An algebra problem by A Former Brilliant Member

General term, $t_n = \dfrac {x^{2^{n - 1}}}{1 - x^{2^{n}}}$ $= \dfrac {1 + x^{2^{n - 1}} - 1}{(1 + x^{2^{n - 1}})(1 - x^{2^{n - 1}})}$ $= \dfrac {1}{1 - x^{2^{n - 1}}} - \dfrac {1}{1 - x^{2^{n}}}$

Now, $S_n = \displaystyle \sum_{n = 1}^n t_n$ $= [\dfrac {1}{1 - x} - \dfrac {1}{1 - x^{2}}] + [\dfrac {1}{1 - x^{2}} - \dfrac {1}{1 - x^{4}}] + ......+[\dfrac {1}{1 - x^{2^{n - 1}}} - \dfrac {1}{1 - x^{2^{n}}}]$

$= \dfrac {1}{1 - x} - \dfrac {1}{1 - x^{2{n}}}$

To Infinite terms, $\displaystyle \lim_{n \rightarrow \infty} S= \dfrac {1}{1 - x} - 1$ $= \dfrac {x}{1 - x}$ . Since, $(\lim_{n \rightarrow \infty} x^{2^{n}} = 0, |x|<1)$

$ANSWER:\boxed{\dfrac {x}{1 - x}}$

Every single term is the sum of an infinite G.P, and expanding each term to the infinite G.P form will eventually give you $x$ to the power of every natural number, thus becoming $x+x^2+x^3+x^4+x^5.....=\frac {x}{1-x}$

( $a+ar+ar^2+ar^3....∞=\frac {a}{1-r}$ , given r<1,>0)