Calculate $\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2n})]$, $|x|<1$ Please help me solving $\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})]$, in the region $|x|<1$.
 A: The product expands out as
$$
(1 + x)(1 + x^2)(1 + x^4)\ldots(1 + x^{2^n}) = \sum_{k=0}^{2^{n+1} -1} x^k = \frac{1 - x^{2^{n+1}}}{1 - x}.
$$
Since $|x| < 1$, this converges to $\frac{1}{1 - x}$ as $n \to \infty$.
A: \begin{align}
& \lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})] \\
&= \lim_{n\to\infty}\frac{(1-x)(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})}{1-x} \\
&= \lim_{n\to\infty}\frac{(1-x^2)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})}{1-x} \\
&= \lim_{n\to\infty}\frac{(1-x^4)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})}{1-x} \\
&= \cdots \cdots \cdots \\
&= \lim_{n\to\infty}\frac{(1-x^{2^n})(1+x^{2^n})}{1-x} \\
&= \lim_{n\to\infty}\frac{1-x^{2^{n+1}}}{1-x} \\
&= \frac{1}{1-x}\lim_{n\to\infty}{(1-x^{2^{n+1}})} \\
&= \frac{1}{1-x}\cdot 1 \\
&= \frac{1}{1-x} .
\end{align}
A: Consider the generating function $$ \prod_{n\geq 1} (1+ X^{2^n})$$
Coefficient of $X^k=\# $ of ways $n$ can be written as sum of distinct powers of $2$ 
But the binary representation of $n \in \mathbb N$ is unique and hence coefficient of $ X^k=1 \ \forall k $
Thus we get  $$ \prod_{n\geq 1} (1+ X^{2^n})= 1 + X + X^2+ X^3+...= \frac {1}{1-X} $$
A: Here is one way. One can write  the product as the telescoping product
$$
\frac{1-x^2}{1-x}\times \frac{1-x^4}{1-x^2}\times \frac{1-x^8}{1-x^4}\times \dotsb\frac{1-x^{2^{n+1}}}{1-x^{2^n}}\times \dotsb.
$$
so that the $n$th partial product (indexed from $0$) equals
$$
\frac{1-x^{2^{n+1}}}{1-x}=1+x+\dotsb+x^{2^{n+1}-1}
$$
which converges in the topology of formal power series to $(1-x)^{-1}$.
