Unique!

Let us rewrite the above series according to:

$\Sigma_{n=1}^{2016} [\frac{2}{n}-\frac{1}{n+1}] \cdot (\frac{1}{2})^{n} = \Sigma_{n=1}^{2016} [\int_{0}^{1} 2x^{n-1}-x^n dx] \cdot (\frac{1}{2})^{n} = \int_{0}^{1} (\frac{2}{x} - 1)[ \Sigma_{n=1}^{2016} (\frac{x}{2})^{n}] dx = \int_{0}^{1} (\frac{2-x}{x}) \cdot \frac{x(1-(x/2)^{2016})}{2-x} dx = \int_{0}^{1} 1 - \frac{x^{2016}}{2^{2016}} dx = x - \frac{x^{2017}}{2^{2016}2017}|_{0}^{1}$

or $\boxed{1-\frac{1}{2^{2016}2017}}.$ Finally, $1 - [1-\frac{1}{2^{2016}2017}] = \frac{1}{2^{2016}2017 } \Rightarrow a = 2016, b = 2017 \Rightarrow b-a = \boxed{1}.$