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Thread: Bolzano-Weirstrauss

  1. #1
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    Bolzano-Weirstrauss

    Suppose the sequence {a_n} converges to L and limsup{b_n} = M < infiniti
    //({b_n} is bounded below and above)// Prove limsup{a_n*b_n} = LM.

    b) also show that the statement is false if {b_n} is not bounded below.


    My Sketch: I have to show that limsup{a_n*b_n} = Q is finite; then use
    subsequences to show Q less than or equal to LM, and that is greater than or equal to LM.
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  2. #2
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    Hang on just a minute.

    If $\displaystyle a_n=-1$ for all $\displaystyle n$ then $\displaystyle \lim a_n=-1$, i.e. $\displaystyle L=-1$.

    If $\displaystyle b_n=\begin{cases}1&\textrm{for odd }n\\-2&\textrm{for even }n\end{cases}$ then $\displaystyle \limsup b_n=1$, i.e. $\displaystyle M=1$. Notice $\displaystyle b_n$ is bounded below.

    Then $\displaystyle a_nb_n=\begin{cases}-1&\textrm{for odd }n\\2&\textrm{for even }n\end{cases}$ and so $\displaystyle \limsup a_nb_n=2\neq LM$.

    As is well known $\displaystyle \limsup(-a_n)=-\liminf a_n$ when both exist.

    The result as stated does not hold, therefore. You might need an extra condition such as $\displaystyle L>0$, but then why would $\displaystyle b_n$ have to be bounded below? I think we need to be told.
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  3. #3
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    [incorrect]
    Last edited by putnam120; Jul 8th 2009 at 03:05 PM. Reason: incorrect proof.
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  4. #4
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    Here's another way to look at it (though I still don't use that $\displaystyle b_n$ is bounded below. and am no longer sure about needing L>0)

    Let $\displaystyle n_k$ be a sequence such that $\displaystyle b_{n_k}\to M$. Then $\displaystyle a_{n_k}b_{n_k}\to\limsup a_nb_n$ since $\displaystyle a_n\to L$ and thus does so along ANY subsequence.
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