Question related to deriving a result based on Stirling s formula

Question

I have been given a problem which I am unable to derive.

Problem-> Using Stirling approximation formula derive that $\Gamma(x +1/n)$ ~ $ x ^{1/n} \Gamma(x) $ when x -> $\infty$ , ~ means asmyptotic to and x belongs to complex numbers and n belongs to positive integers.

Attempt-> I used the formula $\Gamma(z+1/n) = {(\frac {2π} {z} )}^{1/2} {(\frac{z} {e}) }^{n} ( 1 +O(1/z)) $

But when put x = x+1/n and tend x to infinity , ${(\frac {2π} {x} )}^{1/2}$ tends to 0 . Do, I am doing some mistake or some other concept need to be used.

I admit that I feel uncomfortable using Stirling approximations.

Can someone please tell the right way to derive it.

Answer 1

Stirling is the asymptotic formula $$\Gamma(x)\sim\sqrt{\frac{2\pi}x}e^{-x}x^x\tag1.$$ Take some positive number $a$ (you want $a=1/n$) and substituting in $(1)$ gives $$\Gamma(x+a)\sim\sqrt{\frac{2\pi}{x+a}}e^{-x-a}(x+a)^{x+a}\tag2.$$ Divide $(2)$ by $(1)$ to give $$\frac{\Gamma(x+a)}{\Gamma(x)}\sim\sqrt{\frac{x}{x+a}}e^{-a} \frac{(x+a)^{x+a}}{x^x}\tag3.$$ As $$\lim_{x\to\infty}\sqrt{\frac{x}{x+a}}=1,$$ $(3)$ simplifies to $$\frac{\Gamma(x+a)}{\Gamma(x)}\sim e^{-a} \frac{(x+a)^{x+a}}{x^x}=e^{-a}(x+a)^a\left(\frac{x+a}{x}\right)^x\tag4.$$ Now $(x+a)^a\sim x^a$ and $$\lim_{x\to\infty}\left(\frac{x+a}{x}\right)^x =\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^x=e^a$$ (a standard limit). Therefore $(4)$ simplifies to $$\frac{\Gamma(x+a)}{\Gamma(x)}\sim x^a.$$

Answer 2

Just use the plain formula $$\log (\Gamma (p))=p (\log (p)-1)+\frac{1}{2} \log \left(\frac{2 \pi }{p}\right)+O\left(\frac{1}{p}\right)$$ Use it twice and continue with Taylor series $$\log \left(\Gamma \left(x+\frac{1}{n}\right)\right)-\log (\Gamma (x))=\frac{\log(x)}n+O\left(\frac{1}{x}\right)$$ Exponentiate both sides to get $$\frac{\Gamma \left(x+\frac{1}{n}\right) } {\Gamma (x) }\sim x^{1/n}$$