16.16

$\begin{aligned} \sum_{n=0}^\infty (-x)^n & = \frac 1{1+x} \quad \quad \small \color{#3D99F6}{\text{Maclaurin series, for } -1 < x < 1} \\ \frac{d}{dx} \sum_{n=0}^\infty (-x)^n & = \frac{d}{dx} \left( \frac 1{1+x} \right) \\ \sum_{n=0}^\infty (-x)^{n-1} n & = - \frac 1{(1+x)^2} \quad \quad \small \color{#3D99F6}{\text{Multiplying both sides by }x} \\ \sum_{n=0}^\infty (-x)^n n & = - \frac x{(1+x)^2} \quad \quad \small \color{#3D99F6}{\text{Putting }x = \frac 1{2016}} \\ \sum_{n=0}^\infty \frac{(-1)^n n}{2016^n} & = -\frac{2016}{(1+2016)^2} \\ \implies S & = -\frac{2016}{(1+2016)^2} \\ \implies \left \lfloor \frac 1S \right \rfloor & = \left \lfloor \frac {(2016+1)^2}{2016} \right \rfloor \\ & = \left \lfloor \frac {2016^2 + 2(2016) + 1}{2016} \right \rfloor \\ & = \boxed{-2019} \end{aligned}$

Here, $S=-\frac{1}{2016}+\frac{2}{2016^2}-\frac{3}{2016^3}+ \ldots$ Now, $\frac{S}{2016}=-\frac{1}{2016^2}+\frac{2}{2016^3}+ \ldots$ Adding these two equations, we get $(1+\frac{1}{2016})S=-\frac{1}{2016}+\frac{1}{2016^2}-\frac{1}{2016^3}- \ldots$ Using the formalism of geometric progression we have $S=-\frac{2016}{2017^2}$

$S=\sum_{n=0}^{ \infty } \dfrac{(-1)^n n}{ 2016^n }=-\left(\frac{1}{2016}+\frac{3}{2016^3}+\frac{5}{2016^5}+\cdots\right)+\left(\frac{2}{2016^2}+\frac{4}{2016^4}+\frac{6}{2016^6}+\cdots\right)$ Clearly, this is the sum of two Arithmetic-Geometric Progressions, which are easily evaluated, giving us the answer -2019.