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We have

$$\Gamma(n+1)=n!,\ \ \ \ \ \Gamma(n+2)=(n+1)!$$

for integers, so if $\Delta$ is some real value with



$$n!\ <\ \Gamma(n+1+\Delta)\ <\ (n+1)!,$$

because $\Gamma$ is monotone there and so there is another number $f$ with


such that

$$\Gamma(n+1+\Delta)=(1-f)\times n!+f\times(n+1)!.$$

How can we make this more precise? Can we find $f(\Delta)$?

Or if we know the value $\Delta$, which will usually be the case, what $f$ will be a good approximation?

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I think your $f$ depends not just on $\Delta$, but on $n$ as well. – froggie May 17 '12 at 15:58
up vote 2 down vote accepted

Asymptotically, as $n \to \infty$ with fixed $\Delta$, $$ f(n,\Delta) = \dfrac{\Gamma(n+1+\Delta)-\Gamma(n+1)}{\Gamma(n+2)-\Gamma(n+1)} = n^\Delta \left( \dfrac{1}{n} + \dfrac{\Delta(1+\Delta)}{2n^2} + \dfrac{\Delta(-1+\Delta)(3\Delta+2)(1+\Delta)}{24n^3} + \ldots \right) $$

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Can't one just pull out the $n$'s of the first expression on all four Gammas and then it cancels? – NikolajK May 17 '12 at 16:32
@NickKidman: I don't understand that comment. How do you "pull out the $n$'s" from $\Gamma(n+1+\Delta)$? – Robert Israel May 17 '12 at 20:29
I ment with the gamma identity $\Gamma(z+n)=\Gamma(z)\times...$, but while the argument of $\Gamma$ would become smaller, the whole expression wouldn't look much simpler. – NikolajK May 17 '12 at 20:47

Well, $\Gamma(1) = \Gamma(2) = 1$, but $\Gamma(\frac{3}{2}) = \frac{\sqrt{\pi}}{2} <1$, so presumably you need $n>1$.

And $f(n, \Delta) = \frac{\Gamma(n+1+\Delta)-\Gamma(n+1)}{\Gamma(n+2)-\Gamma(n+1)}$.

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Well yeah, okay this involves $\Gamma(n+1+\Delta)$ itself. What would be a good practical approximation? I can't seem to expand this expression for $f(n,\Delta)$ in a series. (That is, I can use Mathematica, but it doesn't tell me the thing in terms of $n$, even if there seems to be a pattern. It'll probably come down to a series expanstion of the polygamma function.). $\textbf{edit}$: Wait, you can factor out $n$ in this thing, can't you. – NikolajK May 17 '12 at 16:25

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