−n ) + 0.09 (10−n ) + · · · = 0.a1 a2 · · · an + 0.9 (10−n ) = 0.a1 a2 · · · an + 10−n 1 − 0.1 = 0.a1 a2 · · · (an + 1) = 0.a1 a2 · · · (an + 1) 0000 · · · 35. The series converges to 1/(1 − x) only if −1 < x < 1. 36. P0 P1 = a sin θ, P1 P2 = a sin θ cos θ, P2 P3 = a sin θ cos2 θ, P3 P4 = a sin θ cos3 θ, . . . (see ﬁgure) Each sum is a geometric series. P1 a sin u cos u P3 a sin u u a sin u cos2 u a sin u cos3 u u u P0 P2 P4 a (a) P0 P1 + P1 P2 + P2 P3 + · · · = a sin θ + a sin θ cos θ + a sin θ cos2 θ + · · · = (b) P0 P1 + P2 P3 + P4 P5 + · · · = a sin θ + a sin θ cos2 θ + a sin θ cos4 θ + · · · a sin θ a sin θ = a csc θ = = 2θ 1 − cos sin2 θ (c) P1 P2 + P3 P4 + P5 P6 + · · · = a sin θ cos θ + a sin θ cos3 θ + · · · a sin θ cos θ a sin θ cos θ = a cot θ = = 2θ 1 − cos sin2 θ P a sin θ 1 − cos θ 371 Chapter 11 37. By inspection, θ θ/2 θθθ −+− + ··· = = θ/3 2 4 8 16 1 − (−1/2) 1 =2 1 − (1/2) 38. A1 + A2 + A3 + · · · = 1 + 1/2 + 1/4 + · · · = 39. (b) 2k 3k+1 − 2k+1 A + 2k 3k − 2k B 2k B 2k A + k+1 = k k+1 −2 3 −2 (3k − 2k ) (3k+1 − 2k+1 ) 3k 3 · 6k − 2 · 22k A + 6k − 22k B (3A + B )6k − (2A + B )22k = (3k − 2k ) (3k+1 − 2k+1 ) (3k − 2k ) (3k+1 − 2k+1 ) = so 3A + B = 1 and 2A + B = 0, A = 1 and B = −2. n (c) sn = k=1 2k+1 2k − k+1 = k − 2k 3 3 − 2k+1 n (ak − ak+1 ) where ak = k=1 3k 2k . − 2k But sn = (a1 − a2 ) +...
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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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