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Math Help - Limit:

  1. #1
    Member great_math's Avatar
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    Limit:

    \lim_{n\to\infty} \left(\frac{(2n)!}{n!n^n}\right)^{1/n}
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  2. #2
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    Quote Originally Posted by great_math View Post
    \lim_{n\to\infty} \left(\frac{(2n)!}{n!n^n}\right)^{1/n}
    We can write,
    \left( \frac{(2n)!}{(n!)^2}\cdot \frac{n!}{n^n} \right)^{1/n} = \left[ {{2n}\choose n} \right]^{1/n} \cdot \tfrac{1}{n}(n!)^{1/n}

    It is a known-fact that \tfrac{1}{n}(n!)^{1/n} \to \tfrac{1}{e}

    Also it is well-known that \sum_{n=0}^{\infty} {{2n}\choose n} x^n = \frac{1}{\sqrt{1-4x}} for |x| < \tfrac{1}{4}
    Now the root test applied to this series must therefore give \left[ {{2n}\choose n}\right]^{1/n} \to 4

    Thus, the limit should (hopefully be if no mistakes were done) \frac{4}{e}.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by great_math View Post
    \lim_{n\to\infty} \left(\frac{(2n)!}{n!n^n}\right)^{1/n}
    Or

    \lim_{x\to\infty}f(x)^{\frac{1}{x}}=\lim_{x\to\inf  ty}\frac{f(x+1)}{f(x)}

    So this limit is equivalent to

    \lim_{n\to\infty}\frac{\frac{(2n+2)(2n+1)(2n)!}{(n  +1)^{n+2}n!}}{\frac{(2n)!}{n!n^n}}=\lim_{n\to\inft  y}\frac{(2n+2)(2n+1)n^n}{(n+1)^{n+2}}=\lim_{n\to\i  nfty}\frac{4n^{n+2}+6n^{n+1}+2n^n}{(n+1)^{n+2}}

    It should be pretty obvious that the second and third terms in the numerator go to zero and as for the other one...it is equivalent ot

    4\lim_{n\to\infty}\frac{1}{\frac{(n+1)^{n+2}}{n^{n  +2}}}=\frac{4}{\lim_{n\to\infty}\left(1+\frac{1}{n  }\right)^{n+2}}= \frac{4}{\lim_{n\to\infty}\left(1+\frac{1}{n}\righ  t)\cdot\lim_{n\to\infty}\left(1+\frac{1}{n}\right)  ^2}=\frac{4}{e\cdot{1}}=\frac{4}{e}

    Or

    n!\sim\sqrt{2\pi{n}}n^ne^{-n}\wedge{(2n)!\sim\sqrt{4\pi{n}}4^n(n^n)^2(e^{-n})^2}

    \therefore\quad\lim_{n\to\infty}\left(\frac{(2n)!}  {n!n^n}\right)^{\frac{1}{n}}\sim\lim_{n\to\infty}\  left(\frac{\sqrt{4\pi{n}}4^n(n^n)^2(e^{-n})^2}{n^n\cdot{\sqrt{2\pi{n}}n^ne^{-n}}}\right)^{\frac{1}{n}} =\lim_{n\to\infty}\frac{\left(4\pi{n}\right)^{\fra  c{1}{2n}}4n^2e^{-2}}{(2\pi{n})^{\frac{1}{2n}}n^2e^{-1}}=\frac{4}{e}\sqrt{\lim_{n\to\infty}2^{\frac{1}{  n}}}=\frac{4}{e}
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  4. #4
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    Quote Originally Posted by great_math View Post
    \lim_{n\to\infty} \left(\frac{(2n)!}{n!n^n}\right)^{1/n}
    You can also use the fact that n!\sim_n \left(\frac{n}{e}\right)^n\sqrt{2\pi n}, well-known as Stirling's approximation.
    This gives you \frac{(2n)!}{n! n^n}\sim_n \frac{\left(\frac{2n}{e}\right)^{2n}\sqrt{4\pi n}}{\left(\frac{n}{e}\right)^n\sqrt{2\pi n}n^n}=\left(\frac{4}{e}\right)^n\sqrt{2}, hence, using the fact that it is possible to take the logarithm of an asymptotic equivalence (this is specific to the logarithm):
    \frac{1}{n}\ln\left(\frac{(2n)!}{n! n^n}\right)\sim_n \ln\left(\frac{4}{e}\right)+\frac{1}{n}\ln \sqrt{2} \to_n \ln\left(\frac{4}{e}\right).
    Using the continuity of the exponential, we deduce:
    \left(\frac{(2n)!}{n!n^n}\right)^{1/n}=\exp\left( \frac{1}{n}\ln\left(\frac{(2n)!}{n! n^n}\right) \right)\to_n \exp\left(\ln\frac{4}{e}\right), hence \lim_{n\to\infty} \left(\frac{(2n)!}{n!n^n}\right)^{1/n}=\frac{4}{e}.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Laurent View Post
    You can also use the fact that n!\sim_n \left(\frac{n}{e}\right)^n\sqrt{2\pi n}, well-known as Stirling's approximation.
    This gives you \frac{(2n)!}{n! n^n}\sim_n \frac{\left(\frac{2n}{e}\right)^{2n}\sqrt{4\pi n}}{\left(\frac{n}{e}\right)^n\sqrt{2\pi n}n^n}=\left(\frac{4}{e}\right)^n\sqrt{2}, hence, using the fact that it is possible to take the logarithm of an asymptotic equivalence (this is specific to the logarithm):
    \frac{1}{n}\ln\left(\frac{(2n)!}{n! n^n}\right)\sim_n \ln\left(\frac{4}{e}\right)+\frac{1}{n}\ln \sqrt{2} \to_n \ln\left(\frac{4}{e}\right).
    Using the continuity of the exponential, we deduce:
    \left(\frac{(2n)!}{n!n^n}\right)^{1/n}=\exp\left( \frac{1}{n}\ln\left(\frac{(2n)!}{n! n^n}\right) \right)\to_n \exp\left(\ln\frac{4}{e}\right), hence \lim_{n\to\infty} \left(\frac{(2n)!}{n!n^n}\right)^{1/n}=\frac{4}{e}.
    Looks like I just beat editing you =P...sorry...It appears are methodology for the actual limit is slightly different...so no harm =D
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  6. #6
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    Quote Originally Posted by Mathstud28 View Post
    \therefore\quad\lim_{n\to\infty}\left(\frac{(2n)!}  {n!n^n}\right)^{\frac{1}{n}}\sim\lim_{n\to\infty}\  left(\frac{\sqrt{4\pi{n}}4^n(n^n)^2(e^{-n})^2}{n^n\cdot{\sqrt{2\pi{n}}n^ne^{-n}}}\right)^{\frac{1}{n}} =\lim_{n\to\infty}\frac{\left(4\pi{n}\right)^{\fra  c{1}{2n}}4n^2e^{-2}}{(2\pi{n})^{\frac{1}{2n}}n^2e^{-1}}=\frac{4}{e}\sqrt{\lim_{n\to\infty}2^{\frac{1}{  n}}}=\frac{4}{e}
    Be careful with this argument: what about 1+\frac{1}{n}\sim 1 hence \left(1+\frac{1}{n}\right)^{1/n}\sim(1)^{1/n}=1? As I wrote, taking the logarithm of an asymptotic equivalence is correct, while taking the exponential is not (except if the argument of the exponential has a limit, like here, in which case the continuity of the exponential function makes it correct).
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