Gamma Function with Big O

Question

I want to prove the following:

$\Gamma (z+1)= \sqrt {2 \pi} {z^{z+ \frac{1}{2}}}\,\,\, e^{-z} \,\,[1+ O(\frac{1}{z})]$ and that

$\Gamma (z+1)= \sqrt {2 \pi} {z^{z+ \frac{1}{2}}}\,\,\, e^{-z} \,\,[1+ \frac{1}{12z}+ O(\frac{1}{z^{2}})]$ where $z$ is real and greater than zero.

I know how to show that

$\Gamma ({n})=n!\approx \sqrt {2 \pi} {n^{n+ \frac{1}{2}}}\,\,\, e^{-n} $

so I write

$\Gamma ({n})=n!= \sqrt {2 \pi} {n^{n+ \frac{1}{2}}}\,\,\, e^{-n}\, \,(1+ \frac{a_{1}}{n}+\frac{a_{2}}{n^{2}}+...)$

and since the formula is valid for $n+1$ too I also write

$\Gamma ({n+1})=(n+1)! = \sqrt {2 \pi} \,{(n+1)^{n+1 + \frac{1}{2}}}\,\,\,\,\, e^{-(n+1)}\, \,\,\,(1+ \frac{a_{1}}{n+1}+\frac{a_{2}}{(n+1)^{2}}\,\,+...)$

and since

$(n+1)!=(n+1)n!$

I got

$(1+ \frac{a_{1}}{n}+\frac{a_{2}}{n^{2}}+...)=(1+\frac{1}{n})^{n+\frac{1}{2}}\,\,\,e^{-1}\,\,(1+\frac{a_{1}}{n+1}+\frac{a_{2}}{(n+1)^{2}}\,\,...)$

when I plugged the right hand side of the $n!$ into the last equation. But I don't know how to get the coefficients $a_{i}$'s from this point. I appreciate for any help. Thanks.

Answer 1

What you wrote is essentially Stirling's formula. The proof of it for the case of $\Gamma$-function is not entirely trivial(for example you can see from my question here), and I suspect not possible in a short answer at here, even though you can take various short cuts here and there in the proof.

You can check Ahlfors, Complex Analysis, around page 200. I think there should be a chapter on this at Stein's book as well.