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Math Help - Real Analysis - Limits and Order #2

  1. #1
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    Real Analysis - Limits and Order #2

    Let (an) be a bounded (not necessarily convergent) sequence, and assume
    lim bn = 0. Show that lim(an*bn) = 0. Why are we not allowed to use the Algebraic Limit Theorem to prove this.
    Last edited by ajj86; October 22nd 2008 at 06:59 PM. Reason: Title Error
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by ajj86 View Post
    Let (an) be a bounded (not necessarily convergent) sequence, and assume
    lim bn = 0. Show that lim(an*bn) = 0. Why are we not allowed to use the Algebraic Limit Theorem to prove this.
    What is the algebraic limit theorem ??

    \lim b_n=0 \Leftrightarrow \forall \epsilon > 0, ~ \exists N \in \mathbb{N}, ~ \forall n > N, ~ |b_n|< \epsilon

    (a_n) \text{ bounded } \Leftrightarrow \exists K>0 \quad |a_n|<K \quad \forall n (in particular for n>N)

    Multiply :
    |a_n| \times |b_n|=|a_n \times b_n|< \epsilon K

    So :
    \forall \delta= \epsilon K, ~ \exists N \in \mathbb{N}, ~ \forall n > N, ~ |a_n| \times |b_n|=|a_n \times b_n|< \delta

    This is the definition for \lim a_n \times b_n=0
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  3. #3
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    Thanks Moo.

    I appreciate your answer to my question. I've been having a lot of trouble understanding real analysis. This class reminds me a lot of abstract algebra. I guess proofs just aren't my forte.
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