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Math Help - Sequence convergence

  1. #1
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    Sequence convergence

    Not sure how to go about this one.

    If pn is a non-decreasing sequence and there is a point, p, to the right of each point of the sequence, then the sequence converges to some point.
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  2. #2
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    Quote Originally Posted by spikedpunch View Post
    If pn is a non-decreasing sequence and there is a point, p, to the right of each point of the sequence, then the sequence converges to some point.
    That is an oddly stated problem.
    Mathematically translated it says “A non-decreasing sequence bound above converges”.
    Which of course is true. The proof follows from the completeness property.
    The sequence converges to its least upper bound.
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by spikedpunch View Post
    Not sure how to go about this one.

    If pn is a non-decreasing sequence and there is a point, p, to the right of each point of the sequence, then the sequence converges to some point.
    Isn't this the Monotone convergence theorem?

    "If a sequence is monotone and bouned, then is converges"

    here is an outline:

    Suppose the sequence (p_{n}) is increasing and bounded above.

    Consider the set of all points in the sequece \{p_{n}|n\in\mathbb{N}\}.

    By the axiom of completeness, sup\{p_{n}|n\in\mathbb{N}\} exists (since the set is a bounded subset of \mathbb{R}.

    Assume lim(p_{n})=s, where s=sup\{p_{n}|n\in\mathbb{N}\}.

    Then, for \epsilon>0, s-\epsilon<s. Thus, there exists a point p_{N} such that s-\epsilon<p_{N}<s. Furthermore, because our sequence is increasing, it follows that, for every n\geq\\N, p_{N}\leq\\p_{n}. Thus, s-\epsilon<p_{N}\leq\\p_{n}<s<s+\epsilon.

    This implies |p_{n}-s|<\epsilon
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  4. #4
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    If I'm trying to prove that pn converges to some point, then I don't think I can assume that lim(pn)=s ......right?
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  5. #5
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    Quote Originally Posted by spikedpunch View Post
    If I'm trying to prove that pn converges to some point, then I don't think I can assume that lim(pn)=s ......right?
    I really don't know how to answer that question.
    Because I don't know what it means.
    It is quite easy to prove this theorem: A bounded non-decreasing sequence converges to its least upper bound.
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  6. #6
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    I'm saying that if I'm trying to prove a sequence converges, i can't assume that it has a limit and converges, because I would be assuming what I'm trying to prove.
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