For $b \gt 2$ , verify that $\sum_{n=1}^{\infty}\frac{n!}{b(b+1)...(b+n-1)}=\frac{1}{b-2}$. For $b \gt 2$ , verify that $$\sum_{n=1}^{\infty}\frac{n!}{b(b+1)...(b+n-1)}=\frac{1}{b-2}$$
This is how I tried..
$$\sum_{n=1}^{\infty}\frac{n!(b-2)}{(b-2)b(b+1)...(b+n-1)}=\sum_{n=1}^{\infty}\frac{n!(b+n-1-2+1-n)}{(b-2)b(b+1)...(b+n-1)}=\sum_{n=1}^{\infty}\frac{n!(b+n-1)-(1+n)}{(b-2)b(b+1)...(b+n-1)}=\sum_{n=1}^{\infty}\frac{n!(b+n-1)}{(b-2)b(b+1)...(b+n-1)}-\frac{n!(1+n)}{(b-2)b(b+1)...(b+n-1)}=\frac{1}{b-2}\left(\frac{b-2}{b}+\sum_{n=2}^{\infty}\frac{n!}{b(b+1)...(b+n-2)}-\frac{(n+1)!}{b(b+1)....(b+n-1)}\right)$$
Now $$S_n=\frac{1}{b-2}\left[\frac{b-2}{b}+\left(\frac{2}{b}-\frac{3!}{b(b+1)}\right)+\left(\frac{3!}{b(b+1)}-\frac{4!}{b(b+1)(b+2)}\right)+...+\left(\frac{n!}{b(b+1)..(b+n-2)}-\frac{(n+1)!}{b(b+1)...(b+n-1)}\right)\right]=\frac{1}{b-2}\left[1-\frac{(n+1)!}{b(b+1)..(b+n-1)}\right]$$
All I need to show is that $$\lim\limits_{n\to \infty}\frac{(n+1)!}{b(b+1)..(b+n-1)}=0$$ which I am unable to show.
Thanks for the help!!
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$\ds{\sum_{n = 1}^{\infty}{n! \over b\pars{b + 1}\ldots\pars{b + n - 1}}
     ={1 \over b-2}}$

\begin{align}
\mbox{Note that}\quad
\Gamma\pars{b}&={\Gamma\pars{b + 1} \over b}
={\Gamma\pars{b + 2} \over b\pars{b + 1}}
={\Gamma\pars{b + 3} \over b\pars{b + 1}\pars{b + 2}}=\cdots
\\[3mm]&={\Gamma\pars{b + n} \over b\pars{b + 1}\pars{b + 2}\ldots\pars{b + n - 1}}
\end{align}
  such that

\begin{align}&\color{#44f}{\large%
\sum_{n = 1}^{\infty}{n! \over b\pars{b + 1}\ldots\pars{b + n - 1}}}
=\sum_{n = 1}^{\infty}{\Gamma\pars{n + 1}\Gamma\pars{b} \over \Gamma\pars{b + n}}
=\sum_{n = 1}^{\infty}n\,{\Gamma\pars{n}\Gamma\pars{b} \over \Gamma\pars{b + n}}
\\[3mm]&=\sum_{n = 1}^{\infty}n\,\int_{0}^{1}t^{n - 1}\pars{1 - t}^{b - 1}\,\dd t
=\int_{0}^{1}\
\overbrace{\pars{\sum_{n = 1}^{\infty}nt^{n - 1}}}^{\ds{=\ \pars{1 - t}^{-2}}}
\ \pars{1 - t}^{b - 1}\,\dd t
=\int_{0}^{1}\pars{1 - t}^{b - 3}\,\dd t
\\[3mm]&=\left.-\,{\pars{1 - t}^{b - 2} \over b - 2}\right\vert_{0}^{1}
=\color{#44f}{\large{1 \over b - 2}}\,,\qquad\qquad b > 2
\end{align}

$\ds{\Gamma\pars{z}}$ is the Gamma Function ${\bf\mbox{6.1.1}}$ and we used well known properties of it.

A: $$\lim_{n\to\infty}\frac{(n+1)!}{b(b+1)\cdot...\cdot(b+n-1)}=\lim_{n\to\infty}\frac{n!n^{b}}{b(b+1)\cdot...\cdot(b+n-1)(b+n)}\cdot\frac{(n+1)(b+n)}{n^{b}}$$
$$=\lim_{n\to\infty}\frac{n!n^{b}}{b(b+1)\cdot...\cdot(b+n)}\cdot\frac{(1+\frac{1}{n})(1+\frac{b}{n})}{n^{b-2}}=\Gamma(b)\lim_{n\to\infty}\frac{(1+\frac{1}{n})(1+\frac{b}{n})}{n^{b-2}}=0$$
where I have used that $b>2$ to justify the existence of $$\lim_{n\to\infty}\frac{(1+\frac{1}{n})(1+\frac{b}{n})}{n^{b-2}}$$
A: $b = 3 \Rightarrow \sum\limits_{n = 1}^\infty  {\frac{{n!}}{{3 \times 4 \times ....\left( {n - 2} \right)}}}  = \sum\limits_{n = 1}^\infty  {\left( {2n\left( {n - 1} \right)} \right)}  =  + \infty 
$
