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Unformatted text preview: Solutions homework 4. Page 22 Problem 1. : By 1.3.8 a n  a and b n  b , so by Theorem 2.5.6 we have a n + ( b n )  a + ( b ), which again by 1.3.8 implies that a n + b n a + b . Page 22 Problem 2. From 0 a n a n +1 and 0 b n b n +1 it follows that a n b n a n +1 b n +1 , so ( a n b n ) is increasing. Also 0 a n a and 0 b n b implies that a n b n ab . It remains to show that ab is the least upper bound of a n b n . To see this we have to consider first the case that a n = 0 for all n 1. In this case a = 0 and a n b n = 0, so 0 = ab = sup a n b n in this case. Similarly we can deal with the case the case that b n = 0 for all n 1. We can therefore assume assume that a n 6 = 0 from some index on and b n 6 = 0 from some index on. In particular this implies that a 6 = 0 and b 6 = 0. Let now s a n b n . Then s b n a n from some index on, which proves that s b n a from some index on. Hence s a b n , which implies...
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This note was uploaded on 02/05/2012 for the course MATH 554 taught by Professor Girardi during the Fall '10 term at South Carolina.
 Fall '10
 Girardi

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