Evaluating $\sum_{n=1}^{\infty}\frac{(n-1)!}{\prod_{r=1}^{n}(x+r)}$ for $x\in\mathbb{R}^{+}$ I am trying to evaluate the following sum. Here $x\in\mathbb{R^{+}}$.
$$f(x)=\sum_{n=1}^{\infty}\frac{(n-1)!}{\prod_{r=1}^{n}(x+r)}$$
At present I do not know how to proceed. Any hints are appreciated. Thanks.
 A: An answer without $\Gamma$-function magic.
Use partial fractions to show that
$$\begin{align}\frac{(n-1)!}{(x+1)(x+2)\cdots(x+n)}&=\sum_{k=1}^{n}(-1)^{k-1}\frac{\binom{n-1}{k-1}}{x+k}\\&=\int_{0}^1 t^x(1-t)^{n-1}\,dt\end{align}\tag 1$$
This is true for $x>-1,$ which we’ll need later.
You can prove this equality directly without partial fractions using integration by parts and induction.
Then switching the integral with the sum, we get:
$$\begin{align}\sum_{n=1}^{\infty} \frac{(n-1)!}{(x+1)\dots(x+n)}&=\int_0^1 t^x\sum_{n=1}^\infty (1-t)^{n-1}\,dt\\&\int_0^1 t^x\frac{1}{1-(1-t)}\,dt\\
&=\int_0^1t^{x-1}\,dt\\&=\frac1x\end{align}$$
You can switch sum and integral because of the Lebesgue Monotone Convergence Theorem.
Without the Monotone Convergence Theorem, you can just take partial sums:
$$\begin{align}
\sum_{n=1}^{N} &=\int_0^1 t^x\frac{1-(1-t)^N}{t}\,dt\\
&=\frac1x-\int_0^1 t^{x-1}(1-t)^N\,dt\\
&=\frac1x-\frac{N!}{x(x+1)\cdots (x+N)}\quad\text{by }(1)\\
&=\frac1x-\frac1{x(1+x/1)(1+x/2)\cdots (1+x/N)}
\end{align}$$
And $x\prod_n (1+x/n)$ diverges to infinity when $x>0$ because $\sum_n x/n$ diverges to infinity.

Telescoping sum
The easiest way to prove this is to prove it is a telescoping series.
The partial sums mean we can rewrite the terms of the series so they telescope:
$$\frac{(n-1)!}{(x+1)\cdots(x+n)}=a_{n-1}-a_{n}\tag 2$$ where $$a_n =\frac{1}{x(1+x/1)(1+x/2)\cdots (1+x/n)}$$
We can write that inductively, $$a_0=\frac1x,\\a_{n}=\frac{a_{n-1}}{1+x/n}$$
We have already shown that $a_n\to 0$ when $x>0.$
We can verify (2) directly: $$\begin{align}a_{n-1}-a_n&= \frac{(1+x/n)-1}{1+x/n}a_{n-1}\\&=\frac{(1+x/n)-1}{x(1+x/1)\cdots(1+x/n)}\\&=\frac{x/n}{x(1+x/1)\cdots(1+x/n)}\\&=\frac{(n-1)!}{(x+1)\cdots (x+n)}\end{align}$$
At this point, the most advanced theorem we are using is that the harmonic series, $\sum_n\frac1n,$ diverges (which allows us to conclude that $a_n\to 0.)$ That is only required for $0<x<1,$ because for $x\geq 1,$ $a_n\leq a_1= \frac{n!}{(n+1)!}=\frac{1}{n+1}.$
For $x>0,$ we have $$\frac{1}{x\log n}\geq a_n\geq \frac{1}{3xn^x}$$ so when $x$ is small, the  convergence to $0$ can be slow.
A: $$\prod _{r=1}^n (r+x)=\frac{\Gamma (n+x+1)}{\Gamma (x+1)}$$
$$f(x)=\sum_{n=1}^{\infty}\frac{\Gamma(n)}{\frac{\Gamma (n+x+1)}{\Gamma (x+1)}}=\sum_{n=1}^{\infty}\frac{\Gamma (n) \Gamma (x+1)}{\Gamma (n+x+1)}=\frac{\Gamma (x)}{\Gamma (x+1)}=\frac{1}{x}$$
