How to approach the divergence of $\sum\frac{n!}{(x-1)(x-2)...(x-n)}$ If $x>0$, prove that the following series is divergent:
$\sum\frac{n!}{(x-1)(x-2)\ldots(x-n)}$
where $n=1,2,3\ldots$
I have proved that the absolute values series is divergent, but I cannot establish an inequality between both series. Help me, please!
 A: If $n\in\mathbb N$, then\begin{align}\frac{n!}{(x-1)(x-2)\ldots(x-n)}&=\frac1{(x-1)\left(\frac x2-1\right)\ldots\left(\frac xn-1\right)}\\&=\frac{(-1)^n}{(1-x)\left(1-\frac x2\right)\ldots\left(1-\frac xn\right)}\end{align}and$$\lim_{n\to\infty}(1-x)\left(1-\frac x2\right)\ldots\left(1-\frac xn\right)=0.$$So,$$\lim_{n\to\infty}\left|\frac{n!}{(x-1)(x-2)\ldots(x-n)}\right|=+\infty.$$
A: In another way
$$
\eqalign{
  & t_n  = {{n!} \over {\left( {x - 1} \right)\left( {x - 2} \right) \cdots \left( {x - n} \right)}} = {{n!} \over {\left( {x - 1} \right)^{\,\underline {\,n\,} } }}  \cr 
  & {{t_{n + 1} } \over {t_n }} = {{n + 1} \over {\left( {x - n - 1} \right)}} = {x \over {x - n - 1}} - 1  \cr 
  & \left| {{{t_{n + 1} } \over {t_n }}} \right| = \left| {{x \over {x - n - 1}} - 1} \right| = \left| {{x \over {n + 1 - x}} + 1} \right| \ge 1\quad \left| \matrix{
  \;0 < x \hfill \cr 
  \;x - 1 < n \hfill \cr}  \right. \cr} 
$$
Now note that
$$
\eqalign{
  & 0 < \left| {{x \over {n + 1 - x}} + 1} \right| < 1\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad  - 1 < {x \over {n + 1 - x}} + 1 < 1  \cr 
  &  \Rightarrow \quad  - 2 < {x \over {n + 1 - x}} < 0  \cr 
  &  \Rightarrow \quad \left[ \matrix{
  \left\{ \matrix{
  0 < n + 1 - x \hfill \cr 
   - 2n - 2 + 2x < x < 0 \hfill \cr}  \right. \hfill \cr 
  \quad  \vee  \hfill \cr 
  \left\{ \matrix{
  n + 1 - x < 0 \hfill \cr 
  0 < x <  - 2n - 2 + 2x \hfill \cr}  \right. \hfill \cr}  \right.\quad  \Rightarrow \quad \left[ \matrix{
  \left\{ \matrix{
  x < n + 1 \hfill \cr 
  x < 0 \hfill \cr}  \right. \hfill \cr 
  \quad  \vee  \hfill \cr 
  \left\{ \matrix{
  n + 1 < x \hfill \cr 
  2\left( {n + 1} \right) < x < 2x \hfill \cr}  \right. \hfill \cr}  \right.\quad  \Rightarrow \quad \left[ \matrix{
  x < 0\left( { \wedge \;0 \le n} \right) \hfill \cr 
  \quad  \vee  \hfill \cr 
  \left( {2 \le } \right)2\left( {n + 1} \right) < x \hfill \cr}  \right. \cr} 
$$
and the second condition is just implying a limited sum.
So we have that for negative $x$ the sum converge, while it doesn't for positive $x$.
For example, for $x=-1$ we have
$$
\left. {t_n } \right|_{\,x =  - 1}  = {{n!} \over {\left( { - 2} \right)^{\,\underline {\,n\,} } }} = \left( { - 1} \right)^{\,n} {{n!} \over {2^{\,\overline {\,n\,} } }} = \left( { - 1} \right)^{\,n} {{n!} \over {1^{\,\overline {\,n + 1\,} } }} = \left( { - 1} \right)^{\,n} {1 \over {n + 1}}
$$
By using the definition of the Gamma function as the limit of the partial Gamma, we can have a better 
overlook on the situation
$$
\eqalign{
  & t_n  = {{n!} \over {\left( {x - 1} \right)^{\,\underline {\,n\,} } }} = \left( { - 1} \right)^{\,n} {{n!} \over {\left( {1 - x} \right)^{\,\overline {\,n\,} } }} = \left( { - 1} \right)^{\,n} {{\left( { - x} \right)} \over {n^{\, - x} }}{{n^{\, - x} n!} \over {\left( { - x} \right)^{\,\overline {\,n + 1\,} } }}  \cr 
  & \mathop {\lim }\limits_{n \to \infty } t_n  = \mathop {\lim }\limits_{n \to \infty } \left( { - 1} \right)^{\,n} {{\left( { - x} \right)} \over {n^{\, - x} }}\Gamma ( - x) = \mathop {\lim }\limits_{n \to \infty } \left( { - 1} \right)^{\,n} n^{\,x} \,\Gamma ( - x + 1) \cr} 
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{x > 0}$:

\begin{align}
&\bbox[10px,#ffd]{\ds{{n! \over \pars{x - 1}\pars{x - 2}\cdots\pars{x - n}}}} =
{n!\over \pars{x - n}^{\large\overline{n}}} =
{n! \over \Gamma\pars{x}/\Gamma\pars{x - n}}
\\[5mm] = &
{1 \over \Gamma\pars{x}}\,n!\,{\pi \over \Gamma\pars{1 - x + n}\sin\pars{\pi\bracks{x - n}}} =
{\pars{-1}^{n}\,\pi \over
\Gamma\pars{x}\sin\pars{\pi x}}\,{n! \over \pars{n - x}!}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,&
\pars{-1}^{n}\,\Gamma\pars{1 - x}\,
{\root{2\pi}n^{n + 1/2}\expo{-n} \over
\root{2\pi}\pars{n - x}^{n - x + 1/2}\expo{-\pars{n - x}}}
\\[5mm] = &
\pars{-1}^{n}\,\Gamma\pars{1 - x}\expo{-x}\,
{n^{n + 1/2} \over n^{n - x + 1/2}\pars{1 - x/n}^{n - x + 1/2}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
{\pars{-1}^{n}\,\Gamma\pars{1 - x} \over n^{-x}}
\end{align}

Note that $\ds{\left.\lim_{n \to \infty}n^{x}\right\vert_{\ x\ >\ 0} = +\infty}$.

