Estimating the behavior for large $n$ I want to find how these coefficients increase/decrease as $n$ increases:
$$ C_n = \frac{1}{n!} \left[(n+\alpha)^{n-\alpha-\frac{1}{2}}\right]$$ 
with $\alpha=\frac{1}{br-1}$ and $0\leq b,r \leq 1$. 
I used the Stirling's Approximation factorial  $n!\sim \sqrt{2\pi n} n^n e^{-n}$ and got:
$$ C_n = \frac{1}{\sqrt{2\pi n} n^n e^{-n}} \left[(n+\alpha)^{n-\alpha-\frac{1}{2}}\right]$$ 
I can't proceed any further. I would greatly appreciate any comment!

Taking the reciprocal of Stirling's Asymptotic expansion as derived in this answer:
$$
n!=\frac{n^n}{e^n}\sqrt{2\pi n}\left(1+\frac{1}{12n}+\frac{1}{288n^2}-\frac{139}{51840n^3}+O\left(\frac{1}{n^4}\right)\right)
$$
we get
$$
\frac1{n!}=\frac{e^n}{n^n}\frac1{\sqrt{2\pi n}}\left(1-\frac{1}{12n}+\frac{1}{288n^2}+\frac{139}{51840n^3}+O\left(\frac{1}{n^4}\right)\right)
$$
Applying this to $\dfrac{(n+\alpha)^{n-\alpha-1/2}}{n!}$ and using the the log and exponential series for $\left(1+\frac\alpha{n}\right)^{n-\alpha-1/2}$ yields
$$
\begin{align}
&\frac{(n+\alpha)^{n-\alpha-1/2}}{n!}\\
&=\frac{e^nn^{-\alpha-1}}{\sqrt{2\pi}}\left(1+\frac\alpha{n}\right)^{n-\alpha-1/2}\left(1-\frac{1}{12n}+\frac{1}{288n^2}+\frac{139}{51840n^3}+O\left(\frac{1}{n^4}\right)\right)\\[4pt]
&=\small\frac{e^{n+\alpha}n^{-\alpha-1}}{\sqrt{2\pi}}\left(1-\frac{1+6\alpha+18\alpha^2}{12n}+\frac{1+12\alpha+144\alpha^2+456\alpha^3+324\alpha^4}{288n^2}+O\left(\frac{1}{n^3}\right)\right)
\end{align}
$$

Approximating $\boldsymbol{\left(1+\frac\alpha{n}\right)^{n-\alpha-1/2}}$
$$
\begin{align}
&\left(n-\alpha-\frac12\right)\log\left(1+\frac\alpha{n}\right)\\
&=\left(n-\alpha-\frac12\right)\left(\frac\alpha{n}-\frac{\alpha^2}{2n^2}+\frac{\alpha^3}{3n^3}+O\left(\frac1{n^4}\right)\right)\\
&=\alpha-\frac{\alpha+3\alpha^2}{2n}+\frac{3\alpha^2+10\alpha^3}{12n^2}+O\left(\frac1{n^3}\right)
\end{align}
$$
Exponentiating, we get
$$
\begin{align}
\left(1+\frac\alpha{n}\right)^{n-\alpha-1/2}
&=e^\alpha\left(1-\frac{\alpha+3\alpha^2}{2n}+\frac{9\alpha^2+38\alpha^3+27\alpha^4}{24n^2}+O\left(\frac1{n^3}\right)\right)
\end{align}
$$
