Proof of $\frac{\Gamma(k+b)}{\Gamma(k)\Gamma(b+1)} = \frac{k^b}{\Gamma(b+1)}\left(1+O\!\left(\frac1k\right)\right)$

Question

How to show $$\frac{\Gamma(k+b)}{\Gamma(k)\Gamma(b+1)} = \frac{k^b}{\Gamma(b+1)}\left(1+O\!\left(\frac1k\right)\right),$$ where $b \in [0,1]$ and $k \in \mathbb{R}$?

Answer 1

Hint:

Multiply both sides by $\Gamma(b+1)$.

By Stirling's approximation:

$$\Gamma(x+1)={\sqrt {2\pi x}}\left({\frac {x}{e}}\right)^{x}\left(1+\mathcal O\left({\frac {1}{x}}\right)\right)$$

And by binomial expansion:

$$(x+y)^n=x^n+nyx^{n-1}+\mathcal O(y^2x^{n-2})$$