Can the product of a sequence of numbers between 0 and 1 converge to positive? Let $x_n \in (0,1)$, is it possible that $\prod_{n=1}^\infty x_n >0$ ? I think it isn't, because such small numbers multiplied together will become smaller and smaller, but I am not sure if there is a positive lower bound for the product. Thanks!
 A: Take your favorite converging series with positive general term 
$$
\sum_{n\geq 1}y_n=S.
$$
Then set 
$$
x_n=e^{-y_n}=\frac{1}{e^{y_n}}.
$$
You have
$$
\prod_{n\geq 1}x_n=e^{-S}=\frac{1}{e^S}.
$$
A: For $n\in\Bbb Z^+$ let $y_n=\prod_{k=1}^nx_k$. Define $x_n$ recursively so that $$y_n=\frac14+\frac1{2^{n+1}}\;.$$
Then $x_1=y_1=\frac14+\frac14=\frac12$, and for $n>1$ we must have $$x_n=\frac{y_n}{y_{n-1}}=\frac{\frac14+\frac1{2^{n+1}}}{\frac14+\frac1{2^n}}=\frac{2^{n-1}+1}{2^{n-1}+2}<1\;.$$
That is, the sequence $$\left\langle\frac{2^{n-1}+1}{2^{n-1}+2}:n\in\Bbb Z^+\right\rangle$$
is a sequence of numbers in $(0,1)$ whose product is
$$\lim_{n\to\infty}y_n=\lim_{n\to\infty}\left(\frac14+\frac1{2^{n+1}}\right)=\frac14>0\;.$$
I picked the target product $\frac14$ arbitrarily and chose to approach it in steps of $2^{-k}$; the same idea can clearly be used to construct by brute force an example with any desired product in $(0,1)$, approached by steps of any reasonable sequence of sizes.
A: For example, set $$a_n = 1+\frac{1}{2^n}$$ 
and then $$b_n = \frac{a_n}{a_{n-1}} < 1,$$
so that  $$\prod_{k=1}^n b_k = \frac{a_n}{2}.$$
Of course, it implies that your $x_n \to 1$. In case $x_n \not\to 1$
it means that there exists $\alpha \in (0,1)$ such that $x_n < \alpha$
infinitely many times, and $\prod_k^n x_k \leq \alpha^{\#\{k \leq n \mid x_k < \alpha\}} \to 0$.
Hope that helps ;-)
A: Well, if $\prod_n x_n = \theta$, and $x_n, \theta >0$, then $\ln(\prod_n x_n ) = \sum_n \ln x_n = \ln \theta$. Then you can use your knowledge of summations to find an example.
Here is one: $\theta = e^{-\frac{1}{2}}$, and $x_n = e^{-\frac{1}{2^{n+2}}}$.
A: It's quite simple to see actually. For any product $\prod_{i=1}^\infty n_i$ where $n_i>1$ that converges (such as $\prod_{i=1}^\infty \frac{1}{i^2}+1 \approx 3.676 $), you can take the reciprocal and then it is obvious that $n_i<1$ but $\prod_{i=1}^\infty{n_i}>0$.
