Proving $ \lim_{n \to \infty} \, \left(\frac{2}{3}\right)^n = 0$ I'm having trouble with a proof of this question:
$$ \lim_{n \to \infty} \, \left(\frac{2}{3}\right)^n = 0$$
I know $N(0)$ is a log of epsilon with base $\frac{2}{3}$ but I can't proceed with a proof.

Since $\{(\frac{2}{3})^n\}$ is strictly decreasing, we only need to show that its inf is $0$.
In fact you can show that if $0<q<1$ , then $$\inf\{q^n|n\in\mathbb{Z}_+\}=0$$. To prove this we assume the contrary, or $i=\inf\{q^n|n\in\mathbb{Z}_+\}>0$(since the inf of the set must exist and it cannot be negative) . By the definition
of inf, exists $n\in\mathbb{Z}_+$ subject to $$i<q^n<\frac{i}{q}$$, which implies $q^{n+1}<i$, a contradiction with the definition of inf.
