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?
 A: 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 $\lim_{n\to\infty} p_n=0$.
A: Your proof doesn't quite work; limits are a bit tricky in that they can fail to exist, which means that there is an unfounded assumption when you write
$$S=\prod_{n=1}^{\infty}\frac{1}2.$$
This assumes there is such a real number, which there might not be.
Note that the fact that the quantity being multiplied has absolute value less than $1$ is never used; you could equally well run your argument on the product of any real number infinitely many times; for instance, if I started with
$$S=\prod_{n=1}^{\infty}2$$
I could multiply by $1/2$ to cancel the first term, exactly as you have done, to find $S=1/2S$ which implies $S=0$. However, when we multiply $2$ by itself over and over, the sum doesn't get closer to $0$ - it just gets bigger and bigger! So your argument does not work. It does establish, by similar reasoning that the product $\prod_{n=1}^{\infty}k$ either diverges or converges to $0$ whenever $k\neq 1$, though.
This said, you have shown that $0$ is the only possible value that the product could converge to - you do need to be careful with arguments like this, since, while your strategy of multiplying $S$ by $2$ and cancelling that $2$ with the first term is perfectly fine, it's easy to make mistakes when you write out infinite products and manipulate them like finite products - for instance, some products converge, but can be rearranged to converge to something else.
It's somewhat unavoidable that you have to do some analysis to prove your result; you are essentially trying to show that the sequence $(1/2)^n$ converges to $0$. This means that, noting that for every $\varepsilon > 0$ there is some $N$ such that $\left|(1/2)^n\right| < \varepsilon$ for every $n > N$. This is not so hard explicitly construct: if you set $N$ to be the ceiling of $\log_{1/2}(\varepsilon)$, you can verify that the sequence really does converge to $0$.
A: You're proof would work if we knew $\lim_{n\to \infty}(\frac 12)^n$ converges.
Then $\lim_{n\to \infty}(\frac 12)^n = L$ would mean $2L = 2\lim_{n\to\infty}(\frac 12)^n =\lim_{n\to \infty}(\frac 12)^{n+1}=\lim_{n-1\to \infty}(\frac 12)^{n}=\lim_{n\to\infty}(\frac 12)^n = L$ so $2L = L$ and $L = 0$.
But we NEED to know that $\lim_{n\to \infty}(\frac 12)^n$ converges first.
(Counterexample is if $\lim_{n\to \infty} 1+ 2 + 4 + .... + 2^n = S$ then $2S+1 = 1+2\lim_{n\to \infty} 1+ 2 + 4 + .... + 2^n = \lim_{n\to \infty} 1+ 2 + 4 + .... + 2^{n+1} = S$ and so $S= -1$ which is clearly false as $\lim_{n\to \infty} 1+ 2 + 4 + .... + 2^n$ doesn't converge.)
I don't know how "intuitive this is but $0 < \frac 12 < 1=(\frac 12)^0$ and if $0 < (\frac 12)^n < (\frac 12)^{n-1}$ then $0 < (\frac 12)^n*\frac 12 < (\frac 12)^{n-1}*\frac 12$ so $0 < (\frac 12)^{n+1} < (\frac 12)^n$ so for $n\to \infty$ we have $(\frac 12)^n$ decreasing but bounded below by $0$.  So $\lim (\frac 12)^n$ must converge.
But I don't know how "intuitive" it is.
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Perhaps the safest both intuitive and valid is:
For any $0 < \epsilon < 1$ then $\frac 1{\epsilon} > 1$ and $N =\log_2{\frac 1\epsilon} > 1$.  If $n > N>\log_2{\frac 1\epsilon}>1$ then $2^n > 2^N > \frac 1\epsilon$.
So $2^N > \frac 1\epsilon$ for $n > N$.
So $0 < \frac 1{2^n} = (\frac 12)^n < \epsilon$.
So $\lim_{n\to \infty} (\frac 12)^n = 0$.
....
Remember  $\lim_{n\to \infty} a_n = L$ means that we can always get $a_n$ and $L$ close enough together by taking $n$ large enough.
So if we want to get $0$ and $(\frac 12)^n$ closer than $\epsilon$ we just have to take $n > \log_2{\frac 1\epsilon}$.
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If that isn't "intuitive" enough
It should be intuitive that as $n\to \infty$ then $2^n \to \infty$ (but I don't like encouraging this approach because just because it is "intuitive" doesnt mean it is true). And it should be intuitive that as $m \to \infty$ that $\frac 1m \to 0$ (but again, ditto).  So  $n\to \infty\implies 2^n \to \infty \implies \frac 1{2^n} \to 0$.  That's intuitive.  And its valid.  But is it allowed?  I don't know.  Depends on your level.  I wouldn't accept it in and of itself; -- at least not the first  time.
