Two Famous Series

Relevant wiki: Telescoping Series - Sum

$\mathbf{Generalisation}$ $\begin{aligned} \displaystyle & \sum_{p=1}^{\infty} \dfrac{1}{p(p+1)(p+2)\ldots(p+q)} \\ = & \dfrac{1}{q} \sum_{p=1}^{\infty}\left[ \dfrac{1}{p(p+1)(p+2)\ldots(p+q-1)}-\dfrac{1}{(p+1)(p+2)(p+3)\ldots(p+q)}\right] \\ = & \dfrac{1}{q}\left[\dfrac{1}{1\cdot2\cdot3\cdot\ldots\cdot q}-\dfrac{1}{2\cdot3\cdot4\cdot\ldots\cdot(q+1)}+\dfrac{1}{2\cdot3\cdot4\cdot\ldots\cdot(q+1)}-\ldots\right] \quad [\small{\text{Telescopic Series}}] \\ = & \dfrac{1}{q}\left[\dfrac{1}{1\cdot2\cdot3\cdot\ldots\cdot q}\right] \\ = & \displaystyle \boxed{\dfrac{1}{q\cdot q!}} \end{aligned}$

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Now we have $\begin{aligned} \displaystyle & \prod_{i=1}^{2017} i \sum_{n=1}^{\infty} \dfrac{1}{n(n+1)(n+2)\ldots(n+2016)} \\ = & \left[2017!\cdot\dfrac{1}{2016\cdot2016!}\right] = 2016!\cdot2017\cdot\dfrac{1}{2016\cdot 2016!} \\ = & \displaystyle \boxed{\boxed{\dfrac{2017}{2016}}} \end{aligned}$