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Math Help - where do you suggest i start with this one (series limit)

  1. #1
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    where do you suggest i start with this one (series limit)

    \lim_{n\to\infty}\frac{1\cdot 3\cdot 5 \cdot \cdots \cdot (2n-1)}{2\cdot 4\cdot 6 \cdot \cdots \cdot (2n)}

    i know the limit should be 0 because the the series in the numerator is larger the the one in the denominator for every n
    but i don't know how to start on proving it

    help?
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  2. #2
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    Re: where do you suggest i start with this one (series limit)

    I believe you mean the denominator is larger than the numerator; therefore, the limit is 0. Wouldn't proving 2n > 2n-1 (and then pointing out that the product of two larger numbers is larger than the product of two smaller numbers) suffice?
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: where do you suggest i start with this one (series limit)

    Quote Originally Posted by idom87 View Post
    \lim_{n\to\infty}\frac{1\cdot 3\cdot 5 \cdot \cdots \cdot (2n-1)}{2\cdot 4\cdot 6 \cdot \cdots \cdot (2n)}

    i know the limit should be 0 because the the series in the numerator is larger the the one in the denominator for every n
    but i don't know how to start on proving it

    help?
    You can start writing the limit as...

    \lim_{n \rightarrow \infty} \frac{1\ 3\ 5\ ...\ (2n-1)}{2\ 4\ 6\ ...\ (2n)}= \lim_{n \rightarrow \infty} \prod_{k=1}^{n} (1-\frac{1}{2k}) (1)

    Now the limit (1) is called infinite product and a basic criterion exstablishes that if the infinite product is written in the form...

    \prod_{k=1}^{\infty} (1 \pm a_{k}) (2)

    ... where all the a_{k}\ge 0, then if the series...

    \sum_{k=1}^{\infty} a_{k} (3)

    ... converges, then the infinite product (2) converges and if the (3) diverges, then the infinite product (2) tends to infinity if in (2) the sign is '+' and the infinite product tends to 0 if the sign in (2) is '-'. In Your case the series...

    \sum_{k=1}^{\infty} \frac{1}{2k} (4)

    ... diverges and the sign is '-', so that...

    Kind regards

    \chi \sigma
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  4. #4
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    Re: where do you suggest i start with this one (series limit)

    can you do it without using products? i haven't gotten there yet
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    Re: where do you suggest i start with this one (series limit)

    your'e right i mixed that up that's what i meant
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: where do you suggest i start with this one (series limit)

    Quote Originally Posted by idom87 View Post
    can you do it without using products? i haven't gotten there yet
    All right!... setting...

    p_{n}= \prod_{k=1}^{n} (1-\frac{1}{2k}) (1)

    ... is...

    \ln p_{n}= \sum_{k=1}^{\infty} \ln (1-\frac{1}{2k}) (2)

    Now if we remember that...

    \ln (1-x)= -x -\frac{x^{2}}{2}-\frac{x^{3}}{3}-... (3)

    ... is also |\ln(1-x)|>|x| so that...

    |\sum_{k=1}^{n} \ln (1-\frac{1}{2k})| > |\sum_{k=1}^{n} \frac{1}{2k}| (4)

    But the series in the right term of (4) diverges and, because for all n is \ln p_{n}<0 is...

    \lim_{n \rightarrow \infty} \ln p_{n}= - \infty \implies \lim_{n \rightarrow \infty}p_{n}=0 (5)

    Kind regards

    \chi \sigma
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