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 $|x|<1$.
Calculate $\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2n})]$, |x|<1
4 Answers
$\begin{align} & \lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})] \\ &= \lim_{n\to\infty}\frac{(1-x)(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})}{1-x} \\ &= \lim_{n\to\infty}\frac{(1-x^2)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})}{1-x} \\ &= \lim_{n\to\infty}\frac{(1-x^4)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})}{1-x} \\ &= \cdots \cdots \cdots \\ &= \lim_{n\to\infty}\frac{(1-x^{2^n})(1+x^{2^n})}{1-x} \\ &= \lim_{n\to\infty}\frac{1-x^{2^{n+1}}}{1-x} \\ &= \frac{1}{1-x}\lim_{n\to\infty}{(1-x^{2^{n+1}})} \\ &= \frac{1}{1-x}\cdot 1 \\ &= \frac{1}{1-x} . \end{align}$
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 $n \to \infty$.
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2The 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 $1$. – 2012-09-10