Intuition about Central limit theorem Suppose $X_{1}, X_{2}, ....$ is a random sample from a probability distribution with mean $\mu < \infty$ and $\sigma^{2} < \infty$. Let $\bar{X}_{n} = \frac{\sum X_{i}}{n}$.
From the Central Limit Theorem we can state that: 
$\sqrt{n}(\bar{X}_{n} - \mu) \xrightarrow {d} N(0,\sigma^{2})$ 
All right, but now, i want to know how to explain (intuitively) the next statements:
$(\bar{X}_{n} - \mu) \xrightarrow {d} N(0,\frac{\sigma^{2}}{n})$ 
$\bar{X}_{n}  \xrightarrow {d} N(\mu,\frac{\sigma^{2}}{n})$
$ \sum X_{i} \xrightarrow {d} N(n\mu,\sigma^{2})$ 
Is it because the Slutsky theorem? (about continuous functions)
Thank's a lot! 
 A: This is a wrong statement
$$(\bar{X}_{n} - \mu) \xrightarrow {d} N\left (0,\frac{\sigma^{2}}{n}\right)$$
because it is translated "the left-hand side converges to a normal random variable as $n$ goes to infinity. But as $n$ goes to infinity,the right hand side acquires zero variance and becomes a degenerate random variable, a constant, not a normal distribution. Let alone that we should write something like 
$$(\bar{X}_{n} - \mu) \xrightarrow {d} N\left (0,\frac{\sigma^{2}}{\lim_{n\to \infty}(n)}\right)$$
The following statement is correct (although notation is not universal)
$$(\bar{X}_{n} - \mu) \sim_{\text{approx}}N\left (0,\frac{\sigma^{2}} {n}\right),\;\;\; n<\infty$$
and how "close" to this normal random variable is it will depend on the sample size in combination with the properties of the distribution $X$ follows. 
A: I was thinking about something:
Suppose we have a suficiently big $n$, so we can state that:
$\sqrt{n}(\bar{X}_{n} - \mu) \sim N(0,\sigma^{2})$
Fix n, and we can ask ourselves the next question:
How is $(\bar{X}_{n} - \mu)$ distributed?
Since is a linear transformation of a normal, it still a normal and the parameters are:
$E[\bar{X}_{n} - \mu] = 0$ and $Var(\bar{X}_{n} - \mu) = \frac{\sigma^{2}}{n}$ 
So finally: $(\bar{X}_{n} - \mu) \sim N(0,\frac{\sigma^{2}}{n})$
For the others random variables is straightforward.
