Is there an infinite product of reals in $(0,1)$ that doesn't converge to $0$? I seem to recall hearing something like this from one of my professors last year. 
Trying to be a little bit more succinct, does there exist a sequence $\{a_n\}_{i=0}^\infty$ such that $(\prod_{n=0}^\infty a_n) \ne 0,$ where $ \forall n,$ $ a_n \in (0,1)$? 
It seems very counterintuitive, since if $\forall n, a_n = 1-\epsilon$, with arbitrarily small $\epsilon > 0$, the product still converges to zero, but I suppose if the sequence converges really quickly to $1$, it might be possible to come up with such an example. Is it?
 A: As another example occurring in number theory, the Euler product for the reciprocal of the Riemann zeta function is also of this sort:
$\frac{1}{\zeta(s)} = \prod\limits_{p\text{ prime}}(1 - p^{-s})$, valid for $\Re(s) > 1$.
When $s$ is real and $s > 1$, each term will be in the interval $(0, 1)$ but $\frac{1}{\zeta(s)}$ is nonzero.
Additionally, given an infinite product converging to a nonzero value with each term larger than 1, we trivially get another infinite product converging to a nonzero value with terms smaller than 1 by taking reciprocals.
A: Yes, such a sequence does exist.  It is easy to come up with such a sequence by thinking instead about the sequence of partial products.  Given any descending sequence $(b_n)$ of elements of $(0,1)$, if you define $a_0=b_0$ and $a_n=b_n/b_{n-1}$, then each $a_n$ will be in $(0,1)$ and $b_n$ will be the $n$th partial product $\prod_{i=0}^n a_i$.  So if you take $(b_n)$ to be any descending sequence of elements of $(0,1)$ converging to some $b>0$, this gives a sequence $(a_n)$ whose product is nonzero.
A: If we choose $a_n = e^{-2^{-n}}$ for all $n \in \mathbb{N}_0$, we have that $0 < a_n < 1$ and
$$\prod_{n = 0}^{\infty}e^{-2^{-n}} = e^{-\sum_{n = 0}^{\infty}{2^{-n}}}=e^{-2} >0.$$
A: In fact, we can generate such a product that evaluates to any number $x\in(0,1)$ as follows:
Since
$$
\frac{k+x-1}{k}\frac{k+1}{k+x}=1-\frac{1-x}{k^2+kx}
$$
the product telescopes
$$
\prod_{k=1}^n\left(1-\frac{1-x}{k^2+kx}\right)=x\,\frac{n+1}{n+x}
$$
Therefore,
$$
\prod_{k=1}^\infty\left(1-\frac{1-x}{k^2+kx}\right)=x
$$
