A 1 x x2 2 x3 3 x4 4 x x3 3 x5 5

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Unformatted text preview: (a2 − a3 ) + (a3 − a4 ) + · · · + (an − an+1 ) which is a telescoping sum, sn = a1 − an+1 = 2 − 2n+1 (2/3)n+1 , lim s = lim 2 − = 2. n+1 n→+∞ n n→+∞ −2 1 − (2/3)n+1 3n+1 EXERCISE SET 11.4 ∞ 1. (a) k=1 ∞ (b) k=1 ∞ k=1 ∞ 2. (a) k=2 ∞ 1/2 1 = 1; = 2k 1 − 1/2 k=1 1/5 1 = 1/4; = 5k 1 − 1/5 1/4 1 = 1/3; = 4k 1 − 1/4 ∞ k=1 1 =1 k (k + 1) k=1 1 1 +k 2k 4 = 1 + 1/3 = 4/3 (Example 5, Section 11.3); 1 1 = 1/4 − 1 = −3/4 − 5k k (k + 1) 1 = 3/4 (Exercise 10, Section 11.3); k2 − 1 ∞ so k=2 ∞ k=2 7/10 7 = 7/9; = 10k−1 1 − 1/10 7 1 − = 3/4 − 7/9 = −1/36 k 2 − 1 10k−1 ∞ (b) with a = 9/7, r = 3/7, geometric, k=1 ∞ with a = 4/5, r = 2/5, geometric, k=1 ∞ ∞ 7−k 3k+1 − k+1 2 k=1 5k 7−k 3k+1 = 9/7 = 9/4; 1 − (3/7) 4/5 2k+1 = 4/3; = k 5 1 − (2/5) = 9/4 − 4/3 = 11/12 3. (a) p = 3, converges (b) p = 1/2, diverges (c) p = 1, diverges (d) p = 2/3, diverges 4. (a) p = 4/3, converges (b) p = 1/4, diverges (c) p = 5/3, converges (d) p = π , converges Exercise Set 11.4 372 1 k2 + k + 3 = ; the series diverges. k→+∞ 2k 2 + 1 2 5. (a) (b) lim lim cos kπ does not exist; (c) (d) k→+∞ lim k→+∞ 1+ 1 k k = e; the series diverges. 1 = 0; no information k→+∞ k ! lim the series diverges. k = 0; no information k→+∞ ek 6. (a) (b) lim 1 (c) lim √ = 0; no information k→+∞ k +∞ (d) 1 1 = lim ln(5x + 2) →+∞ 5 5x + 2 7. (a) 1 +∞ 1 √ lim √ k→+∞ k = 1; the series diverges. k+3 = +∞, the series diverges by the Integral Test. 1 1 1 tan−1 3x dx = lim 2 →+∞ 3 1 + 9x (b) lim ln k = +∞; the series diverges. k→+∞ = 1 1 π /2 − tan−1 3 , 3 the series converges by the Integral Test. +∞ x 1 ln(1 + x2 ) dx = lim →+∞ 2 1 + x2 8. (a) 1 +∞ (b) = +∞, the series diverges by the Integral Test. 1 √ = 1/ 6, √ (4 + 2x)−3/2 dx = lim −1/ 4 + 2x →+∞ 1 1 the series converges by the Integral Test. ∞ 9. k=1 ∞ 10. k=1 1 = k+6 3 = 5k ∞ k=1 12. ∞ k=1 3 5 1 , diverges because the harmonic series diverges. k 1 1 √ , diverges because the p-series with p = 1/2 ≤ 1 diverges. = k + 5 k=6 k 1 lim k→+∞ e1/k +∞ 13. = 1, the series diverges because lim uk = 1 = 0. k→+∞ 3 (2x − 1)2/3 →+∞ 4 (2x − 1)−1/3 dx = lim 1 14. k=7 1 , diverges because the harmonic series diverges. k ∞ √ 11. ∞ +∞ ln x is decreasing for x ≥ e, and x 3 = +∞, the series diverges by the Integral Test. 1 ln x 1 = lim (ln x)2 →+∞ 2 x = +∞, 3 so the series diverges by the Integral Test. 15. lim k→+∞ +∞ 16. 