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Unformatted text preview: MAT 137Y, 2009–2010 Winter Session, Solutions to Supplementary Problem Set #3 1. (SHE 12.5) 4. Given a k = ( 1 ) k 1 k ln k , we know that ∑  a k  = ∑ 1 k ln k diverges by the integral test (see 11.2 #21), but ∑ a k converges by the alternating series test, so the series converges conditionally and not absolutely. 6. As lim k → ∞ k ln k H = ∞ , the alternating series test does not apply and the series diverges. 12. The values of sin k π 4 alternate between 0 , ± 1 / √ 2, and ± 1. The limit as k → ∞ clearly does not exist, so again the series diverges by the basic divergence test. 16. Since lim k → ∞ 1 p k ( k + 1 ) = 0 and the series alternate in sign, the series converges by the alternating series test. Using the limit comparison test with b k = 1 k , the series ∑ ( 1 ) k √ k ( k + 1 ) = ∑ 1 √ k ( k + 1 ) diverges, so the series in question conditionally converges (and not absolutely). 28. Since every absolutely convergent series converges, it is sufficient to show that the series absolutely con verges. Note that  a k  = sin ( k π / 2 ) k √ k < 1 k 3 / 2 , so by the comparison test, the series converges absolutely. 32. With any alternating series, the error estimate of the actual series and the partial sum s n is approximately a n + 1 (Equation 11.4.5). Therefore the error estimate between ∑ ( 1 ) k + 1 1 k and s 20 is a 21 = 1 21 . 42. If the sequence of terms { a n } is nonincreasing instead of decreasing, the alternating series still converges. To see this, we can make slight changes of the proof of Theorem 11.4.4. The even partial sums s 2 m are now nonnegative. Since s 2 m + 2 ≤ s 2 m , the sequence of even terms converges, so s 2 m → L . Since s 2 m + 1 = s 2 m a 2 m + 1 and a 2 m + 1 → 0, we have s 2 m + 1 → L . Hence s n → L . 2. (SHE 12.8) 6. Applying the ratio test, lim k → ∞ 2 k + 1 x k + 1 k 2 k 2 2 k x k = lim k → ∞ 2 x k k + 1 2 = 2  x  . Hence the series converges when 2  x  < 1, or  x  < 1 2 and diverges when  x  > 1 2 . (Therefore the radius of convergence is 1 2 .) To find the interval of convergence, we check the points  x  = 1 2 . If x = 1 2 , then we have ∑ 2 k k 2 · 1 2 k = ∑ 1 k 2 , which is a convergent pseries. if x = 1 2 , then we have ∑ ( 1 ) k k 2 which converges by the alternating series test. Therefore the interval of convergence is [ 1 2 , 1 2 ] . 8. Applying the ratio test, lim k → ∞ ( 1 ) k + 1 x k + 1 √ k + 1 · √ k ( 1 ) k x k = lim k → ∞ r k k + 1  x  =  x  ....
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This note was uploaded on 10/13/2011 for the course MATHEMATIC MAT 137 taught by Professor Brainpigott during the Fall '10 term at University of Toronto.
 Fall '10
 BrainPigott
 Calculus

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