Problem
I have been working through a probability question, where I am supposed to show that the following claim is not true. I believe I am supposed to use the Weak Law of Large Numbers in order to induce a contradiction, however, I'm not sure how to make progress here.
Claim: For $X_n$ following the Chi Square distribution with $n$ degrees of freedom, $\frac{X_n}{n}$ converges in distribution to $X$ with $P(X=0)=1$. We want to show that this claim is false.
Attempt
As $X_n$ follows the Chi Square distribution, we know that $$X_n=Z_1^2+Z_2^2+\cdots+Z_n^2$$ where each $Z_i$ follows the standard normal distribution.
I don't think each $X_i$ will be independent because $$X_i=X_{i-1}+Z_i^2$$ However, without assuming independence, I'm completely unsure of how to to proceed from here.
If we then assume independence, then the Weak Law of Large Numbers states:
Weak Law of Large Numbers: for independent $X_1, X_2, X_3$... with common mean and variance, then for all $\epsilon >0$, the probability $$P \Big{(} \Bigl{|} \frac{\sum_{1}^nX_k}{n}-\mu \Bigl{|} \ge \epsilon \Big{)} \rightarrow 0$$ $$\text{or alternatively}$$ $$P \big{(}\bigl{|}X_n-X \bigl{|} \ge \epsilon \big{)} \rightarrow 0$$
We can rewrite $$P\big{(}\bigl{|}X_n-X \bigl{|} \ge \epsilon \big{)} = P \big{(} \bigl{|}X_n \bigl{|} \ge \epsilon \big{)}$$ as $X=0$ with probability $1$.
I am unsure of how to proceed from here (or if what I have done is correct). I would be grateful for any further guidance as I am not sure if my assumption of independence is correct.