triclino Posted February 9, 2010 Posted February 9, 2010 How do we -prove that: The axiom (theorem ) of the sequence of the nested interval in real Nos implies that every monotone sequence in real nos ,bounded from above, has a limit
triclino Posted April 1, 2010 Author Posted April 1, 2010 What is meant by "monotone sequence"? Is a sequence increasing or decreasing
Amr Morsi Posted April 2, 2010 Posted April 2, 2010 Since the sequence is bounded from above, therefore it reaches a certain number (Ao), as n tends to infinity. Then, lim A(n), as n tends infinity, equals Ao. Therefore, it has a limit.
triclino Posted April 2, 2010 Author Posted April 2, 2010 Since the sequence is bounded from above, therefore it reaches a certain number (Ao), as n tends to infinity. Then, lim A(n), as n tends infinity, equals Ao. Therefore, it has a limit. On what theorem or axiom you base such a conclusion.
triclino Posted April 2, 2010 Author Posted April 2, 2010 It uses the definition of a Bounded Sequence. So, how does the definition of a bounded sequence help to prove that the sequence has a limit . Read the question again ,it says to prove that the sequence has a limit by using the theorem (axiom) of nested intervals
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