Explanation Given: The sequence is a 1 = 2 , a n + 1 = 1 3 − a n (1) Theorem used: Monotonic Sequence Theorem “If the sequence is bounded and monotonic, then the sequence is convergent.” Proof: Induction Method Claim: Let P n be the statement that a n + 1 ≤ a n and 0 < a n ≤ 2 . (2) Base case: n = 1 Substitute 1 for n in equation (2), a 1 + 1 ≤ a 1 and 0 < a 1 ≤ 2 a 2 ≤ a 1 and 0 < a 1 ≤ 2 Obtain the value of a 2 . Substitute 1 for n in equation (1), a 1 + 1 = 1 3 − a 1 a 2 = 1 3 − 2 [ ∵ a 1 = 2 ] = 1 1 = 1 Thus, the value of a 2 = 1 . Since the value of a 1 = 2 and a 2 = 1 , a 2 ≤ a 1 and 0 < a 1 ≤ 2 . Therefore, the claim is true for n = 1 . Induction hypothesis: n = k Assume that the claim is true when n = k . If P k is the statement, then a k + 1 ≤ a k and 0 < a k ≤ 2 is true. Inductive step: n = k + 1 To prove that the claim is true when n = k + 1 That is, if P k + 1 is the statement, then a ( k + 1 ) + 1 ≥ a k + 1 and 0 < a k + 1 ≤ 2 is true. Substitute k + 1 for n in equation (1), a ( k + 1 ) + 1 = 1 3 − a k + 1 (3) From induction hypothesis, a k + 1 ≤ a k − a k + 1 > − a k Add by 3 on both sides of the inequality. 3 − a k + 1 > 3 − a k 1 3 − a k + 1 ≤ 1 3 − a k (4) Substitute the inequality (4) in the equation (3) and then plug in 1 3 − a k for a k + 1 . a ( k + 1 ) + 1 ≤ 1 3 − a k a ( k + 1 ) + 1 ≤ a k + 1 (5) Consider a k + 1 , then apply equation (1) and the induction hypothesis 0 < a k ≤ 2

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ISBN: 9781305266636

Author: James Stewart

Publisher: Cengage Learning

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Sequence and Series

A sequence is defined as a collection of things. Series is defined to sum the things one by one in the sequence. It was invented by German mathematician Carl Friedrich Gauss.

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Chapter 11.1, Problem 82E

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To show: The sequence {an} is increasing and an<3 for all n . Also, deduce that the sequence is convergent and then obtain the limit of the sequence.

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Chapter 11.1, Problem 81E

Chapter 11.1, Problem 83E

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Chapter 11 Solutions

Single Variable Calculus

Ch. 11.1 - (a) What is a sequence? (b) What does it mean to...Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Prob. 6E Ch. 11.1 - Prob. 7E Ch. 11.1 - Prob. 8E Ch. 11.1 - Prob. 9E Ch. 11.1 - Prob. 10E

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