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Math Help - A Sequential Limit

  1. #1
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    A Sequential Limit

    Hi all, I was wondering if someone could help me show

    \lim_{n\rightarrow\infty}{2n \choose n}\left(\frac{1}{4}\right)^n=0

    I was able to do some algebra to show that this limit is

    \lim_{n\rightarrow\infty} \frac{1}{2}\frac{3}{4}\frac{5}{6}\dots \frac{2n-1}{2n}

    which must converge since the sequence is decreasing and bounded by 0. But I haven't been able to figure out how to show the limit is actually 0 using the epsilon/delta definition of convergence. Is there an alternative way that I might prove the limit is 0, or can someone give me a hint with regard to the epsilon/delta definition?

    Thanks
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  2. #2
    MHF Contributor chisigma's Avatar
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    The 'infinite product' ...

    \displaystyle \prod_{k=1}^{\infty} (1+a_{k}) (1)

    ... converges or diverges is converges or diverges the 'infinite series'...

    \displaystyle \sum_{k=1}^{\infty} a_{k} (2)

    In Your case is \displaystyle a_{k} = -\frac{1}{2k} so that the series (2) and the product (1) both diverge. The series tends to -\infty so that the product tends to 0...

    Kind regards

    \chi \sigma
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  3. #3
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    Thanks for the help chisigma, but I'm a little confused still. Why does the sum diverging to -\infty imply that the product goes to 0? I assume that it has something to do with the logarithmic relationship between sums and products and the fact that \lim_{x\rightarrow -\infty} log(x)=0, but I'm not sure how you derive that theorem between the product and the series.
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  4. #4
    MHF Contributor chisigma's Avatar
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    A 'good enough' tutorial can be found in...

    Infinite product - Wikipedia, the free encyclopedia

    Kind regards

    \chi \sigma
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