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.
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.