$p \in \mathbb{N^*}$, $u_n = \frac{1}{n(n + 1)...(n + p)}$
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 $v_n - v_{n+1}$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}
Hint $$\frac{1}{n(n-1)...(n-p)}=\frac{1}{p} \left(\frac{1}{(n-1)...(n-p)}-\frac{1}{n(n-1)...(n-p+1)} \right)$$
