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 $$|x|<1$$.

• Hint: uniqueness of binary representation of a natural number. – Ignorant Mathematician Apr 13 at 19:59
• @SoumikGhosh i dont get it. Can you provide more explanations please? – friendlyuser Apr 13 at 20:50

\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}

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 $$n \to \infty$$.

• The quick and easy way to see this expansion is to note that every number $k$ has a unique binary representation. If the $n$th binary term is $1$, you choose $x^{2^n}$ from the $(1+x^{2^n})$ term, otherwise you choose $1$. – abnry Sep 10 '12 at 20:39

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 $$\prod_{n\geq 1} (1+ X^{2^n})= 1 + X + X^2+ X^3+...= \frac {1}{1-X}$$

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 $$(1-x)^{-1}$$.