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Math Help - Stirling's approximation

  1. #1
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    Stirling's approximation

    Hi, I was just checking out the proof of this in William Feller's book, and it jumps from the arithmetic mean of (nlog(n)-n) and ((n+1)log(n+1)-n) to ((n+(1/2))log(n)-n). I'm not seeing it. The book is certainly a good read, but nonetheless I'm missing out. Any hints?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Stirling's approximation

    Prove that L=\dfrac{1}{2}\displaystyle\lim_{n \to{+}\infty}{\frac{n\log n+(n+1)\log (n+1)-2n}{(n+1/2)\log n-n}}=\ldots=1


    Hint: Divide numerator and denominator by (n+1)\log (n+1) .
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  3. #3
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    Re: Stirling's approximation

    Alright.
    It's not equal, it's only an approximation of the approximation.
    And this is where the ~ comes from then?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Stirling's approximation

    Quote Originally Posted by eulcer View Post
    And this is where the ~ comes from then?
    Right.
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