Finite sum involving Stirling number of first kind and Pochhammer symbol I'm trying to find a closed form of
$$\sum_{m=0}^n\,s(n,m)\,(\alpha)_m\,z^m$$
where $(\alpha)_m$ means pochhammer symbol and $s(n,m)$ are the Stirling numbers of first kind.
I've had a look in related books but I have not luck.
Any help is welcomed.
 A: We can rewrite your sum as
$$
\eqalign{
  & F(z,n,\alpha ) = \sum\limits_{k = 0}^n {\left( { - 1} \right)^{n - k}
 \left[ \matrix{  n \cr  k \cr}  \right]\alpha ^{\,\overline {\,k\,} } z^{\,k} }
  = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{n - k}
 \left[ \matrix{  n \cr  k \cr}  \right]z^{\,k} \alpha ^{\,\overline {\,k\,} } }  =   \cr 
  &  = \left( { - 1} \right)^n \sum\limits_{0\, \le \,k}
 {\left[ \matrix{  n \cr   k \cr}  \right]z^{\,k} \left( { - \alpha } \right)^{\,\underline {\,k\,} } }
  = \left( { - 1} \right)^n \sum\limits_{0\, \le \,k}
 {k!\left[ \matrix{  n \cr   k \cr}  \right]z^{\,k} \left( \matrix{   - \alpha  \cr  k \cr}  \right)}  \cr} 
$$
where

*

*$\alpha ^{\,\overline {\,k\,} }$ represents the Rising factorial;

*$ \left[ \matrix{  n \cr  k \cr}  \right]$ represents the unsigned Stirling N. 1st kind.

I can't provide at the moment a closed formula for that, but considering the well known
e.g.f. for the Stirling N. 1st kind
$$
{1 \over {m!}}\left( {\ln \left( {{1 \over {1 - x}}} \right)} \right)^{\,m}
  = \sum\limits_{0\, \le \,k} {\left[ \matrix{  k \cr   m \cr}  \right]\,{{x^{\,k} } \over {k!}}} 
$$
we can get this e.g.f.
$$
\eqalign{
  & G(z,y,\alpha ) = \sum\limits_{0\, \le \,n} {F(z,n,\alpha ){{y^{\,n} } \over {n!}}}  =   \cr 
  &  = \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,k} {{{k!} \over {n!}}
\left[ \matrix{  n \cr  k \cr}  \right]\left( { - y} \right)^{\,n} z^{\,k} \left( \matrix{   - \alpha  \cr   k \cr}  \right)} }  =   \cr 
  &  = \sum\limits_{0\, \le \,k} {\left( {\sum\limits_{0\, \le \,n}
 {\left[ \matrix{  n \cr  k \cr}  \right]{{\left( { - y} \right)^{\,n} } \over {n!}}} } \right)k!z^{\,k}
 \left( \matrix{   - \alpha  \cr   k \cr}  \right)}  =   \cr 
  &  = \sum\limits_{0\, \le \,k}
 {\left( \matrix{   - \alpha  \cr  k \cr}  \right)\left( {\ln \left( {{1 \over {1 + y}}} \right)} \right)^{\,k} z^{\,k} }  =   \cr 
  &  = \left( {1 + z\ln \left( {{1 \over {1 + y}}} \right)} \right)^{\, - \alpha }
  = {1 \over {\left( {1 - z\ln \left( {1 + y} \right)} \right)^{\,\alpha } }} \cr} 
$$
which means
$$
F(z,n,\alpha ) = \left. {{\partial^n  \over {\partial y^n}}
{1 \over {\left( {1 - z\ln \left( {1 + y} \right)} \right)^{\,\alpha } }}\;} \right|_{\,y\, = \,0} 
$$
