Be Positive in 2017

The expression can be rewritten as:

$\begin{aligned} S & = \sum_{n=2}^{2017} f(x) + \sum_{n=2}^{2017} f \left(\frac 1x \right) \\ & = \sum_{n=2}^{2017} \left(f(x) + f \left(\frac 1x \right) \right) \\ & = \sum_{n=2}^{2017} \left(\frac x{1-x} + \frac {\frac 1x}{1-\frac 1x} \right) \\ & = \sum_{n=2}^{2017} \left(\frac x{1-x} + \frac 1{x-1} \right) \\ & = \sum_{n=2}^{2017} \frac {x-1}{1-x} \\ & = \sum_{n=2}^{2017} -1 \\ & = \boxed{-2016} \end{aligned}$

We can manipulate the function.

$f(a) + f(\frac{1}{a}) = ( \frac{a}{1-a}) + ( \frac{\frac{1}{a}}{1- \frac{1}{a}} )$

$= \frac{a}{1-a} + \frac{1}{a-1} = \frac{a}{1-a} - \frac{1}{1-a} = -1$ .

So we have $f(a) + f(\frac{1}{a}) = -1$ .

Now, let $S$ is the value of the given expression.

$S = [ f(2017) + f(\frac{1}{2017}) ] + [ f(2016) + f(\frac{1}{2016}) ] + ... + [ f(2) + f(\frac{1}{2}) ]$

$S = (-1) + (-1) + (-1) + ... + (-1) \rightarrow 2016$ times.

Hence, $S= -2016$ .