Some months ago, me and a friend tried to solve the following $"~natural~"$ question:
Given weights $p_{1},\ldots,p_{m}$ and distinct points in $S_{0} := \left\{\, x_{1},\ldots,x_{n}\,\right\}$ of ${\mathbb R}^{k}$, we constructed sets $\left(\, S_{n}\,\right)_{\, n\ \in\ {\mathbb N}}$ recursively, so that $S_{n}$ contains all possible weighted means of $S_{n - 1}$.
It turned out that $\lim S_{n}$ is dense in the convex hull of $S_{0}$, provided that $n,m \ge 2$.
At point, we asked about a stronger property, the equidistribution, and the following limit is related to such question$\ldots$, being an application of Levy's continuity theorem.
Given a irrational number $x \in \left(\, 0,1\,\right)$, show that
$$ \lim_{n\ \to\ \infty}\prod_{k\ =\ 1}^{\infty}\ \prod_{0\ \le\ j\le n} \left\{\, 1 - \left[\, {{n \choose j}\,{x^{j}\left(\, 1 - x\,\right)^{n - j} \over k}} \,\right]^{\,2} \,\right\}\ =\ 0 $$
Interchange the two products.
$$ \prod_{k=1}^\infty (1 - p^2/k^2) = \dfrac{\sin(\pi p)}{\pi p} $$
So you're asking about the limit as $n \to \infty$ of
$$ \prod_{j=0}^n \dfrac{\sin(\pi p_j(n))}{\pi p_j(n)} $$ where $p_j(n) = {n \choose j} x^j (1-x)^{n-j}$.
Note that $\sin(t)/t = 1 - t^2/6 + O(t^4) = \exp(-t^2/6 + O(t^4))$. Now $p_j(n) = P(X=j)$ where $X$ is a binomial random variable with parameters $n$ and $x$. The logarithm of your product is thus approximately $$ -\dfrac{\pi^2}{6} \sum_{j=0}^n p_j(n)^2 $$ But since $0 < p_j(n) < 1$ and $\sum_{j=0}^n p_j(n) = 1$ we have $$ -\dfrac{\pi^2}{6} \sum_{j=0}^n p_j(n)^2 \ge - \dfrac{\pi^2}{6}$$ So I don't believe your limit can be $0$.