Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Please help me solving $\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2n})]$, $|x|<1$

share|improve this question
1  
Is the last term supposed to be $(1 + x^{2^n})$? –  Tunococ Sep 10 '12 at 20:34
add comment

2 Answers

up vote 4 down vote accepted

$\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2n})]$=$\lim_{n\to\infty}\frac{(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2n})(1-x)}{1-x}$=$\lim_{n\to\infty}\frac{(1-x^2)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2n})}{1-x}$=$\lim_{n\to\infty}\frac{(1-x^4)(1+x^4)\cdot\cdot\cdot(1+x^{2n})}{1-x}$=$\cdot\cdot\cdot$=$\lim_{n\to\infty}\frac{(1-x^{2n})(1+x^{2n})}{1-x}$=$\lim_{n\to\infty}\frac{1-x^{4n}}{1-x}$=$\frac{1}{x-1}$$\lim_{n\to\infty}{(1-x^{4n})}$=$\frac{1}{x-1}$$\cdot 1$=$\frac{1}{x-1}$.

share|improve this answer
    
Why do you flip the sign? (And the last term is $1 + x^{2^n}$, right?) –  Tunococ Sep 10 '12 at 20:45
add comment

Assuming the last term is $(1 + x^{2^n})$, 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 $n \to \infty$.

share|improve this answer
2  
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 $1$. –  nayrb Sep 10 '12 at 20:39
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.