SampleMidtermwithSolutions1

# SampleMidtermwithSolutions1 - MATH 262 Sample Midterm with...

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Unformatted text preview: MATH 262: Sample Midterm with Solutions Problem 1. (a) Find the sum of the series ∞ X n =1 3 n 6 2 n +1 . (b) Find out for which values of x the series ∞ X n =1 (- 1) n- 1 ( x- 5) 2 n n · 3 n converges absolutely, converges conditionally, diverges. Solution. (a) ∞ X n =1 3 n 6 2 n +1 = 3 6 3 + 3 2 6 5 + ··· = 3 6 3 1 + 3 6 2 + 3 2 6 4 + ··· ¶ = 3 6 3 ˆ 1 + 3 6 2 ¶ + 3 6 2 ¶ 2 + ··· ! = 3 6 3 · 1 1- 3 6 2 = 1 66 . (b) At first we test the series for absolute convergence using the Ratio Test. lim n →∞ fl fl fl fl a n +1 a n fl fl fl fl = lim n →∞ fl fl fl fl (- 1) n ( x- 5) 2( n +1) ( n + 1) 3 n +1 ÷ (- 1) n- 1 ( x- 5) 2 n n 3 n fl fl fl fl = lim n →∞ fl fl fl fl ( x- 5) 2( n +1) n 3 n ( x- 5) 2 n ( n + 1) 3 n +1 fl fl fl fl = lim n →∞ fl fl fl fl ( x- 5) 2 3 · n n + 1 fl fl fl fl = ( x- 5) 2 3 . Hence, the series converges absolutely if and only if ( x- 5) 2 3 < 1, or equivalently, if | x- 5 | < √ 3. Assume | x- 5 | > √ 3. Then a = ( x- 5) 2 3 > 1 and the limit lim n →∞ (- 1) n- 1 ( x- 5) 2 n n 3 n = lim n →∞ (- 1) n- 1 a n n does not exist since a n n → ∞ as n → ∞ . Hence, the series can’t converge if | x- 5 | > √ 3. Finally, we check the endpoints of the interval of convergence, that is, when | x- 5 | = √ 3, equivalently, x = 5 ± √ 3. We have ∞ X n =1 (- 1) n- 1 ( x- 5) 2 n n · 3 n = ∞ X n =1 (- 1) n- 1 ( ± √ 3) 2 n n · 3 n = ∞ X n =1 (- 1) n- 1 3 n n · 3 n = ∞ X n =1 (- 1) n- 1 1 n which converges conditionally by the Alternating Series Test.which converges conditionally by the Alternating Series Test....
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## This note was uploaded on 01/15/2010 for the course MATH MATH 262 taught by Professor Gregrix during the Spring '09 term at McGill.

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SampleMidtermwithSolutions1 - MATH 262 Sample Midterm with...

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