Convergence of an infinite product $\prod_{k=1}^{\infty }(1-\frac1{2^k})$?

Question

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.

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. — Sangchul Lee, Oct 23 '12 at 01:36
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 01:36
@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 01:48
@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 01:49
@C.Lambda, i'd like to, but I'm currently outside and I'm writing with my iPad. Typing TeX formulas with iPad is not far from a mental torture... And already many peoples are starting to post their own nice answers. I will post my answer when I'm available and still no satisfying answer at that moment. — Sangchul Lee, Oct 23 '12 at 01:58
See also: Is the product $\prod_{k=1}^\infty \frac{2^k-1}{2^k}$ necessarily $0$? — Martin Sleziak, Jun 12 '15 at 07:54
This is related, too: What is the value of $\prod_{i=1}^\infty 1-\frac{1}{2^i}$? (Although that question asks about the exact value of the product, not only whether it diverges to zero.) — Martin Sleziak, Jun 12 '15 at 08:03

score 3 · Accepted Answer · answered Oct 23 '12 at 02:29

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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.

answered Oct 23 '12 at 02:29

marty cohen

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I think you're missing a term... $(a+x^{n+1})(1-x^{n+1})$ should be $a+x^{n+1}(1-a-x^{n+1})$ – Grinsekotze Jun 20 '22 at 14:53

score 2 · Answer 2 · edited Apr 13 '17 at 12:21

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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.

edited Apr 13 '17 at 12:21

Community

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answered Oct 23 '12 at 01:52

Erick Wong

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Will Jagy · Answer 3 · 2012-10-23T02:17:24.787

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}. $

Convergence of an infinite product $\prod_{k=1}^{\infty }(1-\frac1{2^k})$?

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