I have also posted this question at https://mathoverflow.net/questions/141552/simple-approximation-to-a-sum-involving-stirling-numbers#141552. I have an exact answer to a problem, which is the function:
$f(x,y)=\frac{1}{y^x}\sum_{i=1}^{x-1} [i\binom{y}{x-i}(x-i)!S(x,x-i)]$ where $S(x,x-i)$ is Stirling number of the second kind. Equivalently, $f(x,y)=\frac{1}{y^x}\sum_{i=1}^{x-1}{\{i\binom{y}{x-i}\sum_{j=0}^{x-i} [(-1)^{x-i-j}\binom{x-i}{j}j^x]}\}$. Equivalently, $f(x,y)=\frac{y!}{y^x}\sum_{i=1}^{x-1}{\{\frac{i}{(y-(x-i))!}\sum_{j=0}^{x-i} [\frac{(-1)^{x-i-j}j^x}{j!(x-i-j)!}]}\}$.
I have noticed that the percent difference between $f(x,y)$ and $g(x,y)$ goes to $0$ for larger values of $x$ and $y$, where $g(x,y)$ is the far more elegant $x-y(1-e^{-\frac{x}{y}})$. How can $f(x,y)$ be approximated by $g(x,y)$? What approximations should be used to make this connection?
I have tried approximations for $S(n,m)$ listed at http://dlmf.nist.gov/26.8#vii, to no avail.
Note that $\binom{y}{x-i}(x-i)!=(y)_{x-i}$ is the Pochhammer symbol or "falling factorial" and the Stirling numbers of the second kind relate falling factorials to monomials by this formula, $$ \sum_{i=0}^x(y)_i\,\begin{Bmatrix}x\\i\end{Bmatrix}=y^x\tag{1} $$ The recurrence relation for Stirling numbers of the second kind is $$ \begin{Bmatrix}n+1\\k\end{Bmatrix}=k\,\begin{Bmatrix}n\\k\end{Bmatrix}+\begin{Bmatrix}n\\k-1\end{Bmatrix}\tag{2} $$ Therefore, $$ \begin{align} \sum_{i=0}^xi(y)_i\,\begin{Bmatrix}x\\i\end{Bmatrix} &=\sum_{i=0}^{x+1}(y)_i\left(\begin{Bmatrix}x+1\\i\end{Bmatrix}-\begin{Bmatrix}x\\i-1\end{Bmatrix}\right)\\ &=\sum_{i=0}^{x+1}(y)_i\begin{Bmatrix}x+1\\i\end{Bmatrix} -\sum_{i=0}^{x+1}y(y-1)_{i-1}\begin{Bmatrix}x\\i-1\end{Bmatrix}\\[8pt] &=y^{x+1}-y(y-1)^x\\[8pt] \sum_{i=0}^xi(y)_{x-i}\,\begin{Bmatrix}x\\x-i\end{Bmatrix} &=\sum_{i=0}^x(x-i)(y)_i\,\begin{Bmatrix}x\\i\end{Bmatrix}\\[8pt] &=(x-y)y^x+y(y-1)^x\tag{3} \end{align} $$ Thus, since the $i=0$ and $i=x$ terms are $0$, that is $\begin{Bmatrix}x\\0\end{Bmatrix}=0$, $$\begin{align} f(x,y) &=\frac1{y^x}\left((x-y)y^x+y(y-1)^x\right)\\ &=x-y+y\left(1-\frac1y\right)^x\\ &\sim x-y\left(1-e^{-x/y}\right)\tag{4} \end{align} $$