The ratio of factorials of sum approximation

Question

I am wondering if the following can be approximated: $$\frac{(k+n)!}{(kN+c+n)!}$$ where all the variables in the above are positive integers. Anyway, I want to separate $c$ from the above so that we can write it as $f(k,n,N).f(c)$ where the '.' is any mathematical operation.

Answer 1

A good approximation is obtained by the Stirling formula $$ n!\approx \sqrt{2\pi n}n^ne^{-n}\qquad n\rightarrow\infty. $$ In your case, we assume $(n+k)\rightarrow\infty$ and this is enforced in $kN+n$ at the denominator. Then, one has $$ \frac{(n+k)!}{(kN+n+c)!}\approx\sqrt{\frac{n+k}{kN+k+c}}\frac{(n+k)^{n+k}}{(kN+n+c)^{kN+n+c}}e^{(N-1)k+c}. $$ Now, $$ (kN+n+c)^{kN+n+c}=e^{(kN+n+c)\ln(kN+n+c)}\approx e^{(kN+n)\ln(kN+n)}\left[1+(\ln(kN+n)+1)c\right], $$ and $$ \frac{1}{\sqrt{kN+n+c}}\approx\frac{1}{\sqrt{kN+n}}\left(1-\frac{1}{2}\frac{1}{kN+n}c\right) $$ This yields, $$ \frac{(n+k)!}{(kN+n+c)!}\approx\sqrt{\frac{n+k}{kN+k}}\frac{(n+k)^{n+k}}{(kN+n)^{kN+n}}\left[1-\left(\ln(kN+n)+1+\frac{1}{2}\frac{1}{kN+n}\right)c\right]e^{(N-1)k+c}\approx\frac{(n+k)!}{(kN+n)!}\left[1-\left(\ln(kN+n)+1+\frac{1}{2}\frac{1}{kN+n}\right)c\right]. $$ Even if the second term multiplying $c$ is neglected, in the considered limit, there is no way to write this exactly in the form $f(k,N,n)\cdot f(c)$.