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Math Help - lim sup, lim inf and cluster points.

  1. #1
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    Cool lim sup, lim inf and cluster points.

    Hi

    So, lets say i have a sequence a(n). A sequence can have a lim superior (greatest cluster point) or lim inferior (lowest cluster point), only if the sequence is bounded, right?

    so, given a(n):= (1 + (-1)^n nē) / (2 + 3n + nē)
    with n >> infinity, lim a(n) diverges between -infinity and infinity therefor it isn't bounded and has no lim superior or lim inferior. correct?

    and a more tricky sequence definied by:
    a(3n-2):= 3 + 1/n
    a(3n-1):= 2/n
    a(3n):= - 1/nē
    my (correct/incorrect) answer:
    the sequence a(n) has two cluster points, a limit inferior of the value 0 and a limit superior of the value 3.

    is everything i mentioned above correct?
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  2. #2
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    Re: lim sup, lim inf and cluster points.

    Quote Originally Posted by nappysnake View Post
    sequence a(n). A sequence can have a lim superior (greatest cluster point) or lim inferior (lowest cluster point), only if the sequence is bounded, right?

    so, given a(n):= (1 + (-1)^n nē) / (2 + 3n + nē)
    with n >> infinity, lim a(n) diverges between -infinity and infinity therefor it isn't bounded and has no lim superior or lim inferior. correct?
    Surely \frac{{1 + ( - 1)^n n^2 }}{{2 + 3n + n^2 }} = \frac{{\tfrac{1}{{n^2 }} + ( - 1)^n }}{{\tfrac{2}{{n^2 }} + \tfrac{3}{n} + 1}}~??

    If I have read the question correctly, you will want to rethink the first part.
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  3. #3
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    Re: lim sup, lim inf and cluster points.

    Oops, the sequence diverges between -1 and 1 therefor possesses a lim superior and a lim inferior..my bad. How about the second sequence, everything alright with it?
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