Gamma functions and limits

Question

How can someone calculate the limit $\lim_{n\rightarrow \infty}\frac{\Gamma(n+p)}{n^p\Gamma(n)}$ ? Is there an article about it? Is $\frac{\Gamma(n+p)}{n^p\Gamma(n)}$ greater than unity?

Answer 1

Stirling's approximation $$\Gamma(z) \sim \sqrt{\frac{2 \pi}{z}} \left( \frac{z}{\mathrm{e}} \right)^z $$ does the trick:

$$\frac{\Gamma(n+p)}{n^p \Gamma(n)} \sim \frac{ \sqrt{\frac{2 \pi}{n+p}} \left( \frac{n+p}{\mathrm{e}} \right)^{n+p}}{n^p \sqrt{\frac{2 \pi}{n}} \left( \frac{n}{\mathrm{e}} \right)^n}= \frac{1}{e^p} \sqrt{\frac{n}{n+p}} \left(\frac{n+p}{n} \right)^{n+p} \to 1, \quad \text{ as } n \to \infty. $$

I don't know about such an article.

Answer 2

You can have quite good approximations for finite values of $n$.

Consider $$A=\frac{\Gamma(n+p)}{n^p \Gamma(n)}\implies \log(A)=\log(\Gamma(n+p))-\log(\Gamma(n))-p\log(n)$$ and use Stirling approximation $$\log(\Gamma(k))=k (\log (k)-1)+\frac{1}{2} \left(\log (2 \pi )-\log(k)\right)+\frac{1}{12 k}+O\left(\frac{1}{k^3}\right)$$ Apply it to each logarithm and continue with Taylor series to get $$\log(A)=\frac{(p-1) p}{2 n}-\frac{(p-1) p (2 p-1)}{12 n^2}+O\left(\frac{1}{n^3}\right)$$ Now, still using Taylor $$A=e^{\log(A)}=1+\frac{(p-1) p}{2 n}+\frac{(p-2) (p-1) p (3 p-1)}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.

For illustration, use $n=100$ and $p=10$; the above expansion would give exactly $A=1.537$ while the exact value would be $A=\frac{2417359996890657}{1562500000000000}\approx 1.54711 $.