How to understand lindeberg CLT, compare with classical CLT For Lindeberg CLT.
Let$\{X_{nj},j=1,2,...k_n\}$ be independent r.v.s with $0<\sigma^2_n=Var(\sum_{j=1}^{k_n}X_{nj})<\infty$, $n=1,2,...$ and $k_n\rightarrow\infty$ as $n\rightarrow\infty$. If
$$\sum_{j=1}^{k_n}E[(X_{nj}-EX_{nj})^2I_{\{|X_{nj}-EX_{nj}|>\epsilon\sigma_n\}}]=o(\sigma^2_n)$$ for any $\epsilon>0$. (Lindeberg condition)
Then,$$\frac{1}{\sigma_n}\sum_{j=1}^{k_n}(X_{nj}-EX_{nj})\rightarrow_dN(0,1)$$
How to understand this Lindeberg condition?
If we consider iid case, with $E(X_{n1})=\mu, Var(X_{n1})=\sigma^2$, the lindeberg condition becomes:
$$\frac{\sum_{j=1}^{k_n}E[(X_{nj}-\mu)^2I_{\{|\frac{X_{nj}-\mu}{\sigma}|>\epsilon\sqrt{k_n}\}}]}{k_n\sigma^2}\rightarrow 0$$ as $k_n\rightarrow\infty$, so
$$\frac{1}{k_n}\sum_{j=1}^{k_n}E[(\frac{X_{nj}-\mu}{\sigma})^2I_{\{(\frac{X_{nj}-\mu}{\sigma})^2>\epsilon^2k_n\}}\rightarrow 0$$ as $k_n\rightarrow\infty$, so
Let $Z_{nj}=\frac{X_{nj}-\mu}{\sigma}$
$$\frac{1}{k_n}\sum_{j=1}^{k_n}E[(Z_{nj}^2I_{\{Z_{nj}^2>\epsilon^2k_n\}}]\rightarrow 0$$ as $k_n\rightarrow\infty$, so
$$E[Z_{nj}^2I_{\{Z_{nj}^2>\epsilon^2k_n\}}]\rightarrow 0$$ as $k_n\rightarrow\infty$.(is this true?)
Since we have $E(X_{nj})=\mu, Var(X_{nj})=\sigma^2$, $E(Z_{nj})=0, Var(Z_{nj})=1$. So $E(Z_{nj})^2=E(Z_{nj})^2+Var(Z_{nj})=1<\infty$. (Is this correct?).
From $E(Z_{nj}^2)<\infty$.
We can already derive $$E[Z_{nj}^2I_{\{Z_{nj}^2>\epsilon^2k_n\}}]\rightarrow 0$$ as $k_n\rightarrow\infty$, by dominated convergence theorem see prove For non-negative r.x. $X$, $E(X)<\infty$ iff $E(XI_{X>x})\rightarrow 0$ as $x\rightarrow\infty$, (is this true?).
Then how this Lindeberg condition be helpful? Are there any mistakes above? Or does it means that for the iid case, Lindeberg's condition always holds?
From Lindeberg CLT's result
$$\frac{1}{\sigma_n}\sum_{j=1}^{k_n}(X_{nj}-EX_{nj})\rightarrow_dN(0,1)$$
For iid case, this result becomes
$$\frac{\sum_{j=1}^{k_n}(X_{nj}-EX_{nj})}{\sqrt{k_nVar(X_{n1})}}\rightarrow_dN(0,1)$$
$$\frac{1}{k_n}\sum_{j=1}^{k_n}(X_{nj}-EX_{n1})\rightarrow_dN(0,Var(X_{n1})),$$
which is the same as classical CLT's result.
For $\{X_{nj}, j=1,2,...,k_n;n=1,2,3,...\}$ is triangle arrow, since $k_n\le n$, and $k_n\rightarrow\infty$ as $n\rightarrow\infty$.  How this "triangle" condition is used in "iid" case?, since this result means that for every r.v. line $\{X_{nj},j=1,2,...k_n\}$, CLT holds
For the classical CLT, we have $$\sqrt{n}(\bar{X}-EX)\to_d N(0,Var(X_1)),$$
Here $\sqrt{n}$ plays a role of converges speed (rate) (Is my understanding correct?) whice is so called "$\sqrt{n}_consistency$" in statistics
If so, what's the converges speed in $$\frac{1}{\sigma_n}\sum_{j=1}^{k_n}(X_{nj}-EX_{nj})\rightarrow_dN(0,1)$$?
Is the converging speed $\frac{1}{\sigma_n}$ or $\frac{1}{\sqrt{k_n}}$?
Can a convergence rate depend on $Var(X_i)$? (I personally don't think so).
If not,  what's the converge rate?
 A: It is true that if for each $n$, the random variables $X_{n,1},\dots,X_{n,k_n}$ have the same distribution, then
$$
\mathbb E\left[Z_{nj}^2I_{\{Z_{nj}^2>\epsilon^2k_n\}}\right]
=\mathbb E\left[Z_{n1}^2I_{\{Z_{n1}^2>\epsilon^2k_n\}}\right].
$$
Moreover, $\sigma_n^2=k_n\operatorname{Var}\left(X_{n,1}\right)$ hence
$$
\frac{1}{\sigma_n^2}\sum_{j=1}^{k_n}\mathbb E\left[(X_{nj}-EX_{nj})^2I_{\{|X_{nj}-EX_{nj}|>\epsilon\sigma_n\}}\right]=
\frac 1{k_n\operatorname{Var}\left(X_{n,1}\right)}k_n\mathbb E\left[Z_{n1}^2I_{\{Z_{n1}^2>\epsilon^2k_n\}}\right]=\frac 1{ \operatorname{Var}\left(X_{n,1}\right)}\mathbb E\left[Z_{n1}^2I_{\{Z_{n1}^2>\epsilon^2k_n\}}\right].
$$
Assuming moreover that the distribution of $X_{n,1}$ is independent of $n$, we indeed find out that Lindeberg's condition holds.
However, in the general case, it will be only true that $\mathbb E\left[Z_{nj}^2I_{\{Z_{nj}^2>\epsilon^2k_n\}}\right]\to 0$ as $n$ goes to infinity, but only for a fixed $j$. This can be problematic when we take the sum over $j$, as this thread shows.
