Let the sequence $x_n = (1-\frac{1}{2})(1-\frac{1}{4})\dots(1-\frac{1}{2^n})$ Let the sequence $x_n$ be given by 
$$x_n = (1-\frac{1}{2})(1-\frac{1}{4})\dots(1-\frac{1}{2^n})$$
Prove that the sequence $x_n$ converges and that the limit is not $0$.
Attempt: 
The convergence part is easy, it is evident that $x_{n+1} < x_n$ and $0<x_n<\frac{1}{2}$. A strictly decreasing sequencing sequence bounded below is convergent hence $x_n$ is convergent. But I am not sure how to prove $x_n $ is not convergent to $0$.

Hint For $x,y \geq 0$ you have 
$$(1-x)(1-y) \geq 1-x-y$$
Then 
$$(1-\frac{1}{4})(1-\frac{1}{8}) \geq 1-\frac{1}{4}-\frac{1}{8} \\
(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16}) \geq (1-\frac{1}{4}-\frac{1}{8})((1-\frac{1}{16}) \geq 1-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}\\
............. \mbox{ ( i.e. by induction)}\\\
(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16}).. (1-\frac{1}{2^n}) \geq 1-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}- ...- \frac{1}{2^n}\\
$$
