sol7 - MATH 138 Calculus 2 Solutions#7 Fall 2010 1 First we...

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MATH 138 Calculus 2 Fall 2010 Solutions #7 1. First, we use induction to show 0 < a n 3 for all n 1. This is true for n = 1 since a 1 = 3. If we assume that it is true for n = k , then we have 0 < a k 3 , 1 4 - a k < 4 1 4 < 1 4 - a k 1 0 < a k +1 3 since a k +1 = 1 / (4 - a k ). We have deduced that 0 < a n 3 is true for n = k + 1. Therefore, the statement that 0 < a n 3 is true for all n 1 by induction. We now show that a n is decreasing, i.e., a n +1 a n for all n 1. This is true for n = 1 since a 2 = 1 4 - a 1 = 1 4 - 3 = 1 3 = a 1 . Assume that it is true for n = k , i.e., a k +1 a k . Then, a k +2 - a k +1 = 1 4 - a k +1 - 1 4 - a k = a k +1 - a k (4 - a k +1 )(4 - a k ) 0 since a k +1 a k , 0 < a k 3 and 0 < a k +1 3. Hence, a n +1 a n is true for n = k + 1. Therefore, a n +1 a n for all n 1 by induction. Since the sequence a n is bounded below and decreasing (or bounded and monotonic), it is convergent by the monotonic sequence theorem. Let
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sol7 - MATH 138 Calculus 2 Solutions#7 Fall 2010 1 First we...

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