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Thread: (kn)! / (kn+k)! = k^k?

  1. #1
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    (kn)! / (kn+k)! = k^k?

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    can anyone explain how why (kn!) / (kn+k)! is K^K in the final part of the answer?
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  2. #2
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    Re: (kn)! / (kn+k)! = k^k?

    Oops. Didn't realize it was a limit at first glance.
    Last edited by Bean; Nov 27th 2012 at 06:34 PM.
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  3. #3
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    Re: (kn)! / (kn+k)! = k^k?

    i just don't know how to write out (kn+k)!.. she hardly taught us how to do factorials
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    Re: (kn)! / (kn+k)! = k^k?

    In the final part of the ratio test, divide the numerator and the denominator by n and take n to infinity which takes all of the constants to zero. The bottom simplifies to k*k for k times. The numerator simplifies to 1^(k) or 1.
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    Re: (kn)! / (kn+k)! = k^k?

    $\displaystyle \frac{kn}{n} + \frac{k}{n} = k$ as n goes to infinity.
    Last edited by Bean; Nov 27th 2012 at 06:30 PM.
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  6. #6
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    Re: (kn)! / (kn+k)! = k^k?

    Within the absolute value bars, disregarding the $\displaystyle x$ what you have are $\displaystyle k$th degree polynomials in the numerator and denominator in $\displaystyle n$, so we know the limit as $\displaystyle n\to\infty$ is the ratio of the leading coefficients, which is:

    $\displaystyle \frac{1}{k^k}$

    And so the limit of the entire expression is:

    $\displaystyle \left|\frac{x}{k^k} \right|$
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