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Math Help - Limit (7)

  1. #1
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    Limit (7)

    Evaluate \lim_{n\to\infty} \sqrt[n]{\binom{n}{\lfloor na \rfloor}}, where 0 < a < 1 is given.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    Evaluate \lim_{n\to\infty} \sqrt[n]{\binom{n}{\lfloor na \rfloor}}, where 0 < a < 1 is given.
    I believe the answer is
    Spoiler:
    \frac{1}{a^a(1-a)^{1-a}}
    but as of right now I have no proof of this conjecture.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Ignore this. I accidentally clicked "Submit" halfway through my proof.
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    Evaluate \lim_{n\to\infty} \sqrt[n]{\binom{n}{\lfloor na \rfloor}}, where 0 < a < 1 is given.
    And here's my proof. Because I don't want to repeatedly type this, and I don't do anything with it for a while, let k=\lfloor na\rfloor. Also, assume there is a \lim_{n\to\infty} in front of each of these.

    Spoiler:
    Rewrite this expression as

    \exp\left[\frac{1}{n}\ln\left(\frac{n!}{(n-k)!k!}\right)\right]=\exp\left[\frac{1}{n}\left(\ln(n!)-\ln((n-k)!)-ln(k!)\right)\right]

    Using Stirling's approximation \ln n!\approx n\ln n-n, the above equals

    \exp\left[\frac{1}{n}\big(n\ln(n)-n-(n-k)\ln(n-k)+(n-k)-k\ln(k)+k\big)\right]= \exp\left[\frac{1}{n}\big(n\ln(n)-n\ln(n-k)+k\ln(n-k)-k\ln(k)\big)\right]=

    \exp\left[\frac{1}{n}\left(n\ln\left(\frac{n}{n-k}\right)+k\ln\left(\frac{n-k}{k}\right)\right)\right]= \exp\left[\ln\left(\frac{n}{n-k}\right)+\frac{k}{n}\ln\left(\frac{n}{k}-1\right)\right]=

    \frac{n}{n-k}\cdot\left(\frac{n}{k}-1\right)^{k/n}=\frac{1}{1-a}\left(\frac{1}{a}-1\right)^a=\frac{(1-a)^a}{(1-a)a^a}=\boxed{\frac{1}{a^a(1-a)^{1-a}}}
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  5. #5
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    Can i use Riemann Sum ?
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  6. #6
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    Quote Originally Posted by redsoxfan325 View Post
    And here's my proof. Because I don't want to repeatedly type this, and I don't do anything with it for a while, let k=\lfloor na\rfloor. Also, assume there is a \lim_{n\to\infty} in front of each of these.

    Spoiler:
    Rewrite this expression as

    \exp\left[\frac{1}{n}\ln\left(\frac{n!}{(n-k)!k!}\right)\right]=\exp\left[\frac{1}{n}\left(\ln(n!)-\ln((n-k)!)-ln(k!)\right)\right]

    Using Stirling's approximation \ln n!\approx n\ln n-n, the above equals

    \exp\left[\frac{1}{n}\big(n\ln(n)-n-(n-k)\ln(n-k)+(n-k)-k\ln(k)+k\big)\right]= \exp\left[\frac{1}{n}\big(n\ln(n)-n\ln(n-k)+k\ln(n-k)-k\ln(k)\big)\right]=

    \exp\left[\frac{1}{n}\left(n\ln\left(\frac{n}{n-k}\right)+k\ln\left(\frac{n-k}{k}\right)\right)\right]= \exp\left[\ln\left(\frac{n}{n-k}\right)+\frac{k}{n}\ln\left(\frac{n}{k}-1\right)\right]=

    \frac{n}{n-k}\cdot\left(\frac{n}{k}-1\right)^{k/n}=\frac{1}{1-a}\left(\frac{1}{a}-1\right)^a=\frac{(1-a)^a}{(1-a)a^a}=\boxed{\frac{1}{a^a(1-a)^{1-a}}}
    your final answer is correct but you need to be more careful with using the approximation \ln n!\approx n\ln n-n.

    simplependulum, you can use whatever you want!
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  7. #7
    Super Member redsoxfan325's Avatar
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    How could I make it more rigorous?
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  8. #8
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    Quote Originally Posted by redsoxfan325 View Post
    How could I make it more rigorous?
    by using n! = \sqrt{2 \pi n} \left(\frac{n}{e} \right)^n (1 + O(1/n)). see Stirling's approximation - Wikipedia, the free encyclopedia.
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  9. #9
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    I got the approximation of  \ln(n!) when  n tends to infinty , using Riemann Sum



     \ln(n!) =   \ln(1) + \ln(2) + ... + \ln(n)


     = n \frac{ \ln(1/n) + \ln(2/n) + ... + \ln(n/n) }{n} + n\ln(n)

     = n \int_0^1 \ln(x) ~dx  + n\ln(n)

     =  n \left [ x\ln(x) - x \right ]_0^1 + n\ln(n)


     = n\ln(n) - n which equals to what redsoxfan325 mentioned .


    I believe the  n\ln(n) - n is correct but i'm not sure if the above is a rigorous proof .
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by simplependulum View Post
    I got the approximation of  \ln(n!) when  n tends to infinty , using Riemann Sum



     \ln(n!) = \ln(1) + \ln(2) + ... + \ln(n)


     = n \frac{ \ln(1/n) + \ln(2/n) + ... + \ln(n/n) }{n} + n\ln(n)

     = n \int_0^1 \ln(x) ~dx + n\ln(n)

     = n \left [ x\ln(x) - x \right ]_0^1 + n\ln(n)


     = n\ln(n) - n which equals to what redsoxfan325 mentioned .


    I believe the  n\ln(n) - n is correct but i'm not sure if the above is a rigorous proof .
    Well the equals sign is pushing it. In your Riemann sum step what you really mean as that as n increases without bound \frac{ \ln(1/n) + \ln(2/n) + ... + \ln(n/n) }{n}\to\int_0^1\ln(x)\text{ }dx. You of course knew that, but in case anyone misinterpreted it.
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