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Unformatted text preview: Math 28 Spring 2008: Final Exam Instructions: This exam is to be turned in no later than Friday May 9th in class (10am). You may use your book and your notes from class, but no additional resources or discussion are permitted. If you need clarification on a problem, please ask me. Each problem will be graded out of 10 points for a total of 100 possible points. Problem 1. Define a 1 = 0 ,a 2 = 1 2 ,a n +1 = 1 3 (1 + a n + a 3 n 1 ). Show that the sequence ( a n ) converges and find its value of convergence. Proof. We will first show by induction that the sequence is increasing. We know a 1 = 0 ,a 2 = 1 2 ,a 3 = 1 2 , so we have established the base case. Assume now that a n 1 ≥ a n 2 and a n ≥ a n 1 . Then, a n +1 = 1 3 ( a n + a 3 n 1 ) ≥ 1 3 ( a n 1 + a 3 n 2 ) = a n . So we have established that ( a n ) is increasing. To show that it is monontone, note that a 1 = 0 and the sequence is increasing so it is bounded below by 0. Now we will show by induction that it is bounded above by 1. The base case is established by a 1 = 0 ,a 2 = 1 2 . So in general a n +1 = 1 3 (1 + a n + a 3 n 1 ) ≤ 1 3 (1 + 1+) = 1 . So the sequence is bounded and montone so it is convergent by the Monotone Convergence Theorem. Let a be the value of convergece of a . Then ( a n ) → a , ( a n 1 ) → a , and ( a n 2 ) → a since all of these sequences are eventually the same. So we have a = 1 3 (1 + a + a 3 ) . So in particular a = 1 , 1 + √ 5 2 , or 1 √ 5 2 Note that the sequence starts at 0 and is increasing, so it converges to the smallest positive value. Since the third is negative and the second is smaller than 1 we have a = 1 + √ 5 2 . Problem 2. In this problem we consider a modified Harmonic Series. Let the ( p,q )Harmonic series, be the Harmonic series with p consecutive positive terms followed by q consecutive negative terms. In other words, the (2 , 3)Harmonic series is given by 1 + 1 2 1 3 1 4 1 5 + 1 6 + 1 7 1 8 1 9 1 10 + 1 11 + ··· . Show that the ( p,q ) Harmonic series converges if and only if p = q . 1 Proof. We first show that if p = q then the series converges. We have S lp = 1 + 1 2 + ··· + 1 p ¶ 1 p + 1 + ··· + 1 2 p ¶ + ··· + ( 1) l +1 1 ( l 1) p + 1 + ··· + 1 lp ¶ Therefore, S lp is the partial sum of an alternating series. Since p is a constant we have fl fl fl fl 1 ( l 1) p + 1 + ··· + 1 lp fl fl fl fl ≤ p p = 1 l → . In fact we have p ( l + 1) p = 1 l + 1 ≤ fl fl fl fl 1 ( l 1) p + 1 + ··· + 1 lp fl fl fl fl ≤ 1 l so the terms of the alternating sequence are decreasing. So this series satisfies the alternating series test, so it is convergent....
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This note was uploaded on 03/02/2010 for the course MATH 28 taught by Professor Staff during the Spring '08 term at UMass (Amherst).
 Spring '08
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