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