Find a closed-form for a power series including binomial coefficients I came across a sequence $\{p_n\}$, which is given by
\begin{equation}
p_n=\frac{1}{n!}\sum_{k=0}^\infty\frac{\mu^k}{k!}(-\delta k)^{(n)},
\end{equation}
where $0<\delta<1$ and $\mu<0$, and the last term is the rising factorial.
I goal is to analyze the monotonicity of this sequence.
By simulation, I found that this sequence is either monotone decreasing or has one peak value. 
Also, I observed that the sequence will never increase once it starts to decrease. If this is the fact, a direct conclusion is, if $p_0>p_1$, i.e., $1+\mu\delta>0$, then the sequence will be monotone decreasing. Therefore, I want to first prove that the sequence is monotone decreasing if $1+\mu\delta>0$.
I have tried to simplify $p_n$ in two ways (getting rid of the infinite summation):
\begin{equation}
\begin{split}
p_n&=\frac{(-1)^ne^\mu}{n!}\sum_{k=0}^ns(n,k)T_k(\mu)\delta^k\\
&=\frac{e^\mu}{n!}\sum_{p=0}^n\frac{(-\mu)^p}{p!}\sum_{k=0}^p(-1)^{k}\binom{p}{k}(-\delta k)^{(n)},
\end{split}
\end{equation}
where $s(n,k)$ denotes the Stirling number of the first kind, and $T_k(\mu)$ is the Touchard polynomial.
It seems that it's difficult to use the ratio to prove the monotonicity, and therefore we calculate
\begin{equation}
\begin{split}
p_n-p_{n+1}&={(-1)^n}\sum_{k=0}^\infty \frac{\mu^k}{k!}\binom{\delta k+1}{n+1}\\
&=\frac{(-1)^ne^\mu}{(n+1)!}\sum_{k=0}^{n+1}\left[s(n,k-1)+s(n,k)\right]T_{k}(\mu)\delta^{k}\\
 &=\frac{(-1)^ne^\mu}{(n+1)!}
\sum_{k=0}^{n+1}\left[T_k(\mu)\sum_{p=k}^{n+1}s(n+1,p)
\binom{p}{k}\right]\delta^k,\\
\end{split}
\end{equation}
where the $\binom{\cdot}{\cdot}$ denotes the binomial coefficients. However, these efforts seem not to help me a lot to derive the monotonicity. Could you please give some idea on it? Thank you so much.
 A: Upon the further information you provided in your post I am reformulating my answer as follows.
The function $P$ 
$$ \bbox[lightyellow] {  
P(n,\delta ,\mu ) = \left( { - 1} \right)^{\,n} \sum\limits_{0\, \le \,k} {{{\mu ^{\,k} } \over {k!}}\left( \matrix{
  \delta k + 1 \cr   n + 1 \cr}  \right)} 
 } \tag{0}$$
can be rewritten in other forms which allow to get some of  its properties.
Finite Sum
$$
\eqalign{
  & P(n,\delta ,\mu ) = \left( { - 1} \right)^{\,n} \sum\limits_{0\, \le \,k} {{{\mu ^{\,k} } \over {k!}}} \left( \matrix{
  \delta k + 1 \cr 
  n + 1 \cr}  \right) =   \cr 
  &  = \left( { - 1} \right)^{\,n} \sum\limits_{0\, \le \,k} {{{\mu ^{\,k} } \over {k!}}} {{\left( {\delta k + 1} \right)^{\,\underline {\,n + 1\,} } } \over {\left( {n + 1} \right)!}} =   \cr 
  &  = {{\left( { - 1} \right)^{\,n} } \over {\left( {n + 1} \right)!}}\sum\limits_{0\, \le \,k} {{{\mu ^{\,k} } \over {k!}}} \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n + 1} \right)} {\left( { - 1} \right)^{\,n + 1 - j} \left[ \matrix{
  n + 1 \cr 
  j \cr}  \right]\left( {\delta k + 1} \right)^{\,j} }  =   \cr 
  &  =  - {1 \over {\left( {n + 1} \right)!}}\sum\limits_{0\, \le \,k} {{{\mu ^{\,k} } \over {k!}}} \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n + 1} \right)} {\left( { - 1} \right)^{\,j} \left[ \matrix{
  n + 1 \cr 
  j \cr}  \right]\sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,j} \right)} {\left( \matrix{
