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I have read $[x^n] (1-x)^\alpha\sim \frac{n^{-\alpha-1}}{\Gamma(-\alpha)}$ as $n\rightarrow\infty$ where $[x^n]$ means coefficient of $x^n$ in what follows. But I have never seen a proof of this assuming can take the symbol $\sim$ to also mean approximately or approaches but I have yet to see any proof. By letting $\alpha \rightarrow -\alpha$ in a series formula I have verified, I use $$(1-x)^\alpha= \frac{1}{\Gamma(-\alpha)} \sum_{n=0}^\infty \frac{\Gamma(n-\alpha)}{n!}x^n,|x|<1$$ So this and the above implies $\frac{\Gamma(n-\alpha)}{n!}\rightarrow n^{-\alpha-1} ,n\rightarrow\infty$ And I certainly cannot reconcile this. So what is the discrepancy ? Is it an error in the text where I read this ?

MPW
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user158293
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  • Isn't the coefficient of $x^n$ really $\frac{\Gamma(n-\alpha)}{n!\Gamma(-\alpha)}$ because of the multiplier out front? – MPW Apr 27 '22 at 13:53
  • yes that is what i wrote – user158293 Apr 27 '22 at 13:55
  • I never said it wasn't – user158293 Apr 27 '22 at 13:56
  • and who is mpw that also would not let me put in the correct title – user158293 Apr 27 '22 at 13:57
  • what I read said it was $\sim \frac {n^{-1-\alpha}}{\Gamma(-\alpha)}$ – user158293 Apr 27 '22 at 14:04
  • I don't mean that is exactly what I wrote but the power series implies that. So what is your point ? – user158293 Apr 27 '22 at 14:07
  • the correct title is 'Discrepancy in that as n approaches inf gives a ratio of $\Gamma $ functions approaches solely n to a negative power.' and who is this mpw who put in the wrong title and would not let me put in the correct one ? – user158293 Apr 27 '22 at 14:11
  • You are asking (I think?) how to derive the Stirling series. One way is to use the integral representation of $\Gamma$ and use Laplace's method to approximate the integral for $n \to \infty$ – Sal Apr 27 '22 at 16:23
  • NO i am not asking that at all. Clearly as written I'm asking to prove $\Gamma(n-\alpha)/n!\rightarrow n^{-1-\alpha}$ as n goes to inf. OR else it is a text error what I read and it's not true. But now I have put in some representative numerical values eg. n:160,$\alpha:1/3,1/2,4/3$ or the negative of these in Maxima and in all 6 cases the |relative error| turned out to be <.01 so I guess it's not a text error and true but have no idea how to prove it. – user158293 Apr 27 '22 at 17:22
  • NOw also I used the lowest order Stirling approximation for$ \frac{\Gamma(n-\alpha)}{n!}$ and get $\exp(\alpha+1)(n+1)^{-n-1/2} (n-\alpha)^{n-\alpha-1/2}$ and using this expression call it $t$ obtain the same low |relative difference| being $|\frac{t-n^{-\alpha-1}}{n^{-\alpha-1}}|<.01$ for those same 6 trial numeric values as prior comment. Someone remarked use integral representation to prove it. Can anyone explain that in some detail for this problem ? – user158293 Apr 27 '22 at 18:07
  • Because otherwise especially with that exponential term $\exp(\alpha+1)$ how on earth it could possibly be true. – user158293 Apr 27 '22 at 18:13
  • I changed the title because what you wrote was improper MathJax. You wrote \Gamma$ which appears exactly as typed (with backslash and single dollar sign). If you really want a capital Greek gamma ($\Gamma$), just change it to $\Gamma$. For what it's worth, one usually speaks of either "the Gamma function" or "$\Gamma$", but not usually "the $\Gamma$ function". Of course, you can put whatever you want. – MPW Apr 28 '22 at 13:32
  • Okay so much for the title,which perhaps better written as Ratio of gamma fct's.... as now I am almost sure $\frac{\Gamma(n-\alpha)}{n!}\rightarrow n^{-\alpha-1}$ as n goes to infinity so it's not a discrepancy . I just need someone to prove it and it is likely similar or uses some of the same ideas as Stirlings derivation or approximation for n! as n becomes very large. – user158293 Apr 30 '22 at 09:13
  • Suppose $\alpha$ were integer then $ \frac{\Gamma(n-\alpha)}{n!}=\frac{(n-\alpha-1)!}{n!}$ which if $\alpha>-1$ is equal $\frac1{n(n-1)...(n-\alpha)}\rightarrow n^{-1-\alpha};as;n\rightarrow\infty$. If $\alpha=-1;then;\frac{\Gamma(n-\alpha)}{n!}=\frac{n!}{n!}=n^{0}=1$!. If $\alpha<-1$ is equal $(n-\alpha-1)(n-\alpha-2)...(n+1)\rightarrow n^{-1-\alpha};as;n\rightarrow\infty$. If $\alpha$ were not an integer then intuitively linearly interpolate between the answers for that of the integer just above and that for the integer just below $\alpha$. Ofcourse this is not a rigorous proof. – user158293 Apr 30 '22 at 12:14
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    Does this answer your question? Proving $\lim_{x\to\infty} \frac{\Gamma(x+a)}{x^a\Gamma(x)}=1$ (stated for $a>0$ there, but holds for any $a\in\mathbb{C}$; actually used here many times, and stated on Wikipedia) – metamorphy May 31 '22 at 04:49
  • with regard to that last comment by metamorphy i think it may answer question. I just need more time to decipher thru it but right now have a vision problem which should hopefully be ok in a few days but not entirely sure when i will be well. thanks and will get back to you – user158293 Jun 06 '22 at 02:15

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