7 - MATH 138 Winter 2011 Assignment 7 Solutions 1. Suppose...

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Winter 2011 Assignment 7 – Solutions 1. Suppose { a n } is the sequence defined recursively by a 1 = 0 a n +1 = 5 + 2 a n for n = 1 , 2 , 3 .... Prove that { a n } converges and find its limit. The first few terms of the sequence are: a 1 = 0 , a 2 = 5 2 . 23 , a 3 = p 5 + 2 5 3 . 07 , a 4 = q 5 + 2 p 5 + 2 5 3 . 34 , a 5 3 . 41 . So we would like to show that the sequence is increasing and bounded above, by say 4 (it is clear that 0 is a lower bound). We will do this by induction. For the base case, we have a 1 = 0 a 2 2 . 23 4 . Then, we assume that the following is true: a n a n +1 4 . Now we would like to show that a n +1 a n +2 4 . Start with the assump- tion: a n a n +1 4 ⇐⇒ 5 + 2 a n 5 + 2 a n +1 5 + 2(4) = 13 ⇐⇒ 5 + 2 a n p 5 + 2 a n +1 13 ⇐⇒ a n +1 a n +2 13 < 4 Therefore, this is true for all n so the sequence is increasing and bounded, and by the Monotonic Sequence Theorem, it converges. If { a n } converges to L , then we must have lim n →∞ a n = lim n →∞ a n +1 = L . Taking the limit of the definition of { a n } gives: lim n →∞ a n +1 = lim n →∞ 5 + 2 a n lim n →∞ a n +1 = q 5 + 2 lim n →∞ a n L = 5 + 2 L Squaring and re-arranging, we get L 2 - 2 L - 5 = 0 which gives L = 1 ± 6 . Since
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7 - MATH 138 Winter 2011 Assignment 7 Solutions 1. Suppose...

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