$\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{1}{k(k+1)…(k+m+1)}=\frac{1}{(m+1)\cdot(m+1)!}$ Can I proof this identity somehow without using the Riemann integral?
I have tried Stolz–Cesàro Theorem and sums of valuations' progressions.
 A: Note that
$$
\frac{1}{k(k+1)\cdots(k+m+1)}=\frac{1}{(m+1)}
\left(\frac{1}{k(k+1)\cdots(k+m)}-\frac{1}{(k+1)(k+2)\cdots(k+m+1)}\right).
$$
Thus
$$
\sum_{k=1}^n \frac{1}{k(k+1)\cdots(k+m+1)}=\frac{1}{(m+1)}
\left(\frac{1}{(m+1)!}-\frac{1}{(n+1)(n+2)\cdots(n+m+1)}\right),
$$
and hence
$$
\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{k(k+1)\cdots(k+m+1)}=\frac{1}{(m+1)\cdot (m+1)!}.
$$
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{\lim_{n \to \infty}\sum_{k = 1}^{n}{1 \over k\pars{k + 1}\ldots\pars{k+m+1}}
     ={1 \over m\,m!}:\ {\Large ?}}$

\begin{align}
&\color{#00f}{\large\sum_{k = 1}^{\infty}{1 \over k\pars{k + 1}\ldots\pars{k+m+1}}}=
\sum_{k = 1}^{\infty}{\Gamma\pars{k} \over \Gamma\pars{k + m + 2}}
=
{1 \over \pars{m + 1}!}\sum_{k = 1}^{\infty}
{\Gamma\pars{k}\Gamma\pars{m + 2} \over \Gamma\pars{k + m + 2}}
\\[3mm]&=
{1 \over \pars{m + 1}!}\sum_{k = 1}^{\infty}{\rm B}\pars{k,m + 2}=
{1 \over \pars{m + 1}!}\sum_{k = 1}^{\infty}
\int_{0}^{1}t^{k - 1}\pars{1 - t}^{m + 1}\,\dd t
\\[3mm]&=
{1 \over \pars{m + 1}!}\int_{0}^{1}\pars{1 - t}^{m + 1}
\pars{\sum_{k = 1}^{\infty}t^{k - 1}}\,\dd t
=
{1 \over \pars{m + 1}!}\int_{0}^{1}\pars{1 - t}^{m + 1}\,{1 \over 1 - t}\,\dd t
\\[3mm]&=
{1 \over \pars{m + 1}!}\bracks{-\,{\pars{1 - t}^{m + 1} \over m + 1}}_{0}^{1}
=\color{#00f}{\large{1 \over \pars{m + 1}\pars{m + 1}!}}
\end{align}

$\ds{{\rm B}\pars{x,y} \equiv \int_{0}^{1}t^{x - 1}\pars{1 - t}^{y - 1}\,\dd t}$
is the Beta Function which satisfies
$\ds{{\rm B}\pars{x,y} = {\Gamma\pars{x}\Gamma\pars{y} \over\Gamma\pars{x + y}}}$. $\Gamma\pars{z}$ is the Gamma Function. Also, $\ds{\Gamma\pars{n + 1} = n!}$
with $\ds{n \in {\mathbb N}}$.
A: The denominator is $k (k+1)_{m+1}$ where the second term is the Pochhammer symbol.   
The summation from $k=1$ to $k=n$ simplifies and write
$$\frac{\frac{1}{\Gamma (m+1)}-\frac{(m+1) \Gamma (n+1)}{\Gamma (m+n+2)}}{(m+1)^2}$$ Going to limit, this seems to give  $$\frac{1}{(m+1)^2 \Gamma (m+1)}$$ which seems to be wrong !  
I would really appreciate if somebody could tell me where I started being wrong.
