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Math Help - show that sup S =1

  1. #1
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    show that sup S =1

    show that sup S =1 where S =  \{ (-1)^{n} - \frac {1}{n^2} ; n \in N \}
    proof:
    1)  \{ (-1)^{n} - \frac {1}{n^2} ; n \in N \}  \to 1 \geq s , \forall s \in S
    2) if v< 1 want to prove  \exists s \in S such that v<s
     1-v>0 \to  \exists N^2  \ such \ that \  \frac {1}{N^2} < 1-v (by using Archimedean property ) "this step true or not ??"
     v < 1- \frac {1}{N^2} , \ \forall  \ N \ is \ even
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  2. #2
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    Quote Originally Posted by flower3 View Post
    show that sup S =1 where S =  \{ (-1)^{n} - \frac {1}{n^2} ; n \in N \}
    proof:
    1)  \{ (-1)^{n} - \frac {1}{n^2} ; n \in N \} \to 1 \geq s , \forall s \in S
    2) if v< 1 want to prove  \exists s \in S such that v<s
     1-v>0 \to \exists N^2 \ such \ that \ \frac {1}{N^2} < 1-v (by using Archimedean property ) "this step true or not ??"
     v < 1- \frac {1}{N^2} , \ \forall \ N \ is \ even

    The first step is wrong: take odd n's and the limit is -1, and take now even n's and the limit is 1 ==> the limit doesn't exist.
    And precisely this gives you the solution: take all the elements of the sequence with even n and show their limit is 1. Now just mention that for odd n's you have negative terms...

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    The first step is wrong: take odd n's and the limit is -1, and take now even n's and the limit is 1 ==> the limit doesn't exist.
    And precisely this gives you the solution: take all the elements of the sequence with even n and show their limit is 1. Now just mention that for odd n's you have negative terms...

    Tonio
    i really don't know any thing about "limits" so i solve this Q by def of sup !!!!
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  4. #4
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    Quote Originally Posted by flower3 View Post
    i really don't know any thing about "limits" so i solve this Q by def of sup !!!!
    Then what does the " \to" in <br />
\{ (-1)^{n} - \frac {1}{n^2} ; n \in N \} \to 1 \geq s , \forall s \in S<br />
mean?
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Then what does the " \to" in <br />
\{ (-1)^{n} - \frac {1}{n^2} ; n \in N \} \to 1 \geq s , \forall s \in S<br />
mean?
    means "implies"
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