There's a pattern!

A fraction of this form

$\frac { n! }{ (n-1)!+ (n-2)! }$

simplifies to

$\frac {n!} { (n-1)(n-2)! + (n-2)!}$

$\frac {n!} { (n-1+1)(n-2)!}$

$\frac {n!} {n (n-2)!}$

$\large = \boxed{n-1}$

Hence, the lengthy expression simplifies to

$\large 2015 + 2014 + 2013 + ... 3 + 2 + 1$

This sum is equal to $\frac {2015}{2} \times 2016$

$\large = \boxed {2031120}$

And the sum of its digits amount to $\large \large \boxed {9}$