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Sterling's Approximation

  1. Nov 10, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the limit of: ##\frac { \Gamma (n+\frac { 3 }{ 2 } ) }{ \sqrt { n } \Gamma (n+1) } ## as ##n\rightarrow \infty ##.

    2. Relevant equations

    ##\Gamma (p+1)=p^{ p }e^{ -p }\sqrt { 2\pi p } ##

    3. The attempt at a solution

    Mathematica and wolfram Alpha gave the limit as 1. My solution was ##\frac{1}{\sqrt{e}}##. My work is here https://plus.google.com/u/0/1096789...6080238892798007794&oid=109678926107781868876 Sorry about the photo, but my school is about to set the alarm, and I need to get out of here. Hopefully it is visible.

    Thanks,
    Chris
     
  2. jcsd
  3. Nov 10, 2014 #2

    Ray Vickson

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    You wrote
    [tex] \Gamma (p+1)=p^{ p }e^{ -p }\sqrt { 2\pi p }\; \Longleftarrow \; \text{FALSE!} [/tex]
    Perhaps you mean ##\Gamma(p+1) \sim p^p e^{-p} \sqrt{2 \pi p}, ##, where ##\sim## means "is asymptotic to". That is a very different type of statement.

    Anyway, I (or, rather, Maple) get a different final answer (=1), using the asymptotic result above. Somewhere you must have made an algebraic error.
     
    Last edited: Nov 10, 2014
  4. Nov 11, 2014 #3
    Sorry, yes I meant asymptotic to. Were you able to see my solution via the google plus link? Mary Boas' manual also gives a limit of one. Our solutions start out the same, but I make some approximations that seem not to be justified in her steps. However, I am unable to follow her steps. I can post that too if you are interested in taking a look at her solution.

    Thanks,
    Chris
     
  5. Nov 11, 2014 #4

    Ray Vickson

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    I don't have her book, and cannot really follow your screenshot. Anyway, if ##r(n)## is your ratio, and using ##\Gamma(p+1) \sim c p^{p+1/2} e^{-p}##, the numerator ##N(n)## has asymptotic form
    [tex] N(n) = \Gamma(n+3/2) \sim c (n+1/2)^{n+1/2 + 1/2} e^{-(n+1/2)} = c (n+1/2)^{n+1} e^{-n} e^{-1/2}[/tex]
    The denominator ##D(n)## has the asymptotic form
    [tex] D(n) = \sqrt{n} \: \Gamma(n+1) \sim n^{1/2} c n^{n+1/2} e^{-n} = c n^{n+1} e^{-n} [/tex]
    Therefore,
    [tex] r(n) \sim \frac{(n+1/2)^{n+1} e^{-n} e^{-1/2}}{n^{n+1} e^{-n}} = \left( 1 + \frac{1}{2} \frac{1}{n} \right)^{n+1} e^{-1/2}[/tex]
    Since ##\lim \,(1 + a/n)^{n+1} = \lim \,(1+a/n)^n = e^a##, we are done: ## \lim r(n) = 1##.
     
  6. Nov 11, 2014 #5
    That is pretty much her solution. However, I can actually follow yours better. I made an approximation that was not justified.

    Thanks,
    Chris
     
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