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!!

  • 1
    $\begingroup$ "$\lim_{n \to \infty}\inf\{\frac{1}{2^n} : n \in \mathbb{N}\}$" That doesn't really make sense $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.