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Unformatted text preview: 92.305 Homework 4 Solutions October 9, 2009 Exercise 2.4.1. Complete the proof of Theorem 2.4.6 by showing that if the series n =0 2 n b 2 n diverges, then so does n =1 b n . Example 2.4.5 may be a useful reference. Proof: We will show that if n =0 2 n b 2 n diverges then n =1 b n does too by exploiting a relationship between the partial sums s m = b 1 + b 2 + + b m and t k = b 1 + 2 b 2 + + 2 k b k . Because n =0 2 n b 2 n diverges, its monotone sequence of partial sums ( t k ) must be unbounded. To show that ( s m ) is unbounded it is enough to show that for all k N , there is term s m satisfying s m t k / 2. This argument is similar to the one for the forward direction, only to get the inequality to go the other way we group the terms in s m so that the last (and hence smallest) term in each group is of the form b 2 k . Given an arbitrary k , we focus our attention on s 2 k and observe that s 2 k = b 1 + b 2 + ( b 3 + b 4 ) + ( b 5 + b 6 + b 7 + b 8 ) + + ( b 2 k 1 +1 + + b 2 k ) b 1 + b 2 + ( b 4 + b 4 ) + ( b 8 + b 8 + b 8 + b 8 ) + + ( b 2 k + + b 2 k ) = b 1 + b 2 + 2 b 4 + 4 b 8 + + 2 k 1 b 2 k = 1 2 (2 b 1 + 2 b 2 + 4 b 4 + 8 b 8 + + 2 k b 2 k ) = b 1 / 2 + t k / 2 . Because ( t k ) is unbounded, the sequence ( s m ) must also be unbounded and cannot converge. Therefore, n =1 b n diverges. squaresolid Exercise 2.4.2. (a) Prove that the sequence defined by x 1 = 3 and x n +1 = 1 4 x n converges. Proof: We will show that this sequence is decreasing and bounded (which will also settle the issue of welldefinedness: since x 1 is 3 and subsequent terms are even smaller, x n never equals 4, so the denominator 4 x n never vanishes). First lets use induction to show that this sequence is decreasing. Observe that x 1 = 3 > 1 = x 2 . We need to prove that if x n > x n +1 then x n +1 > x n +2 . Note that x n > x n +1 implies that x n < x n +1 . Adding 4 to both sides of the inequality gives 4 x n < 4 x n +1 . It follows that 1 4 x n > 1 4 x n +1 , 1 which is precisely what we need to conclude x n +1 > x n +2 . Thus by induction, ( x n ) is decreasing. The argument above shows that ( x n ) is bounded above by 3, so now well use this fact in proving by induction that ( x n ) is bounded below, and more specifically, that x n > 0 for all n . Clearly x 1 > 0. Now assume x n > 0. Because ( x n ) is decreasing, we know that x n x 1 = 3, which implies that x n +1 = 1 4 x n is positive. By induction, ( x n ) is bounded below by 0 for all positive integers N ....
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This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.
 Spring '10
 MetCalfe

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