Easy algebra...(3)

$\begin{aligned} S & = \frac 3{1!+2!+3!} + \frac 4{2!+3!+4!} + \cdots + \frac {2017}{2015!+2016!+2017!} \\ & = \sum_{k=1}^{2015} \frac {k+2}{k!+(k+1)!+(k+2)!} \\ & = \sum_{k=1}^{2015} \frac {k+2}{k!(1+k+1+(k+1)(k+2))} \\ & = \sum_{k=1}^{2015} \frac {k+2}{k!(k+2)^2} = \sum_{k=1}^{2015} \frac 1{k!(k+2)} = \sum_{k=1}^{2015} \frac {k+1}{(k+2)!} = \sum_{k=1}^{2015} \frac {k+2-1}{(k+2)!} \\ & = \sum_{k=1}^{2015} \left(\frac 1{(k+1)!} - \frac 1{(k+2)!} \right) \\ & = \frac 1{2!} - \frac 1{2017!} = \frac 12 - \frac 1{2017!} \end{aligned}$

Therefore, $n=2017$ and the sum of its digits is $2+0+1+7 = \boxed{10}$ .

$\begin{aligned}\displaystyle\sum_{i=3}^{2017}\dfrac{i}{(i-2)!+(i-1)!+(i)!}&=\displaystyle\sum_{i=3}^{2017}\dfrac{i}{(i-2)!\{1+(i-1)+(i)(i-1)\}}\\& =\displaystyle\sum_{i=3}^{2017}\dfrac{i}{(i-2)!(i^2)}\\& =\displaystyle\sum_{i=3}^{2017}\dfrac{(i-1)}{(i)!}\\& =\displaystyle\sum_{i=3}^{2017}\left(\dfrac{i}{(i)!}-\dfrac{1}{(i)!}\right)=\displaystyle\sum_{i=3}^{2017}\left(\dfrac{1}{(i-1)!}-\dfrac{1}{(i)!}\right)\end{aligned}$

$\begin{aligned}\therefore \displaystyle\sum_{i=3}^{2017}\dfrac{i}{(i-2)!+(i-1)!+(i)!}&=\dfrac{1}{(2)!}-\dfrac{1}{(3)!}+\dfrac{1}{(3)!}-\dfrac{1}{(4)!}+\ldots+\dfrac{1}{(2016)!}-\dfrac{1}{(2017)!}\\& =\dfrac{1}{2}-\dfrac{1}{(2017)!}=\dfrac{1}{2}-\dfrac{1}{(n)!}\end{aligned}$

$\therefore n=2017\implies \text{the sum of the digits of} ~n ~=\boxed{10}$