Inspired by Sharky Kesa

First note that $\displaystyle \prod_{n=1729}^{2017} \sum_{k=1}^{\infty} \dfrac{n}{\binom{k+n}{k}}=\prod_{n=1729}^{2017} n\cdot \color{#3D99F6}{\sum_{k=1}^{\infty} \dfrac{1}{\binom{k+n}{n}}}\quad \quad \quad \quad \quad \left[\because \binom{\color{magenta}{a}}{\color{#D61F06}{b}}= \binom{\color{magenta}{a}}{\color{magenta}{a}-\color{#D61F06}{b}}\right].$

Denoting the $\color{#3D99F6}{\text{highlighted}}$ series by $\color{#3D99F6}{\mathcal{S}}$ ,we have

$\begin{aligned} \color{#3D99F6}{\mathcal{S}}=&\displaystyle \sum_{k=1}^{\infty} \dfrac{1}{\binom{k+n}{n}}=\sum_{k=1}^{\infty} \dfrac{1}{\frac{(k+1)(k+2)\ldots(k+n)}{n!}} \\ =& n! \displaystyle \sum_{k=1}^{\infty} \dfrac{1}{(k+1)(k+2)\ldots (k+n)} \\ =& \dfrac{n!}{n-1} \displaystyle \sum_{k=1}^{\infty} \left(\dfrac{1}{(k+1)(k+2)\ldots(k+n-1)}-\dfrac{1}{(k+2)(k+3)\ldots(k+n)}\right) \\ =& \dfrac{n!}{n-1}\left(\dfrac{1}{2\cdot3\cdot\ldots\cdot n}-\dfrac{1}{3\cdot4 \cdot\ldots\cdot (n+2)}+ \dfrac{1}{3\cdot4 \cdot\ldots\cdot (n+2)}- \ldots\right) \quad \quad [\small{\color{forestgreen}{\text{Telescopic Series}}}] \\ =&\dfrac{n!}{n-1}\cdot \dfrac{1}{n!} = \dfrac{1}{n-1} \end{aligned}$ $\large{\therefore \displaystyle \prod_{n=1729}^{2017} \sum_{k=1}^{\infty} \dfrac{n}{\binom{k+n}{k}}=\prod_{n=1729}^{2017} \dfrac{n}{n-1}} \\ \small{\color{forestgreen}{\text{Telescopic Series}}} \\ =\dfrac{\color{turquoise}{1729}}{\color{#EC7300}{1728}}\cdot \dfrac{\color{#E81990}{1730}}{\color{turquoise}{1729}}\cdot \ldots \cdot \dfrac{\color{#20A900}{2017}}{\color{#624F41}{2016}}\\ =\large{\boxed{\dfrac{\color{#20A900}{2017}}{\color{#EC7300}{1728}}}}$