Finite sum involving Stirling number of first kind and Pochhammer symbol

Question

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.

Answer 1

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 $$ F(z,n,\alpha ) = \left. {{\partial^n \over {\partial y^n}} {1 \over {\left( {1 - z\ln \left( {1 + y} \right)} \right)^{\,\alpha } }}\;} \right|_{\,y\, = \,0} $$