Counting on Top, Factorials on Bottom

This is a Telescoping Series .

$\dfrac{k}{(k+1)!}=\dfrac{(k+1)-1}{(k+1)!}=\dfrac{1}{k!}-\dfrac{1}{(k+1)!} \\ \implies \dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\ldots+\dfrac{2015}{2016!} \\ =\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{3!}+\ldots+\dfrac{1}{2015!}-\dfrac{1}{2016!} \\ =1-\dfrac{1}{2016!}.$

Perhaps this is a better presentation:

Let the sum be $S$ , then we have:

$\begin{aligned} S & = \sum_{n=1}^{2015} \frac{n}{(n+1)!} \\ & = \sum_{n=1}^{2015} \left( \frac{1}{n!} - \frac{1}{(n+1)!} \right) \\ & = \sum_{n=1}^{2015} \frac{1}{n!} - \sum_{n=2}^{2016} \frac{1}{n!} \\ & = \frac{1}{1!} - \frac{1}{2016!} \\ \Rightarrow a & = \boxed{2016!} \end{aligned}$