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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 $|x|<1$.

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

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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 $n \to \infty$.

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    $\begingroup$ 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 $1$. $\endgroup$ – abnry Sep 10 '12 at 20:39
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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 $$ \prod_{n\geq 1} (1+ X^{2^n})= 1 + X + X^2+ X^3+...= \frac {1}{1-X} $$

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

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