<s>[INST] Infinite sequence series. Limit If $0<x<1$ and
$$A_n=\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\ldots+\frac{x^{2^n}}{1-x^{2^{n+1}}},$$
then $\lim_{n\to\infty} A_n$ is
$$\text{a) }\ \dfrac{x}{1-x} \qquad\qquad \text{b) }\ \frac{1}{1-x} \qquad\qquad \text{c) }\ \frac{1}{1+x} \qquad\qquad \text{d) }\ \frac{x}{1+x}$$
How to do this? Not able to convert in any standard form.
 [/INST] Using the factorization $1 - x^{2^{k+1}} = (1 - x^{2^k})(1 + x^{2^k})$ we decompose
\begin{align}&\frac{x}{1 - x^2} + \frac{x^2}{1 - x^4} + \cdots + \frac{x^{2^n}}{1 - x^{2^n}}\\
&= \left(\frac{1}{1 - x} - \frac{1}{1 - x^2}\right) + \left(\frac{1}{1 - x^2} - \frac{1}{1 - x^4}\right) + \cdots + \left(\frac{1}{1 - x^{2^n}} - \frac{1}{1 - x^{2^{n+1}}}\right)\\
&=\frac{1}{1 - x} - \frac{1}{1 - x^{2^{n+1}}}.
\end{align}
Since $0 < x < 1$, $x^{2^{n+1}} \to 0$, so the last expression converges to 
$$\frac{1}{1 - x} - 1 = \frac{x}{1 - x}.$$
Therefore, the answer is a).
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