I've encountered the following sum:
$$
s\_n = \sum\_{j=1}^n {n \brace j}(\alpha n)\_j \beta^j.
$$ 
Here $\alpha$ and $\beta$ are positive constants, $(\alpha n)\_j$ is a falling power, and ${n \brace j}$ is the Stirling number of second kind. My computations suggests that $s\_n$ grows like $n^n C^n$ (where the constant $C$ depends on $\alpha$ and $\beta$); specifically, $\sqrt[n]{(s\_n/n^n)}$ seems to converge to a limit as $n \rightarrow \infty$. 


Are there good approaches to figuring out these kinds of limits? A similar situation was discussed in this [question](https://mathoverflow.net/questions/126900/relations-involving-stirling-numbers-of-second-kind), but the sum there had $(\alpha)\_j$ rather than $(\alpha n)\_j$; the extra dependence on $n$ is causing me headaches! 



Asymptotics of sum involving Stirling numbers
The sum in question can be written (using Latin letters instead of Greek) as 
$$ S\_n(a,b):= \sum\_{k=0}^n {n \atopwithdelims \{ \} k} k! \binom{na}{k} b^k $$
where the round brackets denotes an ordinary binomial. It will be shown as $n \to \infty$
$$ (1)\quad S\_n(a,b) \sim \frac{n!}{2}\,\exp{\Big(n\big( h(u\_0) + \frac{h''(u\_0)}{2}\,u\_0^2\big)\Big)} \,
 \text{erfc}\big(\sqrt{\frac{n}{2} h''(u\_0)} \,\,u\_0 \big) $$
where 
$$h(u) = \frac{\log(1+b(e^u-1))^a}{u}, \quad u\_0=\frac{1}{a} + \text{W}\big(\frac{1-b}{a\,b}\,e^{-1/a}\big) $$
and $W()$ is the Lambert W function.
This follows from an asymptotic analysis of an alternative form,


$$(2) \quad S\_n(a,b)=n!\,[u^n] (1+b\,(e^u-1))^{n \,a}$$


where $[u^n]$ is the 'coefficient of' operator. Note that when $b=1,$ the sum can be solved exactly, and gives the form the OP suspects. (This summation is known.) To derive the alternative form, note that sum is the Hadamard product of


$$ \omega\_n(z) = \sum\_{k=0}^\infty {n \atopwithdelims \{ \} k} k! z^k \quad \text{& } \quad (1+z)^{na}=\sum\_{k=0}^\infty \binom{na}{k} z^k $$
The expression for $\omega\_n$ can be written in terrms of Eulerian polynomials, but I prefer the polylogarithm at negative argument,
$$ \omega\_n(t) = \frac{\text{Li}\_{-n}(t/(1+t))}{1+t} $$
Taking the Hadamard product,


$$S\_n(a,b) = \frac{1}{2 \pi \,i} \oint \frac{dt}{t(1+t)} \text{Li}\_{-n}(\frac{t}{1+t}) \big(1+\frac{b}{t}\big)^{na} \, dt$$
where the integration path surrounds the origin.
Use $\text{Li}\_{-n}(t/(1+t))=(-1)^{n+1}\text{Li}\_{-n}(1+1/t).$ Let $t \to 1/t$ in the integral and check residue at $\infty.$ Then 
$$S\_n(a,b) = \frac{(-1)^n}{2\pi i}\oint \frac{ \text{Li}\_{-n}(1+t)}{1+t} \big(1+b\,t\big)^{n\,a} \, dt$$
Now use the known generating function
$$\sum\_{n=0}^\infty \frac{u^n}{n!} (-1)^n \frac{ \text{Li}\_{-n}(1+t)}{1+t} =
\frac{1}{e^u-1-t}$$ Putting this equation in the penultimate and using Cauchy's theorem results in equation (2).


For the asymptotic analysis, write (2) as a countour integral
$$S\_n(a,b)=\frac{n!}{2 \pi \,i} \oint \Big( \frac{(1+b(e^u-1))^a}{u} \Big)^n \frac{du}{u} $$
Classic saddle point analysis tells us to write this as 
$$S\_n(a,b)=\frac{n!}{2 \pi \,i} \oint \exp{\big(n h(u)\big)} \frac{du}{u} $$
where $h$ has been given in (1), then expand $h$ in a power series about $u\_0,$ which has also been found explicitly and given in (1). Run the contour through $u\_0$ and open the loop so that it becomes a vertical line. Then, so long as certain conditions are satisfied (I have not checked them, other than $h''(u\_0)>0$, a necessity) then 
$$ S\_n(a,b) \sim \exp{(n\ h(u\_0))} \frac{n!}{2 \pi} \int\_{-\infty}^\infty \frac{dt}{u\_0+i\,t} \exp{\big(-\frac{n}{2}\ h''(u\_0) t^2 \big)} $$
The integral is solvable in terms of the complementary error function. For $n$ sufficiently large, it will simplify in the asymptotic limit to an exponential, as long as $h''(u\_0)$ isn't too small. A potential problem with this analysis is that if $h''(u\_0)$ does become small, and much smaller than $h^{(3)}(u\_0)$ then a quadratic expansion of $h$ is not sufficient. 


For examples of the accuracy of the approximation, a table is shown. 
$$
\begin{array}{c|lcr}
n & a & b & \text{true} & \text{asym} & \text{% err}\\
\\
\hline
20 & 1/2 & 1/2 & 2.7717\ 10^{17} & 2.6867\ 10^{17} & 3.07\\
{} & 1/2 & 3/2 & 3.7421\ 10^{21} &3.5745\ 10^{21} & 4.48 \\
{} & 3/2 & 1/2 & 2.1498\ 10^{25} & 2.0862\ 10^{25} &2.96\\
{} & 3/2 & 3/2 & 1.6663\ 10^{23} & 1.5884\ 10^{23} & 4.68\\
200 & 1/2 & 1/2 & 6.634\ 10^{374} & 6.617\ 10^{374} & 0.34\\
{} & 1/2 & 3/2 & 3.336\ 10^{415} & 3.319\ 10^{415} & 0.51\\
{} & 3/2 & 1/2 & 6.158\ 10^{453} & 6.138\ 10^{453} & 0.33\\
{} & 3/2 & 3/2 & 8.720\ 10^{521} & 8.673\ 10^{521}& 0.53
\end{array}
$$


It's really surprising that the approximation for a problem with 2 parameters is so easily characterized. 




---


Added 6/3/2020


There is a much simpler derivation of (2) that bypasses the need for the Hadamard product and the polylogs with negative index. Here it is:


It is known that
$$ {n \atopwithdelims \{ \} k} k! = n![u^n](e^u-1)^k $$
Then
$$ \sum\_{k=0}^n {n \atopwithdelims \{ \} k} k! \binom{na}{k} b^k =$$
$$ n! [u^n] \sum\_{k=0}^\infty \binom{na}{k} b^k (e^u-1)^k = 
n! [u^n] \big(1+b(e^u-1) \big)^{n \ a} $$
by the binomial theorem.