1 k 1 = lim = +∞, the series diverges because lim uk = 0. k→+∞ ln(k + 1) k→+∞ 1/(k + 1) 2 2 1 xe−x dx = lim − e−x →+∞ 2 = e−1 /2, the series converges by the Integral Test. 1 373 17. 18. Chapter 11 lim (1 + 1/k )−k = 1/e = 0, the series diverges. k→+∞ k2 + 1 = 1 = 0, the series diverges. k→+∞ k 2 + 3 lim +∞ tan−1 x 1 tan−1 x dx = lim →+∞ 2 1 + x2 19. 1 2 = 3π 2 /32, the series converges by the Integral Test, since 1 1 − 2x tan−1 x d tan−1 x = < 0 for x ≥ 1. 2 dx 1 + x 1 + x2 +∞ √ 20. 1 21. 1 dx = lim sinh−1 x →+∞ x2 + 1 = +∞, the series diverges by the Integral Test. 1 lim k 2 sin2 (1/k ) = 1 = 0, the series diverges. k→+∞ +∞ 22. 1 3 3 1 x2 e−x dx = lim − e−x →+∞ 3 = e−1 /3, 1 the series converges by the Integral Test (x2 e−x is decreasing for x ≥ 1). 3 ∞ 23. 7 k −1.01 , p-series with p > 1, converges k=5 +∞ 24. 25. = 1 − tanh(1), the series converges by the Integral Test. sech2 x dx = lim tanh x →+∞ 1 1 +∞ 1 dx is decreasing for x ≥ e−p , so use the Integral Test with to get p p x(ln x) e−p x(ln x) +∞ if p < 1 1−p (ln x) = +∞ if p = 1, lim = lim ln(ln x) −1 →+∞ →+∞ 1 − p e−p if p > 1 e−p (−1)p pp (p − 1) Thus the series converges for p > 1. 26. Set g (x) = x(ln x)[ln(ln x)]p , g (x) = (1 + ln x) ln(ln x) + p, so for fixed p there exists A > 0 such +∞ dx that g (x) > 0, 1/g (x) is decreasing for x > A; use the Integral Test with x(ln x)[ln(ln x)]p A to get +∞ if p < 1, [ln(ln x)]1−p lim ln[ln(ln x)] = +∞ if p = 1, lim = 1 →+∞ →+∞ 1−p if p > 1 A A (p − 1)[ln(ln A)]p−1 Thus the series converges for p > 1. ∞ 27. (a) 3 k=1 ∞ (c) k=2 1 − k2 ∞ k=1 1 = (k − 1)4 1 = π 2 /2 − π 4 /90 k4 ∞ k=1 1 = π 4 /90 k4 ∞ (b) k=1 1 1 − 1 − 2 = π 2 /6 − 5/4 2 k 2 Exercise Set 11.4 374 28. (a) Suppose Σ(uk + vk ) converges; then so does Σ[(uk + vk ) − uk ], but Σ[(uk + vk ) − uk ] = Σvk , so Σvk converges which contradicts the assumption that Σvk diverges. Suppose Σ(uk − vk ) converges; then so does Σ[uk − (uk − vk )] = Σvk which leads to the same contradiction as before. (b) Let uk = 2/k and vk = 1/k ; then both Σ(uk + vk ) and Σ(uk − vk ) diverge; let uk = 1/k and vk = −1/k then Σ(uk + vk ) converges; let uk = vk = 1/k then Σ(uk − vk ) converges. ∞ 29. (a) diverges because ∞ (2/3)k−1 converges and 1/k diver...
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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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