Simplify $\frac{\Gamma(n)}{\Gamma(n+a)}$ with $a\in\mathbb C$.

Question

How can simplify the following expression? $$\frac{\Gamma(n)}{\Gamma(n+a)}\sim \cdots\text{ ?}$$ Where $a\in\mathbb C$, $n\in \mathbb N$. Any suggestions please?

I propose the following. We have the classical Stirling's approximation formula for the Gamma-Function in the form: $$ \Gamma(z)=\sqrt{2\pi}e^{-z}z^{z-1/2}\left(1+O\left(\frac{1}{|z|}\right)\right) $$

for $|\arg(z)|<\pi$ as $|z|\to\infty$.

And, from another question here there is also the shifted Stirling's approximation for the Gamma-Function due to C. Rowe that says: $$ \Gamma(z+a)=\sqrt{2\pi}e^{-z}z^{z+a-1/2}\left(1+O\left(\frac{1}{|z|}\right)\right) $$

Therefore

$$ \frac{\Gamma(z)}{\Gamma(z+a)}=\frac{\sqrt{2\pi}e^{-z}z^{z-1/2}\left(1+O\left(\frac{1}{|z|}\right)\right)}{\sqrt{2\pi}e^{-z}z^{z+a-1/2}\left(1+O\left(\frac{1}{|z|}\right)\right)}=\frac{\left(1+O\left(\frac{1}{|z|}\right)\right)}{z^a\left(1+O\left(\frac{1}{|z|}\right)\right)} $$ and: $$ \frac{\Gamma(n)}{\Gamma(n+a)}\sim n^{-a} \ \ \ (n\rightarrow \infty) $$

I would appreciate some corrections on this procedure. Thanks.

Answer 1

Hint: $$B(n,a) = \int_{0}^{1} t^{n-1}(1-t)^{a-1} dt = \frac{\Gamma(n)\Gamma(a)}{\Gamma(n + a)}$$

Where $B$ is the Beta Function and $\mathcal {Re} (a) > 0$.

Answer 2

$$ \Gamma(x+1)=x\Gamma(x). $$ So \begin{align} \Gamma(5+a) & = (4+a)\Gamma(4+a) \\[8pt] & = (4+a)(3+a)\Gamma(3+a) \\[8pt] & = (4+a)(3+a)(2+a)\Gamma(2+a) \\[8pt] & = (4+a)(3+a)(2+a)(1+a)\Gamma(1+a) \\[8pt] & = (4+a)(3+a)(2+a)(1+a)a\Gamma(a) \end{align} and then cancel a factor from the numerator and denominator. And similarly for $n\ne4$.