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.

Answer 1

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.

Answer 2

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.

Answer 3

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}. $