# Prove $\lim_{n\to\infty}\inf\{\frac{1}{2^n} : n \in \mathbb{N}\}=0$

I'm trying to prove the sequence $\{\frac{1}{2^n}\}_{n=1}^{\infty}$ converges to 0. So far what I have done is said notice, $$\frac{1}{2^n} > \frac{1}{2} \times \frac{1}{2^n}=\frac{1}{2^{n+1}}$$ Thus the sequence is monotone decreasing. Also $\frac{1}{2^n}$ is never negative so it is bounded below by zero. Therefore we conclude that the sequence converges to, $$A= \lim_{n \to \infty}\inf\{\frac{1}{2^n} : n \in \mathbb{N}\}$$

I need to show this is $0$. Thus I need to show that if $b \in A$ then $b \geq 0$ and that $b \leq 0$. The first one comes easy as it is bounded below by zero, now for the second one I'm a little confused because we haven't formally learned logarithms in this class yet and so I cannot use them. Any help is appreciated, thanks!!

• "$\lim_{n \to \infty}\inf\{\frac{1}{2^n} : n \in \mathbb{N}\}$" That doesn't really make sense – zhw. Sep 30 '17 at 17:42

The problem you're having is that there is no $b\in A$ such that $b\le 0$. Indeed, every element of $A$ is positive. Since you've shown that the sequence is bounded below and decreasing, we have that $$\lim_{n\to\infty}\frac{1}{2^n} = \inf\left\{\frac{1}{2^n} : n\in\mathbb{N}\right\}.$$ Therefore we only have to show that $0=\inf\{\frac{1}{2^n} : n\in\mathbb{N}\}$. Well, as you noted, $0$ is a lower bound. Thus it only remains to prove that $0$ is the greatest lower bound. This amounts to showing that for every $\varepsilon>0$, the element $0+\varepsilon$ is not a lower bound. This means that for every $\varepsilon>0$ there exists $n\in\mathbb{N}$ such that $$0+\varepsilon > \frac{1}{2^n}.$$ For this you can use the Archimedian property. Given $\varepsilon>0$, the Archimedian property says that there exists $n\in\mathbb{N}$ such that $1/\varepsilon < n$. Hence $1/\varepsilon < n < 2^n$, so that $1/2^n < \varepsilon=0+\varepsilon$, as desired.