$2018!$

Obviously,we don't have to evaluate the expression. We need to search for a method, which will ease our work. So,my solution is

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The expression can be written as

$\large \sum_{n=3}^{2018}=\dfrac{n}{(n-2)!+(n-1)!+n!}$ $=\dfrac{n}{(n-2)!(1+n-1+n(n-1))}$ $=\dfrac{n}{n^2(n-2)!}=\dfrac{1}{n(n-2)!}$ $=\dfrac{n-1}{n!}$ $=\dfrac{1}{(n-1)!}-\dfrac{1}{n!}$

Therefore,the sum becomes $\large \frac{1}{2!}-\cancel{\frac{1}{3!}}+\cancel{\frac{1}{3!}}-\cancel{\frac{1}{4!}}+\cancel{\cdots}-\frac{1}{2018!}$ $=\frac{1}{2!} - \frac{1}{2018!}$

$\implies$ $a=2$ and $b=2018$ $\implies ab=4036$ , hence the sum of its digits is $\boxed{13}$

$\begin{aligned} S & = \sum_{k=1}^{2016} \frac {k+2}{k!+(k+1)!+(k+2)!} \\ & = \sum_{k=1}^{2016} \frac {k+2}{k! \left(1+k+1+(k+1)(k+2)\right)} \\ & = \sum_{k=1}^{2016} \frac {k+2}{k! (k+2)^2} \\ & = \sum_{k=1}^{2016} \frac 2{k! (k+2)} \\ & = \sum_{k=1}^{2016} \frac {k+1}{(k+2)!} \\ & = \sum_{k=1}^{2016} \frac {k+2-1}{(k+2)!} \\ & = \sum_{k=1}^{2016} \left(\frac 1{(k+1)!} - \frac 1{(k+2)!}\right) \\ & = \frac 1{2!} - \frac 1{2018!} \end{aligned}$

Therefore, $ab = 2(2018) = 4036$ and its sum of digits is $\boxed{13}$ .