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Math Help - Sequences and limits

  1. #1
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    Sequences and limits

    Suppose that a(n) goes to a and that a(n) is greater than or equal to b for each n. Prove that a is greater than or equal to b.

    I began by supposing for a contradiction that a is less than b.

    After this I am not quite sure how to proceed. Any help is appreciated.
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  2. #2
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    Re: Sequences and limits

    Quote Originally Posted by Shapeshift View Post
    Suppose that a(n) goes to a and that a(n) is greater than or equal to b for each n. Prove that a is greater than or equal to b.
    Suppose that a<b then let \epsilon=b-a .
    The contradiction is almost right there.
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  3. #3
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    Re: Sequences and limits

    Ok so I get

    a - b < a(n) - a < b - a which implies that a(n) < b, which is a contradiction to what we are given. Is this correct? Also just a quick question, why do we let epsilon = b - a? In other words, what is the reasoning behind that?
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  4. #4
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    Re: Sequences and limits

    well, the idea is that in order to prove a can't be less than b, we have to show that a < b will force some a(k)< b. but that doesn't tell us "which" k we need to find.

    however, we DO know that a(n) converges to a, so, given any postive number, ε, we can always find an N for which a(k) is within ε of a for all k > N.

    if a < b, then b-a is positive, so we can use it as our epsilon. so now we have an N for which |a - a(k)| < b - a, for all k > N.

    so for all these k, a(k) is closer to a than b is. since b is "above" a, a(k) is "below" b, our desired contradiction.

    in other words, we know the a(n) are getting closer to a. if a < b, how close to do we need to be? closer to a than b, that is less than b-a.

    this will force a(n) < b, which is what we're trying to show (to get our contradiction).
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