Let the sequence $x_n = (1-\frac{1}{2})(1-\frac{1}{4})\dots(1-\frac{1}{2^n})$

Question

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 $0$.

Answer 1

Let $0<t\le1/2$. Then $$\ln(1-t)=-t-\frac{t^2}2-\frac{t^3}3-\cdots\ge-t-t^2-t^3-\cdots =-\frac{-t}{1-t}\ge -2t.$$ So $$\ln x_n\ge -1-\frac12-\cdots-\frac{1}{2^{n-1}}>-2$$ etc.

Answer 2

Hint For $x,y \geq 0$ you have $$(1-x)(1-y) \geq 1-x-y$$

Then $$(1-\frac{1}{4})(1-\frac{1}{8}) \geq 1-\frac{1}{4}-\frac{1}{8} \\ (1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16}) \geq (1-\frac{1}{4}-\frac{1}{8})((1-\frac{1}{16}) \geq 1-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}\\ ............. \mbox{ ( i.e. by induction)}\\\ (1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16}).. (1-\frac{1}{2^n}) \geq 1-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}- ...- \frac{1}{2^n}\\ $$