$\frac{n!}{(n+\frac{1}{x})!} = ?$ for some positive real $x$

Question

Is there a simple way to express : $\frac{n!}{(n+\frac{1}{x})!} = ?$ where $x, n$ are positive real numbers ?

Answer 1

The answer you are looking for is found in the identity

$$\frac{\Gamma(n+x)}{\Gamma(n)}=(x)_n$$

where $(x)_n$ is the Pochhammer polynomial.

Ref: K. Oldham, J. Myland, & J. Spanier, An Atlas of Functions, Springer.

Answer 2

$n!\sim\sqrt{2n\pi}\cdot(\frac{n}{e})^n$.

$\left(n+{1\over x}\right )!\sim\sqrt{2\left(n+{1\over x}\right ) \pi}\cdot\left (\frac{\left(n+{1\over x}\right ) }{e}\right )^\left(n+{1\over x}\right ) $.

$\frac{n!}{\left(n+{1\over x}\right )!} \sim \frac{e^{1 \over x}}{\sqrt{\left(1+{1\over nx}\right )}} \frac{1}{n^{1\over x}}\frac{1}{\left(1+{1\over nx}\right )^{1\over x}} \frac{1}{\left(1+{1\over nx}\right )^n} $

$\frac{n!}{\left(n+{1\over x}\right )!} \sim \frac{1}{\sqrt{\left(1+{1\over nx}\right )}} \frac{1}{\left(n+{1\over x}\right )^{1\over x}} $