Going an year ahead

Let $S(x)=\displaystyle \sum_{n=0}^{\infty} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}\quad;\quad S_{N}(x)=\displaystyle \sum_{n=0}^{N} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}.$ The series in the problem is $S(1/2017)$ .Now note that $\dfrac{x^{2^n}}{1-x^{2^{n+1}}} =\dfrac{1}{1-x^{2^n}}-\dfrac{1}{1-x^{2^{n+1}}}$ Using this identity,the partial sums telescope: $S_{N}(x)=\left(\dfrac{1}{1-x}-\dfrac{1}{1-x^2}\right)+\left(\dfrac{1}{1-x^2}-\dfrac{1}{1-x^4}\right)+\ldots+\left(\dfrac{1}{1-x^{2^N}}-\dfrac{1}{1-x^{2^{N+1}}}\right) \\ = \dfrac{1}{1-x}-\dfrac{1}{1-x^{2^{N+1}}}.$ So that $S(x)=\lim_{n \to \infty} S_{N}(x)=\dfrac{1}{1-x}-1=\dfrac{x}{1-x}.$

$\therefore S (1/2017)= (1/2017)/(1 - 1/2017) = \boxed{\dfrac{1}{2016}}.$