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Thread: Divergence of limit superior

  1. #1
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    Divergence of limit superior

    $\displaystyle \lim\sup x_n=-\infty\implies \lim x_n=-\infty,$ does the converse hold?

    How to prove this?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Connected View Post
    $\displaystyle \lim\sup x_n=-\infty\implies \lim x_n=-\infty,$ does the converse hold?

    How to prove this?
    You tell me, I've helped you with a bunch. Consider maybe the fact that $\displaystyle \lim x_n\leqslant \limsup x_n$ assuming that $\displaystyle \lim x_n$ converges to some element of $\displaystyle \mathbb{R}\cup\{-\infty,\infty}$
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  3. #3
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    If you understand the definition then this is basically trivial.

    $\displaystyle \limsup x_{n} = \lim_{n\to\infty} \sup \{x_{i}\ : i \geq n\}$

    Then use the definition of $\displaystyle \lim_{n\to\infty}x_{n} = -\infty$
    Last edited by Beaky; Mar 29th 2011 at 06:27 PM.
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