  j \hfill \cr 
  l \hfill \cr}  \right)\delta ^{\,l} k^{\,l} } }  =   \cr 
  &  =  - {1 \over {\left( {n + 1} \right)!}}\sum\limits_{0\, \le \,k} {{{\mu ^{\,k} } \over {k!}}} \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n + 1} \right)} {\left( { - 1} \right)^{\,j} \left[ \matrix{
  n + 1 \cr 
  j \cr}  \right]\sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,j} \right)} {\left( \matrix{
  j \hfill \cr 
  l \hfill \cr}  \right)\delta ^{\,l} \sum\limits_{\left( {0\, \le } \right)\,m\,\left( { \le \,l} \right)} {\left\{ \matrix{
  l \cr 
  m \cr}  \right\}k^{\,\underline {\,m\,} } } } }  =   \cr 
  &  =  - {1 \over {\left( {n + 1} \right)!}}\sum\limits_{\matrix{
   {0\, \le \,k}  \cr 
   {\left( {0\, \le } \right)\,m\,\left( { \le \,n + 1} \right)}  \cr 
 } } {\left( {\sum\limits_{\matrix{
   {\left( {0\, \le } \right)\,j\,\left( { \le \,n + 1} \right)}  \cr 
   {\left( {0\, \le } \right)\,l\,\left( { \le \,j} \right)}  \cr 
 } } {\left( { - 1} \right)^{\,j} \left[ \matrix{
  n + 1 \cr 
  j \cr}  \right]\left( \matrix{
  j \hfill \cr 
  l \hfill \cr}  \right)\delta ^{\,l} \left\{ \matrix{
  l \cr 
  m \cr}  \right\}} } \right){{\mu ^{\,k} } \over {k!}}\;} k^{\,\underline {\,m\,} }  \cr} 
$$
where $x^{\,\underline {\,m\,} } $ denotes the Falling Factorial, $[\, ]$ the (unsigned) Strirling N. 1st kind and
$\{\,\}$ the Stirling N. 2nd kind.
But the sum in $k$ is clearly $\mu^m e^{\mu}$, therefore
$$ \bbox[lightyellow] {  
P(n,\delta ,\mu ) =  - {{e^{\,\mu } } \over {\left( {n + 1} \right)!}}\sum\limits_{\matrix{
   {\left( {0\, \le } \right)\,j\,\left( { \le \,n + 1} \right)}  \cr 
   {\left( {0\, \le } \right)\,l\,\left( { \le \,j} \right)}  \cr 
   {\left( {0\, \le } \right)\,m\,\left( { \le \,l} \right)}  \cr  } }
 {\left[ \matrix{n + 1 \cr  j \cr}  \right]\left( { - 1} \right)^{\,j} 
\left( \matrix{  j \hfill \cr   l \hfill \cr}  \right)\delta ^{\,l}
 \left\{ \matrix{  l \cr  m \cr}  \right\}\mu ^{\,m} } \quad \left| {\; - 1 \le n} \right.\quad 
 } \tag{1}$$
which has the clear advantage of including a finite sum.
Hypergeometric
When $1 \le n$, we can reformulate $P$ as
$$
\eqalign{
  & P(n,\delta ,\mu )\quad \left| {\;1 \le n} \right.\quad  =   \cr 
  &  = \left( { - 1} \right)^{\,n} \sum\limits_{0\, \le \,k} {{{\mu ^{\,k} } \over {k!}}\left( \matrix{
  \delta k + 1 \cr 
  n + 1 \cr}  \right)}  = \left( { - 1} \right)^{\,n} \sum\limits_{1\, \le \,k} {{{\mu ^{\,k} } \over {k!}}\left( \matrix{
  \delta k + 1 \cr 
  n + 1 \cr}  \right)}  =   \cr 
  &  = \left( { - 1} \right)^{\,n} \mu \sum\limits_{0\, \le \,k} {{{\mu ^{\,k} } \over {\left( {k + 1} \right)!}}\left( \matrix{
  \delta k + \delta  + 1 \cr 
  n + 1 \cr}  \right)}  \cr} 
$$
and therefrom, since the coefficients of $\mu ^k$ are rational functions of $k$,  reach to express it
$$ \bbox[lightyellow] {  
\eqalign{
  & P(n,\delta ,\mu )\quad \left| {\;1 \le n} \right.\quad  =   \cr 
  &  = \left( { - 1} \right)^{\,n} \mu \left( \matrix{
  \delta  + 1 \cr 
  n + 1 \cr}  \right)\;{}_{n + 2}F_{n + 2} \left( {1,2 + {{1 - j} \over \delta };\;2,1 + {{1 - j} \over \delta };\;\mu } \right)\quad \left| {\;0 \le j \le n} \right. \cr} 
 } \tag{2}$$
but here $n$ is also determining the number of upper and lower parameters.
