Does the multiplication of countably infinite many numbers between $0$ and $1$ equal $0$? Suppose every term of a countably infinite sequence $x_1,x_2,\dots$ is between $0$ and $1$, i.e. $0<x_i<1$ for every $i$. Does $\mathop {\lim }\limits_{n \to \infty } \prod\limits_{i = 1}^n {{x_i}} $ equal $0$ or there is counterexample that it might not be $0$? Thank you!
 A: Start with the decreasing sequence 
$$p_1 = 1 - \frac{1}{4}, \quad p_2 = 1 - \frac{3}{8}, \quad p_3 = 1 - \frac{7}{16}\quad, \quad ...\quad , \quad p_n = 1 - \frac{2^n-1}{2^{n+1}}
$$
whose limit equals $\frac{1}{2}$. Then set
$$x_1 = p_1, \quad x_2 = p_2/p_1, \quad x_3 = p_3/p_2,\quad \ldots
$$
The partial product $\prod_{i=1}^n x_i$ equals $p_n$, and so the limit is $\frac{1}{2}$.
A: Take $x_i = 1-\frac{1}{(i+1)^2}$. Then, we have
$$
\ln \prod_{i=1}^n x_i = \sum_{i=1}^n \ln\left(1-\frac{1}{(i+1)^2}\right)
$$
and as $\ln\left(1-\frac{1}{(i+1)^2}\right) \sim_{i\to\infty} -\frac{1}{i^2}$, the series $\sum_{i=1}^n \ln\left(1-\frac{1}{(i+1)^2}\right)$ converges (by comparison) to some real number $\ell < 0$. But this means that
$$
\prod_{i=1}^n x_i\xrightarrow[n\to\infty]{} e^\ell > 0.
$$
A: Good result to know: Suppose $a_n\in (0,1)$ for all $n.$ Then $\prod_{n=1}^\infty (1-a_n) > 0$ iff $\sum a_n < \infty.$ The proof is a nice exercise in taking logs and using $\log (1+u) = u + o(u).$
A: Let $(a_i)$ be a strictly increasing sequence of positive numbers which is bounded above. Such a sequence necessarily converges, call the limit $a$. Now set $x_i = \frac{a_i}{a_{i+1}}$. Then $0 < x_i < 1$ and 
$$\prod_{i=1}^nx_i = x_1\dots x_n = \frac{a_1}{a_2}\dots\frac{a_n}{a_{n+1}} = \frac{a_1}{a_{n+1}}.$$
Therefore, 
$$\prod_{i = 1}^{\infty}x_i = \lim_{n\to\infty}\prod_{i=1}^n x_i = \lim_{n\to\infty}\frac{a_1}{a_{n+1}} = \frac{a_1}{a}$$
which is non-zero as $a_1 \neq 0$. It follows from this construction that any $p \in (0, 1)$ arises as such an infinite product.
A: $$\frac{1}{2}=\frac{1}{2^{1/2}}\cdot \frac{1}{2^{1/4}}\cdot \frac{1}{2^{1/8}}\cdot \frac{1}{2^{1/16}}\cdot \frac{1}{2^{1/32}}\cdot \frac{1}{2^{1/64}} \cdots$$ 
