<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] If you expand each of the geometric series in the $A_n$, and combine each of the series together as one sum (you can since these are each absolutely convergent), then you can demonstrate that this is tending to $x(1-x)^{-1}$.
What you need to make sure of though, is that $x^k$ can appear only once between each of the terms $x^{2^n}(1-x^{2^{n+1}})^{-1}$ for each $k \in \mathbb{N}$.
This amounts to showing that each integer $k$ can be expressed uniquely as $2^n+m\cdot 2^{n+1}$. Or in other words that each integer can be written as $2^n(2m+1)$ (a power of 2 times an odd number). This representation is certainly unique.

More explicitly consider, $A_0$, $A_1$ and $A_2$:
$$A_0 = \frac{x}{1-x^2} = x+x^3+x^5+x^7+\cdots$$
Here we have all of the odd numbers.
$$A_1 = A_0 + \frac{x^2}{1-x^4} = (x+x^3+x^5+x^7+\cdots) + (x^2 + x^6 + x^{10} + \cdots)$$
$$=x+x^2+x^3+\underline{ }+x^5+x^6+x^7+\underline{ }+x^9+x^{10}+x^{11}+\underline{ } +\cdots$$
Notice that we are slowly filling in the spaces for the geometric series.
$$A_2 = A_1 + \frac{x^{4}}{1-x^8} = A_1 + (x^4+x^{12}+x^{20}+\cdots)$$
$$=x+x^2+x^3+x^4+x^5+x^6+x^7+\underline{ }+x^9+x^{10}+x^{11}+x^{12} +\cdots$$
With $A_2$ we see that we are still missing $x^8$, but that will appear with $A_3$. The uniqueness at the top of my answer tells us that we will never have $2x^k$ since each $x^k$ will appear only once. Thus $A_n$ tends to the geometric series (times $x$) and $$A_n \to \frac{x}{1-x}.$$
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