Recursion
We consider first the partial derivative of $P$ wrt $\mu$, which we need in the following.
$$ \bbox[lightyellow] {  
\eqalign{
  & Q(n,\delta ,\mu ) = {\partial  \over {\partial \mu }}P(n,\delta ,\mu ) = \left( { - 1} \right)^{\,n} \sum\limits_{0\, \le \,k} {{{k\mu ^{\,k - 1} } \over {k!}}} \left( \matrix{
  \delta k + 1 \cr 
  n + 1 \cr}  \right) =   \cr 
  &  = \left( { - 1} \right)^{\,n} \sum\limits_{0\, \le \,k} {{{\mu ^{\,k} } \over {k!}}} \left( \matrix{
  \delta k + 1 + \delta  \cr 
  n + 1 \cr}  \right) = \sum\limits_{0\, \le \,j} {\left( { - 1} \right)^{\,n} \sum\limits_{0\, \le \,k} {{{\mu ^{\,k} } \over {k!}}} \left( \matrix{
  \delta k + 1 \cr 
  n + 1 - j \cr}  \right)\left( \matrix{
  \delta  \cr 
  j \cr}  \right)}  =   \cr 
  &  = \sum\limits_{0\, \le \,j\, \le \,n + 1} {\left( { - 1} \right)^{\,j} \left( \matrix{
  \delta  \cr 
  j \cr}  \right)P(n - j,\delta ,\mu )} \quad \left| {\;0 \le n} \right. \cr} 
 } \tag{3.a}$$
it is to note that the last  summation includes $P(-1,\,d,\, \mu)$.
Then let's consider identity (1): applying to the Binomial and to the Stirling Numbers
their known recursive identities, through simple algebraic passages (which shall be omitted here), we can easily
reach to
$$ \bbox[lightyellow] {  
P(n,\delta ,\mu ) = {{n - 1} \over {\left( {n + 1} \right)}}P(n - 1,\delta ,\mu ) - {{\delta \mu } \over {\left( {n + 1} \right)}}{\partial  \over {\partial \mu }}P(n - 1,\delta ,\mu )\quad \left| {\;0 \le n} \right.
 } \tag{3.b}$$
And combining the two
$$ \bbox[lightyellow] {
\eqalign{
  & P(n,\delta ,\mu ) = {{n - 1} \over {n + 1}}P(n - 1,\delta ,\mu ) - {{\delta \,\mu } \over {n + 1}}\sum\limits_{0\, \le \,j\, \le \,n} {\left( { - 1} \right)^{\,j} \left( \matrix{
  \delta  \cr 
  j \cr}  \right)P(n - 1 - j,\delta ,\mu )}  =   \cr 
  &  = {{n - \left( {1 + \delta \,\mu } \right)} \over {n + 1}}P(n - 1,\delta ,\mu ) + {{\delta \,\mu } \over {n + 1}}\sum\limits_{0\, \le \,j\, \le \,n - 1} {\left( { - 1} \right)^{\,j} \left( \matrix{
  \delta  \cr 
  j + 1 \cr}  \right)P(n - 2 - j,\delta ,\mu )} \quad \left| {\;0 \le n} \right. \cr} 
 } \tag{3.c}$$
The sum has alternating sign, so it remains always to reply to the fundamental question 
of whether (and under which conditions) does $P$ remain positive for all $n$.
However I wish that the recurrence might provide a good leverage to prove that.
