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Math Help - Limit Proof

  1. #1
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    Limit Proof

    I am having a bit of trouble with what should be a simple proof:

    Let (an) be a convergent sequence of real numers, an >= 0 for all n in the natural number set. If the limit of an is a (an->a) show that a >= 0.

    I understand that I need to use an epsilon neighbourhood to show a contradiction when a is < 0 but am having a hard time finding an epislon that produces a contradiction.

    Thanks for the help
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  2. #2
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    Suppose that a<0 then choose \varepsilon  = \frac{{ - a}}{2} > 0.
    Because of sequence convergence we have \left( {\exists N} \right)\left[ {n \ge N \Rightarrow \left| {a_n  - a} \right| < \varepsilon } \right].

    But that means: a - \varepsilon  < a_n  < a + \varepsilon  \le a + \frac{{ - a}}{2} = \frac{a}{2} < 0.

    Do you see the contradiction?
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  3. #3
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    \le a + \frac{{ - a}}{2} = \frac{a}{2} < 0


    I am sure I am just being very dense but where did the rhs of the inequality come from?
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  4. #4
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    Quote Originally Posted by jlindbo View Post
    \le a + \frac{{ - a}}{2} = \frac{a}{2} < 0


    I am sure I am just being very dense but where did the rhs of the inequality come from?
    \epsilon = - \frac a2, so a + \epsilon = a - \frac a2 = \frac a2

    since - \frac a2 > 0, we have \frac a2 < 0

    now continue to show the contradiction
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  5. #5
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    Doh ;P

    And since a_n > 0 that produces the contradiction.

    Thanks
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jlindbo View Post
    Doh ;P

    And since a_n > 0 that produces the contradiction.

    Thanks
    correct
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  7. #7
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    This is a hugely important idea in sequence convergence that has a similar application to continuous functions.

    If the limit of a sequence is positive then almost all the terms of that sequence must be positive. There is a similar statement of a negative limit.
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