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Thread: Convergent Sequence product Rule Proof.

  1. #1
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    Convergent Sequence product Rule Proof.

    Not sure how we get from the left to right right hand side of this equation. The idea being that $\displaystyle a_nb_n - lm$ is being expressed in terms of $\displaystyle a_n - l $ and $\displaystyle b_n - m$

    $\displaystyle a_nb_n - lm = (a_n - l)(b_n - m) +m(a_n - l) +l(b_n - m)$

    Thanks.
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  2. #2
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    Re: Convergent Sequence product Rule Proof.

    Quote Originally Posted by AlyoshaKaz View Post
    Not sure how we get from the left to right right hand side of this equation. The idea being that $\displaystyle a_nb_n - lm$ is being expressed in terms of $\displaystyle a_n - l $ and $\displaystyle b_n - m$

    $\displaystyle a_nb_n - lm = (a_n - l)(b_n - m) +m(a_n - l) +l(b_n - m)$

    Thanks.
    You just need to show that

    $\displaystyle |a_nb_n-ml|< \epsilon$

    So just add and substract $\displaystyle -a_nb+a_nb$ which is equal to zero to get

    $\displaystyle |a_nb_n-mb_n+mb_n-ml|=|(a_n-m)b_n+(b_n-l)m|$



    You still need to bound $\displaystyle b_n$ and finish this up.

    Best wishes

    IF you get stuck post back your workings and where your are stuck.
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  3. #3
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    Re: Convergent Sequence product Rule Proof.

    Quote Originally Posted by AlyoshaKaz View Post
    Not sure how we get from the left to right right hand side of this equation. The idea being that $\displaystyle a_nb_n - lm$ is being expressed in terms of $\displaystyle a_n - l $ and $\displaystyle b_n - m$

    $\displaystyle a_nb_n - lm = (a_n - l)(b_n - m) +m(a_n - l) +l(b_n - m)$

    If you carefully multiply the right side out, term by term, and combine then you will see how it all works.
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