I'm studying Law of Large Numbers and came across the next question: so, I know that if I have $ X_1 \ldots X_n $ independent random variables with $ X_i \sim N(\mu, \sigma^2),\ \forall i = 1, \ldots n \ $ then
$$ \overline{X_n} = \dfrac{X_1 + \ldots + X_n}{n} \sim N(\mu, \dfrac{\sigma^2}{n}) $$
for any given value $n$.
I also know, because of the Law of Large Numbers, that the arithmethic mean of the random variables $\overline{X_n}$ converges to the expected value when $n \rightarrow \infty$ i.e.
$$\overline{X_n} = \dfrac{X_1 + \ldots + X_n}{n} \rightarrow \mu \ \ \ \ \ \ \text{for} \ \ \ \ \ \ n \rightarrow \infty $$
But I'm wondering if the first statement is a contradiction to the Law of Large Numbers.
Edit: I was wondering that if, in the first statement, when $n \rightarrow \infty$ we can assume that the limit $\lim_{n\rightarrow\infty}\overline{X_n} \sim N(\mu, 0) $? That would mean that our distribution has no standard deviation so, all the values in the random variables will be $\mu$, but what's causing me conflict is that how can we say that the arithmetic mean of random variables converges to a distribution and because of the Law of Large Numbers converges, also, to a number.
The normal distribution $\mathcal{N}(\mu,0)$ with variance zero does not exists in that way. With the right understanding what "$\mathcal{N}(\mu,0)$" has to be, yes, you can assume $\lim_{n\rightarrow\infty}\overline{X_n} \sim \mathcal{N}(\mu,0)$. So what is $\mathcal{N}(\mu,0)$?
We better understand this as limit $\lim_{n\rightarrow\infty}\mathcal{N}(\mu,\frac{\sigma^2}{n})$. Roughly speaking the gaussian density function concentrates all mass at point $\mu$. With a bit more theory we can determine
$\mathcal{N}(\mu,\frac{\sigma^2}{n}) \rightarrow \delta_\mu$
where $\delta_\mu$ is the Dirac-Distribution of $\mu$, i.e. $\delta_\mu ([a,b]) = \begin{cases} 1, & \mu \in [a,b] \\ 0, & \text{otherwise} \end{cases}$
This means nothing else that, if $\lim_{n\rightarrow\infty}\overline{X_n} \sim \delta_\mu$, then $\lim_{n\rightarrow\infty}\overline{X_n} = \mu$ almost surely (this is the same result of LLN)