Study the convergence of the following series I have to study the convergence of the following series:
$$\sum_{n\ge1} \frac{ n! } { p (p+1) \cdots (p + n - 1) }\text{ where }p > 0.$$
I tried d'Alembert criterion but $\lim_{n\to\infty} \frac {a_{n+1}} {a_n} = 1$ (where $a_n = \frac{ n! } { p  (p+1) \cdots (p + n - 1) }$).
Because that limit is $1$ the nature of the series is inconclusive.
Intuitively I can say that the series is convergent because when $p\in\{1,2,n\}$ the sum becomes:
For $p = 1$, $\sum_{n\ge1} \frac{ n! } {1\cdot2\cdots(1+n-1)} = \sum_{n\ge1} \frac{n!}{n!} = n$ is convergent
and
For $p = 2$, $\sum_{n\ge1} \frac{ n! } {2\cdot3\cdots(2+n-1)} = \sum_{n\ge1} \frac{n!}{2\cdot3\cdots(n+1)} =  \sum_{n\ge1} \frac{n!}{(n+1)!} =\sum_{n\ge1} \frac{1}{n+1}$ is convergent
and
For $p = n$, $\sum_{n\ge1} \frac{ n! } {n(n+1)\cdots(2n-1)} = \sum_{n\ge1} \frac{(n-2)!}{(n+2)(n+3)\cdots(2n-1)}$ is convergent (d'Alembert)
How do I proof my intuition is a rigorous mathematical way?

First, for $p=1$ the series is $\sum_{n\ge 1}\frac{n!}{n!}=\sum_{n\ge 1}1$, which is divergent. You’re right that for $p=2$ the series is $\sum_{n\ge 1}\frac1{n+1}$, but this series is also divergent: it’s essentially just the harmonic series. 
For $p\ge 3$ we have 
$$\frac{n!}{p(p+1)\dots(p+n-1)}\le\frac{n!}{3\cdot4\cdot\ldots\cdot(n+2)}=\frac2{(n+1)(n+2)}<\frac2{n^2}\;,$$
so the series converges by comparison with the $p$-series for $p=2$.
For $0<p\le2$, 
$$\frac{n!}{p(p+1)\dots(p+n-1)}\ge\frac{n!}{(n+1)!}=\frac1{n+1}\;,$$
so the series diverges by comparison with the harmonic series.
Assuming, as your work suggests, that $p$ is required to be an integer, this covers all possibilities; if not, it leaves open the question for $2<p<3$.
