Intuitive way to prove that $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \cdots = 0$ I’m trying to come up with an intuitive way to prove that $$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \cdots = 0$$ That is, I want to prove that $$\prod_{n=1}^\infty \frac {(1)^n}{2}=0$$
I thought of this way. $$S= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \cdots$$ Multiply both sides by 2 to give $$2S= 1 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \cdots$$ But $$2S=S$$ So $S=0$. Is this proof valid?

Let $p_n = \prod_{i=1}^n \frac12 = \left(\frac12\right)^n$. Then $\frac{p_{n+1}}{p_n} = \frac12<1$ so $\{p_n\}$ is decreasing, and the product of positive numbers is a positive number, so $p_n>0$ for all $n$. Since $p_n$ is decreasing and bounded below, it converges to its infimum. Now, given $\varepsilon>0$, we may choose a positive integer $N$ such that $N>\frac{\log \varepsilon}{\log\frac12}$. It follows that for $n\geqslant N$ we have
$$
p_n= \left(\frac12\right)^n <\varepsilon,
$$
so that $\lim_{n\to\infty} p_n=0$.
