How to calculate sum for $k\ge 1\in\mathbb N\quad\sum\limits_{i=1}^\infty\frac1{i(i+1)(i+2)...(i+k)}$ $$\forall k\ge 1\in\mathbb N\\\displaystyle\sum\limits_{i=1}^\infty\frac1{i(i+1)(i+2)...(i+k)}=?$$
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.
 A: Hint. One may observe that
$$
\begin{align}
\frac{k}{i(i+1)(i+2)...(i+k)}&=\frac{(i+k)-i}{i(i+1)(i+2)...(i+k)}
\\\\&=\frac{1}{i(i+1)(i+2)...(i+k-1)}-\frac{1}{(i+1)(i+2)...(i+k)}
\end{align}
$$ then one may see that terms telescope.
A: Since
$$
\frac1{i(i+1)(i+2)\cdots(i+k)}=\frac1k\left(\frac1{i(i+1)(i+2)\cdots(i+k-1)}-\frac1{(i+1)(i+2)\cdots(i+k)}\right)
$$
we can use Telescoping Series to get
$$
\sum_{i=1}^\infty\frac1{i(i+1)(i+2)\cdots(i+k)}=\frac1{k\,k!}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
A: In the answer I referred to
when I voted to close,
I showed that
$\sum_{j=1}^m \dfrac{1}{\prod_{k=0}^{n} (j+k)}
= \dfrac1{n}\left(\dfrac1{n!}-\dfrac1{\prod_{k=0}^{n-1} (m+1+k)}\right)
$.
This follows from
$\begin{array}\\
\dfrac1{\prod_{k=0}^{n-1} (x+k)}-\dfrac1{\prod_{k=0}^{n-1} (x+1+k)}
&=\dfrac1{\prod_{k=0}^{n-1} (x+k)}-\dfrac1{\prod_{k=1}^{n} (x+k)}\\
&=\dfrac1{\prod_{k=1}^{n-1} (x+k)}\left(\dfrac1{x}-\dfrac1{x+n}\right)\\
&=\dfrac1{\prod_{k=1}^{n-1} (x+k)}\left(\dfrac{n}{x(x+n)}\right)\\
&=\dfrac{n}{\prod_{k=0}^{n} (x+k)}\\
\end{array}
$
