Sum of the series with Stirling numbers of the first kind. Yesterday I worked on one problem in discrete math and in the process of decision I came across this series. Try to do it with generating functions, but there is no success for me.
So, what do you think about this sum? s(n - 1,k) - Stirling number of the first kind.
$$\sum\limits_{k = 0}^{n - 2}\frac{s(n - 1,k)\cdot k\cdot2^k}{n!}$$
 A: Let $(x)_n$ be the failing factorial i.e. :
$$(x)_n=x(x-1)...(x-n+1) $$
Then from the definition of the Stirling numbers of the first kind :
$$(x)_n=\sum_{k=0}^ns(n,k)x^k $$
Then :
$$\frac{d}{dx}(x)_n=\sum_{k=0}^ns(n,k)kx^{k-1} $$
Then, taking $x=2$ :
$$\frac{d}{dx}(x)_n(2)=\sum_{k=0}^ns(n,k)k2^{k-1}$$
So :
$$\sum_{k=0}^{n-2}s(n,k)k2^k=2\sum_{k=0}^ns(n,k)k2^{k-1}-s(n,n-1)(n-1)2^{n-1}-s(n,n)n2^n$$
$$=2\frac{d}{dx}(x)_n(2)-\begin{pmatrix}n\\2\end{pmatrix}(n-1)2^{n-1}-n2^n $$
Now we want to compute $\frac{d}{dx}(x)_n(2)$, if $n=2$ then $(x)_2=x^2-x$ and then :
$$\frac{d}{dx}(x)_2(2)=3$$
if $n=3$ then $(x)_3=x(x-1)(x-2)$ then :
$$\frac{d}{dx}(x)_3=(x-1)(x-2)+x(x-2)+x(x-1)$$
Then, we see that it is easy to evaluate in $2$ all but one polynomials in the sum will be null hence :
$$\frac{d}{dx}(x)_3(2)=2$$
For $n\geq 3$ we will have (expand the derivative, evaluation in $2$ will kills all but one of the polynomials in the sum):
$$\frac{d}{dx}(x)_n(2)=2\times (-1)^{n-3}(n-3)!$$
Now this gives you a formula for your number. 
