The CLT says that given finite variance of iid RVs, we have $$\sqrt{n}( \bar{X} - \mu) \rightarrow \mathcal{N}(0,\sigma^2),$$ but if this is true, then $\bar{X} - \mu$ should converge to $\mathcal{N}(0,\sigma^2 /n)$, right? And if this is true, then $\bar{X}$ should converge to $\mathcal{N}(\mu, \sigma^2/n)$, right?
My questions are: are above two statements true (i.e, can we just multiply and subtract constants like we'd normally do with a normal distribution), and if so, why isn't the latter $\left( \bar{X} \rightarrow \mathcal{N}(\mu, \sigma^2/n) \right) $statement the way we state the CLT which seems much more intuitive, since it clearly says that the mean of our RVs are almost normally distributed with the proper mean and a diminishing variance, while it's not immediately obvious what the other statement is on about?
Let's write this out: The CLT says that if $X_j$ are iid with finite expected value $\mu$ and finite variance $\sigma^2$, then the sequence
$$Y_n = \sqrt{n}\cdot\left({\frac{X_1+\ldots+X_n}{n}-\mu}\right)$$
converges in distribution to $N(0, \sigma^2)$.
Now $n$ here is not actually a variable, it is merely an index. So saying that $Y_n/n$ converges in distribution to $N(0, \sigma^2/n)$ really doesn't make any sense.
Instead, since $Y_n/\sqrt{n}$ converges almost surely to $0$ by the law of large numbers, $Y_n/n$ does too.