TITLE: How to calculate sum for $k\ge 1\in\mathbb N\quad\sum\limits_{i=1}^\infty\frac1{i(i+1)(i+2)...(i+k)}$
QUESTION [6 upvotes]: $$\forall k\ge 1\in\mathbb N\\\displaystyle\sum\limits_{i=1}^\infty\frac1{i(i+1)(i+2)...(i+k)}=?$$
Try$(1)$I tried to apply vieta If it be considered like this ;
$$\frac1{i(i+1)..(i+k)}=\dfrac{A_{0}}{i}+\dfrac{A_{1}}{i+1}+...+\dfrac{A_{k}}{i+k}$$
Try$(2)$
$$\frac1{i(i+1)..(i+k)}=\dfrac{(i-1)!}{(i+k)!}=\dfrac{(i-1)!}{(i+k)!}\dfrac{(k+1)!}{(k+1)!}=\dfrac{1}{(k+1)!}\dfrac{1}{\dbinom{i+k}{i-1}}$$
But I couldn't get any usefull equalition.I want to calculate this series but how?
By the way, I exactly know that this series is convergence.

REPLY [3 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[15px,#ffe]{\ds{\left.\sum_{i = 1}^{\infty}{1 \over i\pars{i + 1}\pars{i + 2}\ldots\pars{i + k}}\right\vert_{\ k\ \in\ \mathbb{N}_{\ \geq\ 1}}}} =
\sum_{i = 1}^{\infty}{1 \over i^{\,\overline{k + 1}}} =
\sum_{i = 1}^{\infty}{1 \over \Gamma\pars{i + k + 1}/\Gamma\pars{i}}
\\[5mm] = &\
{1 \over k!}\sum_{i = 1}^{\infty}{\Gamma\pars{i}\Gamma\pars{k + 1} \over \Gamma\pars{i + k + 1}} =
{1 \over k!}\sum_{i = 1}^{\infty}\int_{0}^{1}t^{i - 1}\pars{1- t}^{k}\,\dd t =
{1 \over k!}\int_{0}^{1}\pars{1- t}^{k - 1}\,\dd t =
\bbx{\ds{1 \over k\ k!}}
\end{align}