Let $X_1, X_2,\ldots $ be a sequence of infinite independent and identically distributed random variables, where $X_i\sim U[0,1]$.
Let $D_n = \sqrt{X_1^2 + X_2^2 + X_3^2 + \ldots + X_n^2}$ be a random variable. What is the the value of the following cumulative distribution function at the point $\frac{1}{\sqrt{3}}$: $$\lim_{n\to \infty}F_{\frac{Dn}{\sqrt{n}}} \left(\frac{1}{\sqrt{3}}\right)= \text{?}$$
Is it $1$ because all CDFs are right continuous or is it $\frac12$ because of the delta like function the distribution is like at $n=\infty$?
It's not the Weak Law of Large Numbers, it's the Central Limit Theorem.
$\mathbb E[X_n^2] = 1/3$. By the Central Limit Theorem, $Y_n = \sqrt{n}(D_n^2/n - 1/3)$ tends in distribution to $\mathcal N(0, \sigma^2)$, where $\sigma$ is the standard deviation of $X_n^2$. In particular
$$ \mathbb F_{\frac{D_n}{\sqrt{n}}}\left(\frac{1}{\sqrt{3}}\right) = \mathbb P\left(\frac{D_n}{\sqrt{n}} \le \frac{1}{\sqrt{3}}\right) = \mathbb P\left(Y_n \le 0\right) \to \mathbb P(Z \le 0) = \frac12$$