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.
 A: 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).
A: 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 $$A_n \to \frac{x}{1-x}.$$
A: Treating this strictly as a multiple-choice question, some simple considerations narrow it down to a) as the only reasonable possibility:  c) and d) are clearly less than $1$ for all $x$, but if $x$ is close to $1$, then $A_n$ is clearly (much) greater than $1$, while if $x\approx0$, then $A_n\approx0$ is not too hard to see, which eliminates b).
A: Hint: Each term of $A_n$ can be expressed in terms of geometric series.
