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Problem:

I want to prove that the infinite product $\prod_{k=1}^{\infty }(1-\frac{1}{2^{k}})$ does not converge to zero. It doesn't matter to find the value to which this product converges, but I am still curious to know if anybody is able (if possible of course) to find the value to which this infinite product converges. I appreciate any help. I tried the following trick: $\prod_{k=1}^{n}(1+a_{k})\geq 1+\sum_{k=1}^{n}a_{k}$ which can be easily proven by inudction, where $a_{k}>-1$ and they are all positive or negative. In this case, $a_{k}=-\frac{1}{2^{k}}$, but I get : the infinite product is greater than or equal to zero.

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You may want to take logarithm and then apply some sort of comparison to conclude that the resulting logarithmic sum converges. Then by exponentiation, you can find that the product converges to a positive number. –  sos440 Oct 23 '12 at 1:36
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You can show via induction that $\prod_{k=1}^n (1-\frac{1}{2^k}) \ge \frac14 + \frac{1}{2^{n+1}}$ and so in particular the infinite product converges to something that's at least $\frac14$. –  Alan Guo Oct 23 '12 at 1:36
    
@Alan Guo: How did you figure out your above inequaliy? I am just curious to see how you proved it. –  C. Lambda Oct 23 '12 at 1:48
    
@sos440: Can you, please, show me what kind of comparison you use to conclude convergence of the sum of the logarithms. –  C. Lambda Oct 23 '12 at 1:49
    
It's pretty straightforward to prove using induction. –  Alan Guo Oct 23 '12 at 1:54
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3 Answers

up vote 3 down vote accepted

Suppose $\prod_{k=1}^n (1-x^k) \ge a + x^{n+1}$ where $0 < x < 1$ and $0 < a < 1$. Then $\prod_{k=1}^{n+1} (1-x^k) \ge (a + x^{n+1})(1-x^{n+1}) = a + x^{n+1}(1-a) $.

To make this $\ge a + x^{n+2}$, we want $1-a \ge x$. For $x = 1/2$, $a = 1/4$ will work.

So, this argument gives a basis for choosing values for $a$ that can makes this inequality true for this inductive proof.

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This question is almost identical to this one.

In particular, while I wouldn't hope for a satisfying "closed form", Euler's pentagonal number theorem provides a simple expression for the binary expansion of this limit.

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From Taylor's Theorem with remainder, for $ 0 \leq x \leq \frac{1}{2}, $ we find $$ -x - \frac{x^2}{2} \geq \log (1-x) \geq -x -2 x^2. $$ because $$ f(x) = f(a) + f'(a) (x-a) + f''(\xi) \frac{(x-a)^2}{2} $$ where $\xi$ is between $x$ and $a.$

Your $x = \frac{1}{2^k}$ and $x^2 = \frac{1}{4^k}.$

Edddiiittt: now that I think of it, we could use Taylor's and stop at the linear term, $ -x \geq \log (1-x) \geq -2x , $ still for $ 0 \leq x \leq \frac{1}{2}. $

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