triclino Posted February 9, 2010 Share 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 Link to comment Share on other sites More sharing options...
Amr Morsi Posted March 28, 2010 Share Posted March 28, 2010 What is meant by "monotone sequence"? Link to comment Share on other sites More sharing options...
triclino Posted April 1, 2010 Author Share Posted April 1, 2010 What is meant by "monotone sequence"? Is a sequence increasing or decreasing Link to comment Share on other sites More sharing options...
Amr Morsi Posted April 2, 2010 Share 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. Link to comment Share on other sites More sharing options...
triclino Posted April 2, 2010 Author Share 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. Link to comment Share on other sites More sharing options...
Amr Morsi Posted April 2, 2010 Share Posted April 2, 2010 It uses the definition of a Bounded Sequence. Link to comment Share on other sites More sharing options...
triclino Posted April 2, 2010 Author Share 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 Link to comment Share on other sites More sharing options...
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