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Posted

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

  • 1 month later...
Posted

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.

Posted
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.

Posted
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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