227_sample_exam_3_soltuionsSp2011

# 227_sample_exam_3_soltuionsSp2011 - Math 227 Sections 4 and...

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Unformatted text preview: Math 227 Sections 4 and 5 Solutions to Sample Exam 3 1. Write out the &rst three terms and &nd the sum of the series P 1 n =1 2( 1 3 ) n . Solution. This is a geometric series with &rst term equal to 2 = 3 and common ratio 1 = 3 : P 1 n =1 2( 1 3 ) n = 2 3 + 2 9 + 2 27 + & & & = 2 3 1 ¡ 1 3 = 1 : 2. Test the following series for convergence/divergence. Show all steps.. (a) 1 X n =1 n 2 n 2 ¡ 1 Solution. The general term of this series is essentially n 2 n 2 = 1 2 n : This suggests using a comparison with the divergent p-series P 1 n : Letting a n = n 2 n 2 ¡ 1 and b n = 1 n we can apply the LCT: lim n !1 a n b n = lim n !1 n 2 n 2 ¡ 1 & n 1 = lim n !1 n 2 2 n 2 ¡ 1 = lim n !1 1 2 ¡ 1 n 2 = 1 2 : Conclusion: P n 2 n 2 ¡ 1 diverges by the LCT. Remark: In this case, Direct comparison is even easier: 1 X n =1 n 2 n 2 ¡ 1 ¢ 1 X n =1 n 2 n 2 = 1 X n =1 1 2 n = 1 : (b) 1 X n =1 ( ¡ 2) n ( n + 1) n n Solution. Apply the Root Test: lim n !1 j a n j 1 =n = lim n !1 & & & & 2 n + 1 & & & & = 0 < 1 : Thus the series CONVERGES BY THE ROOT TEST....
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227_sample_exam_3_soltuionsSp2011 - Math 227 Sections 4 and...

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