Tricky 2015 Sum

The series can be written as

$\displaystyle\sum_{n=0}^{\infty} \dfrac{x^{2^{n}}}{1 - x^{2^{n+1}}} = \sum_{n=0}^{\infty} \left(\dfrac{1}{1 - x^{2^{n}}} - \dfrac{1}{1 - x^{2^{n+1}}}\right),$

with $x = \dfrac{1}{2015}.$ This is now a telescoping series with sum

$\dfrac{1}{1 - x} - \displaystyle \lim_{n \rightarrow \infty} \left(\dfrac{1}{1 - x^{2^{n+1}}}\right) = \dfrac{1}{1 - x} - 1 = \dfrac{x}{1 - x}.$

Plugging in our value for $x$ gives us the sum $\dfrac{\frac{1}{2015}}{1 - \frac{1}{2015}} = \dfrac{1}{2014}.$

Thus $b - a = 2014 - 1 = \boxed{2013}.$

I know it's a little bit overkill but I didn't see the telescoping initially so I did it in a more involved way:

$\displaystyle\frac{1}{X^1-X^{-1}}+\frac{1}{X^2-X^{-2}}+\frac{1}{X^4-X^{-4}}+\frac{1}{X^8-X^{-8}}+\cdots$

$\displaystyle=\frac{X}{X^2-1}+\frac{X^2}{X^4-1}+\frac{X^4}{X^8-1}+\frac{X^8}{X^{16}-1}+\cdots$

$\displaystyle=\frac{1}{2}\left(\frac{1}{X^1+1}+\frac{1}{X^1-1}+\frac{1}{X^2+1}+\frac{1}{X^2-1}+\frac{1}{X^4+1}+\frac{1}{X^4-1}+\frac{1}{X^8+1}+\frac{1}{X^8-1}+\cdots\right)$

Now let $A=\displaystyle\frac{1}{X^1+1}+\frac{1}{X^2+1}+\frac{1}{X^4+1}+\frac{1}{X^8+1}+\cdots$ and $B=\displaystyle\frac{1}{X^1-1}+\frac{1}{X^2-1}+\frac{1}{X^4-1}+\frac{1}{X^8-1}+\cdots$

Then the answer to the original problem is thus $\displaystyle \frac{A+B}{2}$

We observe that $\displaystyle B=\frac{1}{X-1}+\frac{1}{2}\left(-\frac{1}{X^1+1}+\frac{1}{X^1-1}-\frac{1}{X^2+1}+\frac{1}{X^2-1}-\frac{1}{X^4+1}+\frac{1}{X^4-1}-\cdots\right)$

$\displaystyle B=\frac{1}{X-1}+\frac{B-A}{2}$

$\displaystyle A+B=\frac{1}{X-1}+\frac{B+A}{2}$

$\displaystyle \frac{B+A}{2}=\frac{1}{X-1}$

plug in $X=2015$ , $\displaystyle \frac{B+A}{2}=\frac{1}{2014}$