The circles intersect when cos t 3 10 a 1 cos2 d

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Unformatted text preview: |, so the interval of convergence is −4 < x < −2, uk 1 1 = (the series diverges for x = −4, −2) 1 + (x + 3) 4+x 5. (a) geometric series, ρ = converges there to lim k→+∞ uk+1 = |x/2|, so the interval of convergence is −2 < x < 2, uk 2 1 = ; (the series diverges for x = −2, 2) 1 + x/2 2+x (b) f (0) = 1; f (1) = 2/3 6. (a) geometric series, ρ = lim k→+∞ converges to uk+1 x−5 , so the interval of convergence is 2 < x < 8, = uk 3 3 1 = (the series diverges for x = 2, 8) 1 + (x − 5)/3 x−2 (b) f (3) = 3, f (6) = 3/4 389 Chapter 11 7. ρ = ∞ k=0 k+1 |x| = |x|, the series converges if |x| < 1 and diverges if |x| > 1. If x = −1, k→+∞ k + 2 lim (−1)k converges by the Alternating Series Test; if x = 1, k+1 ∞ k=0 1 diverges. The radius of k+1 convergence is 1, the interval of convergence is [−1, 1). 8. ρ = lim 3|x| = 3|x|, the series converges if 3|x| < 1 or |x| < 1/3 and diverges if |x| > 1/3. If k→+∞ ∞ ∞ (−1)k diverges, if x = 1/3, x = −1/3, k=0 (1) diverges. The radius of convergence is 1/3, the k=0 interval of convergence is (−1/3, 1/3). |x| = 0, the radius of convergence is +∞, the interval is (−∞, +∞). k→+∞ k + 1 9. ρ = lim 10. ρ = lim k→+∞ k+1 |x| = +∞, the radius of convergence is 0, the series converges only if x = 0. 2 5k 2 |x| = 5|x|, converges if |x| < 1/5 and diverges if |x| > 1/5. If x = −1/5, k→+∞ (k + 1)2 ∞ 11. ρ = lim ∞ 1/k 2 converges. Radius of convergence is 1/5, interval of convergence is converges; if x = 1/5, k=1 [−1/5, 1/5]. 12. ρ = ∞ k=2 k=1 (−1)k k2 ln k |x| = |x|, the series converges if |x| < 1 and diverges if |x| > 1. If x = −1, k→+∞ ln(k + 1) lim (−1)k converges; if x = 1, ln k ∞ 1/(ln k ) diverges (compare to (1/k )). Radius of convergence k=2 is 1, interval of convergence is [−1, 1). k |x| = |x|, converges if |x| < 1, diverges if |x| > 1. If x = −1, 13. ρ = lim k→+∞ k + 2 ∞ if x = 1, k=1 14. ρ = lim 2 k→+∞ k+1 |x| = 2|x|, converges if |x| < 1/2, diverges if |x| > 1/2. If x = −1/2, k+2 diverges; if x = 1/2, is (−1/2, 1/2]. √ 15. ρ = lim √ ∞ x = 1, k=1 (−1)k converges; k (k + 1) 1 converges. Radius of convergence is 1, interval of convergence is [−1, 1]. k (k + 1) ∞ k→+∞ ∞ k=0 ∞ k=0 −1 2(k + 1) (−1)k converges. Radius of convergence is 1/2, interval of convergence 2(k + 1) k |x| = |x|, converges if |x| < 1, diverges if |x| > 1. If x = −1, k+1 ∞ −1 √ diverges; if k k=1 (−1)k−1 √ converges. Radius of convergence is 1, interval of convergence is (−1, 1]. k k=1 Exercise Set 11.8 390 |x|2 = 0, radius of convergence is +∞, interval of convergence is (−∞, +∞). k→+∞ (2k + 2)(2k + 1) 16. ρ = lim |x|2 = 0, radius of convergence is +∞, interval of convergence is (−∞, +∞). k→+∞ (2k + 3)(2k + 2) 17. ρ = lim ∞ k 3/2 |x|3 1 = |x|3 , converges if |x| < 1, diverges if |x| > 1. If x = −1, converges; 3/2 3/2 k→+∞ (k + 1) k k=0 18. ρ = lim ∞ if x = 1, (−1)k converges. Radius of convergence is 1, interval of convergence is [−1, 1]. k 3/2 k=0 19. ρ = lim k→+∞ 3|x| = 0, radius of convergence is +∞, interval of convergence is (−∞, +∞). k+1 k (ln k )2 |x| = |x|, converges if |x| < 1, diverges if |x| > 1. If x = −1, then, by k→+∞ (k + 1)[ln(k + 1)]2 20. ρ = lim ∞ Exercise 11.4.25, k=2 −1 converges; if x = 1, k (ln k )2 ∞ k=2 (−1)k+1 converges. Radius of convergence k (ln k )2 is 1, interval of convergence is [−1, 1]. 1 + k2 |x| = |x|, converges if |x| < 1, diverges if |x| > 1. If x = −1, k→+∞ 1 + (k + 1)2 ∞ 21. ρ = lim ∞ converges; if x = 1, [−1, 1]. k=0 k=0 1 converges. Radius of convergence is 1, interval of convergence is 1 + k2 1 1 |x − 3| = |x − 3|, converges if |x − 3| < 2, diverges if |x − 3| > 2. If x = 1, k→+∞ 2 2 ∞ (−1)k 22. ρ = lim ∞ diverges; if x = 5, (−1)k 1 + k2 k=0 1 diverges. Radius of convergence is 2, interval of convergence is (1, 5). k=0 k |x + 1| = |x + 1|, converges if |x + 1| < 1, diverges if |x + 1| > 1. If x = −2, k→+∞ k + 1 ∞ 23. ρ = lim ∞ diverges; if x = 0, (−2, 0]. 24. ρ = k=1 k=1 −1 k (−1)k+1 converges. Radius of convergence is 1, interval of convergence is k (k + 1)2 |x − 4| = |x − 4|, converges if |x − 4| < 1, diverges if |x − 4| > 1. If x = 3, k→+∞ (k + 2)2 lim ∞ ∞ 1/(k +1)2 converges; if x = 5, k=0 of convergence is [3, 5]. (−1)k /(k +1)2 converges. Radius of convergence is 1, interval k=0 391 Chapter 11 25. ρ = lim (3/4)|x + 5| = k→+∞ 3 |x + 5|, converges if |x + 5| < 4/3, diverges if |x + 5| > 4/3. If 4 ∞ x = −19/3, ∞ (−1)k diverges; if x = −11/3, k=0 1 diverges. Radius of convergence is 4/3, interval k=0 of convergence is (−19/3, −11/3). (2k + 3)(2k + 2)k 3 |x − 2| = +∞, radius of convergence is 0, k→+∞ (k + 1)3 series converges only at x = 2. 26. ρ = lim k2 + 4 |x + 1|2 = |x + 1|2 , converges if |x + 1| < 1, diverges if |x + 1| > 1. If x = −2, k→+∞ (k + 1)2 + 4 27. ρ = lim ∞ k=1 (−1)3k+1 converges; if x = 0, k2 + 4 ∞ k=1 (−1)k co...
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