227_sample_exam_3_soltuionsSp2011

227_sample_exam_3_soltuionsSp2011 - Math 227 Sections 4 and...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

Page1 / 3

227_sample_exam_3_soltuionsSp2011 - Math 227 Sections 4 and...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online