Determine $\sum_{n = 1}^{+ \infty} \frac{1}{n(n+1)...(n+p)} $ 
$p \in \mathbb{N^*}$,  $u_n = \frac{1}{n(n + 1)...(n + p)}$
Determine $\sum_{n=1}^{+ \infty}u_n$
Hint: write $pu_n$ in the form $v_n - v_{n+1}$

I couldn't see how to use the hint, I tried to use $ln$ but I did not proceed much to write it in the from of $v_n - v_{n+1}$
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}
{1 \over n\pars{n + 1}\ldots\pars{n + p}}} =
\sum_{n = 1}^{\infty}{1 \over n^{\,\overline{p + 1}}} =
\sum_{n = 1}^{\infty}{1 \over \Gamma\pars{n + p + 1}/\Gamma\pars{n}}
\\[5mm] = &\
{1 \over p!}\sum_{n = 1}^{\infty}{\Gamma\pars{n}\Gamma\pars{p + 1} \over \Gamma\pars{n + p + 1}} =
{1 \over p!}\sum_{n = 1}^{\infty}\int_{0}^{1}t^{n - 1}\pars{1 - t}^{\, p}
\,\dd t
\\[5mm] = &\
{1 \over p!}\int_{0}^{1}\pars{\sum_{n = 1}^{\infty}t^{n - 1}}
\pars{1 - t}^{\, p}\,\dd t =
{1 \over p!}\int_{0}^{1}{1 \over 1 - t}\,\pars{1 - t}^{\, p}\,\dd t
\\[5mm] = &\
{1 \over p!}\int_{0}^{1}t^{\, p - 1}\,\dd t =
\bbx{{1 \over p}\,{1 \over p!}}
\end{align}
A: Hint
$$\frac{1}{n(n-1)...(n-p)}=\frac{1}{p} \left(\frac{1}{(n-1)...(n-p)}-\frac{1}{n(n-1)...(n-p+1)} \right)$$
