$\prod_{k=1}^\infty (1-1/2^k)$ converge to zero?

Solution 1:

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 $\in (0,1)$