227-04_exam_3_solnsSp2011

227-04_exam_3_solnsSp2011 - Math 227.04 Exam 3 Solutions...

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Math 227.04 Exam 3 Solutions April 28, 2011 Name Instructions. You must show your work in order to receive full credit. Please display your answers and erase or cross out any work you wish to be ignored. P 1 n =2 3( 1 5 ) n . P 1 n =2 3( 1 5 ) n = 3 25 3 125 + 3 625 = 3 = 25 1 1 5 ± = 1 10 : 2. (10 pts each) Test the following series for convergence/divergence. In each case, state clearly which test you are using. If you are applying a comparison test, declare your benchmark series. (a) 1 X n =1 ln n n 3 Note that ln n < n for n > 0 : Thus ln n n 3 ² n n 3 = 1 n 2 so P ln n n 3 converges by the Direct Comparison Test using benchmark series P 1 n 2 : (b) 1 X n =2 n n 3 1 Use the Limit Comparison Test with the benchmark series P 1 n 2 : Letting a n = n n 3 1 and b n = 1 n 2 ; lim n !1 a n b n = lim n !1 n n 3 1 ± n 2 1 = lim n !1 n 3 n 3 1 = 1 : Since P b n converges, the series P a n
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227-04_exam_3_solnsSp2011 - Math 227.04 Exam 3 Solutions...

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