Unformatted text preview: k . This implies k 2 ≥ m 2 + 2 m + 1 > m 2 + 1 = n + 1 = k 2 , which is a contradiction. Page 29 Problem 5. a. We prove by induction that a n < a n +1 . The claim is true for n = 1 as we have that a 1 = √ 2 < p 2 + √ 2 = a 2 . Assume now that a n < a n +1 (induction hypothesis), then to prove a n +1 < a n +2 . Now a n +1 = √ 2 + a n < √ 2 + a n +1 = a n +2 . Hence ( a n ) is strictly increasing. b. Again we prove this by induction. The case n = 1 follows from a 1 = √ 2 < √ 4 = 2. Assume now a n < 2 (induction hypothesis), Then a n +1 = √ 2 + a n , √ 2 + 2 = 2 and the claim follows by induction. c. Let a = sup a n . Then also a 2 n ↑ a 2 , since a n and a are positive. Since a 2 n +1 = a n + 2, so we conclude a 2 = a + 2. Now a 2a2 = ( a2)( a + 1) = 0 implies a =1 or a = 2. From a > 0 we conclude that a = 2 is the only possibility. 1...
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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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