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Math Help - limit of xth root etc etc

  1. #1
    Senior Member polymerase's Avatar
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    limit of xth root etc etc

    Determine if \lim_{x\to\infty} \frac{\sqrt[x]{x!}}{x} exist (and value) and if it doesn't explain why.
    (in case thats too small, it reads the xth root of x! all over x)

    Potentially VERY easy series questions?!?

    Thanks
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by polymerase View Post
    Determine if \lim_{x\to\infty} \frac{\sqrt[x]{x!}}{x} exist (and value) and if it doesn't explain why.
    (in case thats too small, it reads the xth root of x! all over x)

    Potentially VERY easy series questions?!?

    Thanks
    use the fact taht n!\sim{\sqrt{2\pi{n}}n^{n}e^{-n}}

    then it should be apparent
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    use the fact taht n!\sim{\sqrt{2\pi{n}}n^{n}e^{-n}}

    then it should be apparent
    Firstly, thank you

    Secondly, so is the answer \frac{1}{e}?

    Thirdly, is there another way to do this, because we never learned the relationship between n! and the stuff you stated above
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by polymerase View Post
    Firstly, thank you

    Secondly, so is the answer \frac{1}{e}?

    Thirdly, is there another way to do this, because we never learned the relationship between n! and the stuff you stated above
    Good job! Indeed the answer is \frac{1}{e}

    and you could also do this y=\lim_{x\to\infty}\frac{(x!)^{\frac{1}{x}}}{x}\Ri  ghtarrow{\ln(y)=\frac{\ln(x!)}{x^2}}

    and this is painful but \ln(x!)\approx{x\ln(x)-x+\frac{\ln(x)}{2}+\frac{\ln(2\pi)}{2}}

    Or you could als use that n!=\int_0^{\infty}x^{n}e^{-x}dx

    The gamma function and go L'hopitals
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  5. #5
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Good job! Indeed the answer is \frac{1}{e}

    and you could also do this y=\lim_{x\to\infty}\frac{(x!)^{\frac{1}{x}}}{x}\Ri  ghtarrow{\ln(y)=\frac{\ln(x!)}{x^2}}

    and this is painful but \ln(x!)\approx{x\ln(x)-x+\frac{\ln(x)}{2}+\frac{\ln(2\pi)}{2}}

    Or you could als use that n!=\int_0^{\infty}x^{n}e^{-x}dx

    The gamma function and go L'hopitals
    REALLY?! thats the alternative! how the %$!@ does our prof. expect us to solve that? we never learn ANY of that!

    LOL is there ANOTHER way?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Good job! Indeed the answer is \frac{1}{e}

    and you could also do this y=\lim_{x\to\infty}\frac{(x!)^{\frac{1}{x}}}{x}\Ri  ghtarrow{\ln(y)=\frac{\ln(x!)}{x^2}}

    and this is painful but \ln(x!)\approx{x\ln(x)-x+\frac{\ln(x)}{2}+\frac{\ln(2\pi)}{2}}

    Or you could als use that n!=\int_0^{\infty}x^{n}e^{-x}dx

    The gamma function and go L'hopitals
    Expanding upon my first thought after simplifying you would have \lim_{n\to\infty}\frac{\ln(x)}{x}-1+\frac{\ln(x)}{2x^2}+\frac{\ln(2\pi)}{2x^2}=-1 therefore \ln(y)=-1\Rightarrow{y=e^{-1}}

    This came from the ln(x!) approximation over x
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by polymerase View Post
    REALLY?! thats the alternative! how the %$!@ does our prof. expect us to solve that? we never learn ANY of that!

    LOL is there ANOTHER way?
    Haha...Uhm..I am surprised I thought of three ways...uhm...I cant think of anymore...maybe if you look at this post above you cna use that...that is not overly complicated? haha
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  8. #8
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Expanding upon my first thought after simplifying you would have \lim_{n\to\infty}\frac{\ln(x)}{x}-\frac{1}{x}+\frac{\ln(x)}{2x^2}+\frac{\ln(2\pi)}{2  x^2}=-1 therefore \ln(y)=-1\Rightarrow{y=e^{-1}}

    This came from the ln(x!) approximation over x
    How do you figure out/calculate that \ln (x!)=above stuff?
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by polymerase View Post
    How do you figure out/calculate that \ln (x!)=above stuff?
    Uhm....well I have the memorized but if you need to see the entire thing...hold on let me find...an ariticle...ah! here we go read here

    Natural logarithm - Wikipedia, the free encyclopedia

    sorry edit:
    Look here
    Factorial - Wikipedia, the free encyclopedia
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  10. #10
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    Quote Originally Posted by Mathstud28 View Post
    Or you could als use that n!=\int_0^{\infty}x^{n}e^{-x}dx
    How will you use this?
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  11. #11
    Math Engineering Student
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    Quote Originally Posted by polymerase View Post

    LOL is there ANOTHER way?
    \begin{aligned}<br />
   \frac{\sqrt[x]{x!}}{x}&=\exp \left[ \ln \left( \prod\limits_{k\,=\,1}^{x}{\left( \frac{k}{x} \right)^{1/x}} \right) \right] \\ <br />
 & =\exp \left[ \frac{1}{x}\sum\limits_{k\,=\,1}^{x}{\ln \frac{k}{x}} \right] \\ <br />
 & =\exp \left[ \int_{0}^{1}{\ln y\,dy} \right],\,\text{as }x\to\infty.<br />
\end{aligned}
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  12. #12
    MHF Contributor arbolis's Avatar
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    Nice limit question. Using my intuition I thought it would tend to positive infinite.
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