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Math Help - limit and convergent subsequence?

  1. #1
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    limit and convergent subsequence?

    If an >= 0 for all n and has no convergent subsequence, show lim n->infinity an=infinity.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cowgirl123 View Post
    If an >= 0 for all n and has no convergent subsequence, show lim n->infinity an=infinity.
    Here's what i am thinking. I don't know if it is 100% correct, but it will at least give you something to think about.

    Assume \{ a_n \} is a sequence which has no convergent subsequence, and assume to the contrary that \lim_{n \to \infty} a_n \ne \infty. Then we have three other possibilities for the infinite limit: (1) \lim_{n \to \infty}a_n = - \infty, (2) \lim_{n \to  \infty}a_n = L, where L is finite, or (3) \lim_{n \to \infty}a_n does not exist, in that the a_n's alternate between certain values.

    Clearly case (1) is impossible, since a_n \ge 0 for all n

    Consider case (2): If \lim_{n \to \infty}a_n = L it means that for some N \in \mathbb{N}, n > N implies |a_n - L|< \epsilon for all \epsilon > 0. In other words, the terms of the sequence all get arbitrarily close to L after a certain point in the sequence. Thus we can construct a subsequence, say, of the even terms of \{ a_n \}. Call this sequence \{ a_{n_k} \} where n_k = 2k for k \in \mathbb{Z}. Clearly, this sequence converges to L also, since all terms of \{ a_n \} are getting close to L. But this contradicts our original assumption, and so, cannot be the case.

    Consider case (3): If the a_n's are alternating, again we can construct a convergent subsequence. This subsequence can consist of any single term that occurs in the sequence repeatedly. Which again, contradicts our assumption that \{ a_n \} has no convergent subsequence.

    Thus, it must be the case that \lim_{n \to \infty}a_n = \infty

    QED


    Now what bothers me is that the argument for case (3) is not as "airtight" as i would like. i feel like i'm missing something. also, i'm not sure if i am neglecting to consider some other case.
    Last edited by Jhevon; September 26th 2007 at 03:23 AM.
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  3. #3
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    For each positive integer K, there are only finitely many terms of a_n < K.
    Otherwise, there would be a convergence subsequence.
    Hence, the sequence must diverge to infinity.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Plato View Post
    For each positive integer K, there are only finitely many terms of a_n < K.
    Otherwise, there would be a convergence subsequence.
    Hence, the sequence must diverge to infinity.
    well, that was easy
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