# A06S - Math 138 Assignment 6 Solutions 1(a Based on the...

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Unformatted text preview: Math 138 Assignment 6 Solutions 1. (a) Based on the given terms we see that a 1 = √ 5 a 2 = q 5 √ 5 = √ 5 a 1 a 3 = r 5 q 5 √ 5 = √ 5 a 2 . . . from which we deduce the pattern a n +1 = √ 5 a n , a 1 = √ 5 (b) This sequence converges. To prove this we show that it is increasing and bounded. Increasing Clearly a 2 > a 1 . Assume a k > a k- 1 . Then we have a k > a k- 1 5 a k > 5 a k- 1 √ 5 a k > p 5 a k- 1 a k +1 > a k and so by induction the sequence is increasing. Bounded Evaluating the first few terms we see that a 1 = 2 . 236067977 a 2 = 3 . 343701525 a 3 = 4 . 088827170 a 4 = 4 . 521519197 a 5 = 4 . 754744576 a 6 = 4 . 875830481 a 7 = 4 . 937524927 a 8 = 4 . 968664271 and so we might suspect that 5 is a good choice for an upper bound. Indeed, a 1 < 5. Assume a k < 5, then a k < 5 5 a k < 25 √ 5 a k < √ 25 a k +1 < 5 and so by induction a n < 5 for all n . Since the sequence is increasing and bounded it must have a limit by the monotonic sequence theorem. The limit is in fact 5 in this question although you are not requiredsequence theorem....
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A06S - Math 138 Assignment 6 Solutions 1(a Based on the...

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