Isn't it still 2016?

Let, $f(x) = e^{kx}$
$\therefore \dfrac{d^{2017}f(x)}{dx^{2017}} = k^{2017}f(x)$
$\therefore k^{2017}f(x) = f(x)$
The values of k which satisfy this relation are the $2017^{th}$ roots of unity.

$\therefore r_{i}$ are $2017^{th}$ roots unity with $r_{2017} = 1$
Now , $\displaystyle \sum_{i=1}^{2017} r_{i} = 0 \rightarrow \displaystyle \sum_{i=1}^{2016}r_{i} = -1$
$\therefore S = 1$
$\displaystyle \lim_{x\rightarrow2017} \dfrac{2017(e^{2017-x}-1)}{e(2017-x)} = \dfrac{2017}{e} \displaystyle \lim_{t\rightarrow0} \dfrac{e^{t}-1}{t} = 2017e^{-1}$
$\therefore A = 2017, a = 1$
$A - a = 2016$

Side note : You guys need to stop making questions with the current year as answers. XD