Relating Gamma and factorial function for non-integer values.

Question

We have

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

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

$$0<\Delta<1,$$

then

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

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

$$0<f<1,$$

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?

Answer 1

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

Answer 2

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

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