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Math Help - factorials

  1. #1
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    factorials

    may i know how to prove the theorem that factorials beat exponentials?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    may i know how to prove the theorem that factorials beat exponentials?
    n!\underset{n\to\infty}{\sim}\sqrt{2\pi n}n^ne^{-n}
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  3. #3
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    sorry, i dont really get what you are trying to hint to me. how do i make use of this formula to show that

    eg the fraction of an exponential over factorial will tend to 0?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    sorry, i dont really get what you are trying to hint to me. how do i make use of this formula to show that

    eg the fraction of an exponential over factorial will tend to 0?
    \lim_{n\to\infty}\frac{a^n}{n!}=\lim_{n\to\infty}\  frac{a^n}{\sqrt{2\pi n}n^n e^{-n}}=\frac{1}{\sqrt{2\pi}}\lim_{n\to\infty}\frac{(a  \cdot e)^n}{n^{n+\frac{1}{2}}}=0

    Why?
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    Here's a much simpler way. We know e^a = \sum_{n=0}^\infty \frac{a^n}{n!} converges for every real number a, which immediately implies \lim_{n \rightarrow \infty} \frac{a^n}{n!} = 0.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Here's a much simpler way. We know e^a = \sum_{n=0}^\infty \frac{a^n}{n!} converges for every real number a, which immediately implies \lim_{n \rightarrow \infty} \frac{a^n}{n!} = 0.
    How do you know the OP has seen series? Really all you tacitly assumed was that if \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<  1 then it's dominated by a convergent geometric sequence.
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  7. #7
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    isit because n(n+0.5) can be taken as n^n thus the fraction becomes (a.e/n)^n where (a.e/n) is less than 1?
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  8. #8
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    in my course that im learning, i have yet to learn the formula for factorials. so assuming that i dont know that formula, is there another method to prove it?

    thanks
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  9. #9
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    Quote Originally Posted by Drexel28 View Post
    How do you know the OP has seen series?
    What does OP stand for?
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  10. #10
    Junior Member Renji Rodrigo's Avatar
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    Proposition[ratio teste for sequences]
    Iff (x_n) is a sequence with x_n >0 forall n in N
    and  \lim \frac{x_{n+1}}{x_{n}}  <1
    then  \lim x_{n} =0 .


    Take x_n =\frac{a^n}{n!} a>0

    then

    \frac{x_{n+1}}{x_{n}} = \frac{a^{n+1} n!}{(n+1)! a^n}=\frac{a}{n+1}
    so  \lim \frac{x_{n+1}}{x_{n}}=0  <1
    then
     \lim \frac{a^n}{n!}=0
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  11. #11
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    How do you know the OP has seen series? Really all you tacitly assumed was that if \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<  1 then it's dominated by a convergent geometric sequence.
    How do you know the OP has seen Stirling's formula?

    You're right about all that I've assumed, which is comparatively quite little.
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  12. #12
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by dwsmith View Post
    What does OP stand for?
    "Original Poster"
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  13. #13
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    by the way, is there a way to use sandwich theorem to prove this?
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  14. #14
    Junior Member Renji Rodrigo's Avatar
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    Given a>0 exists  n_{0}\in N with n_{0}>2a ,

    \frac{n_{0}}{a}>2 so for n\geq n_{0} we have
    \frac{n}{a}\geq \frac{n_{0}}{a}>2

    taking the product in \frac{t}{a}>2 from t=n_{0} to n we have


    \prod^{n}_{t=n_{0}}\frac{t}{a}>\prod^{n}_{t=n_{0}}  2=2^{n+1-n_{0}}

    taking the product with \prod\limits^{n_{0}-1}_{t=1}\frac{t}{a}=p on the both sides off the inequality


    \prod\limits^{n_{0}-1}_{t=1}\frac{t}{a}\prod^{n}_{t=n_{0}}\frac{t}{a}=  \prod^{n}_{t=1}\frac{t}{a}=\frac{n!}{a^{n}}>p.2^{n  +1-n_{0}}


    \frac{n!}{a^{n}}>p.2^{n+1-n_{0}}

    the limit on the right goes to infinity so the limit on the left too
    , so
    \lim \frac{n!}{a^{n}}=\infty,\lim \frac{a^{n}}{n!}=0.
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