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227_sample_exam_3_soltuions_spring_2009[1]

# 227_sample_exam_3_soltuions_spring_2009[1] - Math 227.04...

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Math 227.04 Solutions to Sample Exam 3 1. (10 ptsWrite out the °rst three terms and °nd the sum of the series P 1 n =1 2( 1 3 ) n . Solution. P 1 n =1 2( 1 3 ) n = 2 3 + 2 9 + 2 27 + ° ° ° = 2 3 1 ± 1 3 = 1 2. (10 pts each) Test the following series for convergence/divergence. Show all steps.. (a) 1 X n =1 ( ± 1) n n 2 n 2 ± 1 Solution. Let u n = n 2 n 2 ± 1 : Then each u n > 0 . Also, d dx x 2 x 2 ± 1 = ° 2 x 2 ± 1 ± ± x ° 4 x (2 x 2 ± 1) 2 = ± 1 ± 2 x 2 (2 x 2 ± 1) 2 < 0 : Thus u n = f ( n ) is decreasing. This means we can apply the Alternating Series Test: lim n !1 n 2 n 2 ± 1 = 0 so the series CONVERGES BY THE ALTERNATING SERIES TEST.. (b) 1 X n =1 ( ± 2) n ( n + 1) n Solution. Apply the Root Test: lim n !1 j a n j 1 =n = lim n !1 ² ² ² ² 2 n + 1 ² ² ² ² = 0 < 1 : Thus the series CONVERGES BY THE ROOT TEST. (c) 1 X n =2 1 n ln n Solution. If you try comparing with any p ± series, you±ll get a "smaller than big" or "bigger than small" comparison, which is inclusive. Thus we must use the test of last resort: the Integral Test. First check the conditions: let f ( x ) = 1 x ln x : Then f is clearly positive and decreasing for x ² 2 :

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