$\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{1}{k(k+1)…(k+m+1)}=\frac{1}{(m+1)\cdot(m+1)!}$

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 $$ \lim_{n\to\infty}\sum_{k=1}^n \frac{1}{k(k+1)\cdots(k+m+1)}=\frac{1}{(m+1)\cdot (m+1)!}. $$

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\lim_{n \to \infty}\sum_{k = 1}^{n}{1 \over k\pars{k + 1}\ldots\pars{k+m+1}} ={1 \over m\,m!}:\ {\Large ?}}$

\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 $\ds{n \in {\mathbb N}}$.