# 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?

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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 $2<p<3$.

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I was editing my question to correct the fact that the series is divergent for p=1. For p=2 I didn't see my mistake. Thanx a lot for correcting me. And yes p is an integer. – Dr.Optix Aug 27 '12 at 21:48
By using the Stirling estimate for $\Gamma(n+p)$, one can show convergence for real $p\gt 2$. – André Nicolas Aug 27 '12 at 22:44