I've been studying statistics for a while now, and honestly, I've been struggling to gain an intuition for both the Central Limit Theorem (CLT) and the Law of Large Numbers (LLN) up to this point. Now, I just came across this paragraph which describes both theorems as first and second-order asymptotic expansions. Unfortunately, the paragraph is very short and quite vague (probably by intention). In particular it doesn't state the underlying asymptotic scale $\{ \phi_k \}$ and stops after the second order.
So, here is my (also very vague) interpretation: Let $(X_1, X_2, \dots)$ be i. i. d. random variables with finite $1$st, $2$nd, ... , $k$th momentum $\mu_1, \dots, \mu_k$. We set $\bar{X}_n = \sum_{i=1}^n X_i$ and we assume that there is an asymptotic expansion $$ \bar{X}_n \sim \sum_{k=1}^{\infty} A_k \,n^{\frac{1}{k}}, $$ where $A_k$ are random variables and convergence is meant in distribution. I thought that $ \left\{ n^{\frac{1}{k}} \right\}_k$ is a reasonable choice for the asymptotic scale. If I made no mistake, in general, we should have $$ A_k = \lim_{n \to \infty} \bar{X}_n^{k}, \qquad \text{where} \quad \bar{X}_n^{k}:=n^{\frac{1}{k(k-1)}} \left( \bar{X}_n^{k-1}-A_{k-1} \right). $$ The LLN then says that $A_1 = \mu_1$, so it is a deterministic random variable. And the CLT says that $A_2$ has law $\mathcal{N}(0,\mu_2)$.
Assuming that I am not completely off-track, can we say something about $$ A_3 = \lim_{n \to \infty} \bar{X}^3_n , \qquad \text{where} \quad \bar{X}^3_n = \frac{\left( \bar{X}_n- n\mu_1 - \sqrt{n}A_2 \right)}{\sqrt[3]{n}} $$ and beyond?