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Math Help - a<c<b

  1. #1
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    a<c<b

    Ok, so {a},{b}, and {c} are sequences such that {a} converges to A, {b} converges to A and a<c<b for all n. prove that {c} converges to A.

    I know I need arbitrary epsilon and arbitrary N, but we have 2 different N's. The 2 different N's confuse me. How do I get started?
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  2. #2
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    Find an N that works for both sequences.
    n \geqslant N \Rightarrow \quad \left| {a_n  - A} \right| < \varepsilon \,\& \,\left| {b_n  - A} \right| < \varepsilon .
    a_n  \leqslant c_n  \leqslant b_n  \Rightarrow \quad  - \varepsilon  < a_n  - A \leqslant c_n  - A \leqslant b_n  - A < \varepsilon
    Carry on.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Find an N that works for both sequences.
    n \geqslant N \Rightarrow \quad \left| {a_n - A} \right| < \varepsilon \,\& \,\left| {b_n - A} \right| < \varepsilon .
    a_n \leqslant c_n \leqslant b_n \Rightarrow \quad - \varepsilon < a_n - A \leqslant c_n - A \leqslant b_n - A < \varepsilon
    Carry on.
    But I thought you ahd to do it with N1 nad N2 since they're different sequences. hence different N's.
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  4. #4
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    Quote Originally Posted by kathrynmath View Post
    But I thought you ahd to do it with N1 nad N2 since they're different sequences. hence different N's.
    Let N=N_1 + N_2.
    Then N works for both sequences.
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  5. #5
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    Quote Originally Posted by Plato View Post
    Find an N that works for both sequences.
    n \geqslant N \Rightarrow \quad \left| {a_n - A} \right| < \varepsilon \,\& \,\left| {b_n - A} \right| < \varepsilon .
    a_n \leqslant c_n \leqslant b_n \Rightarrow \quad - \varepsilon < a_n - A \leqslant c_n - A \leqslant b_n - A < \varepsilon
    Carry on.
    Ok, so I assume a<c<b and show it works?
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  6. #6
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    Quote Originally Posted by kathrynmath View Post
    Ok, so I assume a<c<b and show it works?
    From  - \varepsilon  < a_n  - A \leqslant c_n  - A \leqslant b_n  - A < \varepsilon
    it comes at once \left| {c_n  - A} \right| < \varepsilon .
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