How to approach the divergence of $\sum\frac{n!}{(x-1)(x-2)...(x-n)}$

Question

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!

Answer 1

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 $$ \eqalign{ & t_n = {{n!} \over {\left( {x - 1} \right)^{\,\underline {\,n\,} } }} = \left( { - 1} \right)^{\,n} {{n!} \over {\left( {1 - x} \right)^{\,\overline {\,n\,} } }} = \left( { - 1} \right)^{\,n} {{\left( { - x} \right)} \over {n^{\, - x} }}{{n^{\, - x} n!} \over {\left( { - x} \right)^{\,\overline {\,n + 1\,} } }} \cr & \mathop {\lim }\limits_{n \to \infty } t_n = \mathop {\lim }\limits_{n \to \infty } \left( { - 1} \right)^{\,n} {{\left( { - x} \right)} \over {n^{\, - x} }}\Gamma ( - x) = \mathop {\lim }\limits_{n \to \infty } \left( { - 1} \right)^{\,n} n^{\,x} \,\Gamma ( - x + 1) \cr} $$

Answer 2

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,$$\lim_{n\to\infty}\left|\frac{n!}{(x-1)(x-2)\ldots(x-n)}\right|=+\infty.$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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