Asymptotic expansion of a sum containing binomial coefficients I'm trying to find an asymptotic expansion for the following sum:
$$\sum_{k=2}^{n}\frac{n!}{k\left(n-k\right)!}=\sum_{k=2}^{n}\left(\begin{array}{c}
n\\
k
\end{array}\right)\left(k-1\right)!$$
for large $n$. According to Maple, this sum is equivalent to:
$$\frac{n^{2}-n}{2}\,_{3}F_{1}\left(1,2,2-n;3;-1\right)$$
where $_{3}F_{1}$ is a special case of the generalized hypergeometric function. But on the internet I cannot find any asymptotic expansion of $_{3}F_{1}$, because most work focused on other cases, e.g. $_{2}F_{1}$. I think it could be easier to work directly on the original sum. But by searching in previous posts I cannot find the specific sum I'm interested in. Moreover, asymptotic approximations of $\left(\begin{array}{c}
n\\
k
\end{array}\right)$ do not look very useful since they do not work for $k\approx n$. Help would be very appreciated!
 A: Can be done in a similar way to this. Let $k = n - k$ for convenience:
$$S = \sum_{k = 2}^n \frac {n!} {k (n - k)!} =
\sum_{k = 0}^{n - 2} \frac {n!} {(n - k) k!} =
(n - 1)! \sum_{k = 0}^{n - 2} \frac 1 {k!} \frac 1 {1 - \frac k n}.$$
The main contribution comes from small values of $k$, and it is sufficient to use an approximation of the summand that holds for large $n$ and fixed $k$. The $i$th order term in the approximation will be simply $(k/n)^i$. Then the summation range in $S$ can be extended to infinity, and we get a complete asymptotic series for $S$:
$$S \approx e (n - 1)! \sum_{i = 0}^\infty \frac {B_i} {n^i},$$
where $B_0 = 1, B_1 = 1,  .\!.\!.$ are the Bell numbers.
A: $$\sum_{k=2}^{n}\binom{n}{k}(k-1)! = \sum_{k=2}^{n}\binom{n}{k}\int_{0}^{+\infty}x^{k-1}e^{-x}\,dx = \int_{0}^{+\infty}\frac{(1+x)^n-nx-1}{x}e^{-x}\,dx $$
is expected to behave like $\int_{0}^{+\infty}x^{n-1}e^{-x}\,dx = (n-1)!$ plus a perturbation due to the fact that $\frac{(1+x)^n-nx-1}{x}$ does not behave like $x^{n-1}$ in a right neighbourhood of the origin. More accurate approximations can be derived from Laplace's method.
A: This is just a numerical confirmation of the result derived by Maxim. The figure below shows that Maxim's asymptotic series (red line, truncated to $i=10$ in this example) converges to the original series (blue line) for large $n$, and that both the series converge to Euler's number $e$. 
