# A different version of central limit theorem?

When I was taking the probability class myself, I remember my professor give us a standard counter example of central limit theorem:

Let $X_{i}\sim X$ be $i.i.d$ random variables. Let $E(X)=0, Var(X)=1$. Now by central limit theorem we have:

$$\sqrt{n}\overline{X}\rightarrow^{d} N(0,1)$$

The counter example begins by assuming we can multiply $\sqrt{n}$ on both sides like a constant. Then we trivially have

$$\sum^{n}_{i=1}X_{i}\sim^{d} N(0,n)$$ which is now nonsense.

Of course the problem is both left and right hand side are changing, and claiming something converge to $N(0,n)$ does not make sense. In general one needs something like Slutsky's theorem. My question is:

1) From pedagogical point of view, is this still a good motivation for central limit theorem, that the shape of $\sum^{n}_{i=1}X_{i}$ is approximately a (rescaled) normal?

2) For the best way to make sense of this statement, I am considering the following: Let $[A,B]\subseteq \mathbb{R}$ be any non-degenerate interval, then $$\forall \epsilon,\exists N, \forall n>N,|P(\sum^{n}_{i}X_{i}\in [A,B])-P(N(0,n)\in [A,B])|\le \epsilon$$ I feel this might be a good re-framing of convergence in distribution. But this is really appropriate?

I ask because my primary research field is not probability and I have never seen anyone approaching central limit theorem heuristically this way.

However, your "reformulation" in 2) holds for far too many laws to be of any use. If you replace $N(0,n)$ by any sequence of laws such that the probability of any segment $[A,B]$ goes to $0$ (e.g. the uniform law on $[-n,n]$), then the same "reformulation" holds. It is in fact much, much weaker than the TCL, to the point of being useless.
You have to start from the formulation of the convergence of $\bar X/\sqrt{n}$, and then rescale, i.e. you can compare the probability that $\bar X$ and respectively $N(0,n)$ gets into $[\sqrt{n} A,\sqrt{n} B]$; or equivalently, say that the probability that $\bar X$ gets into $[\sqrt{n} A,\sqrt{n} B]$ tends to the probability that $N(0,1)$ gets into $[A,B]$.