$\prod_{k=1}^\infty (1-1/2^k)$ converge to zero? I know this product converge
$$\prod_{k=1}^\infty (1-1/2^k),$$
but I don't know how to prove that this limite is different to zero. This is equivalent to prove that 
$$\sum^{\infty}_{k=1} \log(1-1/2^k)$$ converge. I can't get it. Could you give me any hint? Thank!
I'm trying to prove that 
$$\frac{|GL_n(\mathbb{F}_2)|}{2^{n^2}} \to \alpha,$$
where $\alpha > 0$ 
 A: for the lower bound, the general result is that for $\delta_k \in (0,1)$
$0\leq \prod_{k=1}^\infty (1-\delta_k)$
and the inequality is strict iff $\sum_{k=1}^\infty \delta_k \lt \infty$ 
the easy direction
comes from using $1 + x \leq e^x$ 
the harder direction
 (your direction) comes from using if $\sum_{k=1}^\infty \delta_k \lt \infty$  then for every $\epsilon \gt 0$ there is some $K$ such that $\sum_{k=K}^\infty \delta_k \lt \epsilon$.  You know this is true e.g. by looking at the partial sums $s_n = \sum_{k=1}^n \delta_k$ and seeing that they are Cauchy.  Select $\epsilon' := \frac{1}{3}$, $K'$ and you can bound the tail of your product as  
$0\lt \frac{2}{3} = 1 -\frac{1}{3}\leq \prod_{k=K'}^\infty (1-\delta_k)$
This is a slightly generalized Bernouli inequality.  This can be proven directly with induction, but the nice approach is to recognize it follows from the union bound in probability i.e.
$1-P\Big(\bigcup_{k=K'}^\infty A_{k}\Big)  = P\Big(\bigcap_{k=K'}^\infty A_{k}^C\Big) = \prod_{k=K'}^\infty (1-\delta_k)$  with independent events $A_k$ 
so the tail is $\in (0,1)$ and necessarily  $\big(\prod_{k=1}^{K'-1}(1-\delta_k)\big) \in (0,1)$ and so their product is $\in (0,1)$ 
A: If $x$ is positive and near $0$ then
$$
-2x \le \log(1-x) \le -x.
$$
This can be seen by observing that the function $x\mapsto\log(1-x)$


*

*is equal to $0$ at $0$, and

*has slope $-1$ at $0$, and

*has a negative second derivative at $0,$ so the graph is concave downward.

A: $\log(1-x)$ is a concave function on $[0,1)$, so for any $x\in[0,1/2]$ we have
$$ \log(1-x)\geq -2\log(2) x $$
immediately implying
$$ \sum_{k\geq 1}\log\left(1-\frac{1}{2^k}\right) \geq -2\log(2)\sum_{k\geq 1}\frac{1}{2^k} = -2\log(2) $$
and
$$ \prod_{k\geq 1}\left(1-\frac{1}{2^k}\right) \geq \frac{1}{4}.$$
Much better approximations can be derived through the Mellin transform, as done by Marko Riedel here.
Creative telescoping is also a chance. Numerically the LHS is $\approx 0.288788095$.
