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sol to supp3

# sol to supp3 - MAT 137Y 20082009 Winter Session Solutions...

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MAT 137Y, 2008–2009 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 p -series. 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 | . Hence the series converges when | x | < 1 and diverges when | x | > 1. If x = 1, then we have ( - 1 ) k k which converges by the alternating series test. If x = - 1, then we have ( - 1 ) k · ( - 1 ) k k = 1 k , which diverges by the p -series test. Therefore the interval of convergence is ( - 1 , 1 ] .

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