Imagine to have a set of random variables $\{ X\_i \}\_{i=1}^{n}$ independent (**Non** identically distributed). In these scenario, if the Lindeberg's condition hold we can extend the result of the CLT, *i.e.*, calling $\mu\_i = \mathbb{E}(X\_i)$ , $s\_n^2 = \sum\_i \left(\mathbb{E}(X\_i^2) - \mathbb{E}(X\_i)^2\right)$ :
$$
\frac{\sum\_{i=1}^n X\_i-\mu\_i}{s\_n} \xrightarrow{n\rightarrow\infty} \mathcal{N}(0,1)
$$
(for more details see for example <http://shannon.cm.nctu.edu.tw/prob/c27s08.pdf> or <https://drive.google.com/file/d/1beeTLVbuXEQabqUHReI3K0eJiZ7xfek3/view>).


Now even if the set of random variable is not identically distributed, assume that the distribution of $X\_i$ respect the CLT hypothesis $\forall i$ (**Hp 1**).


Is this a sufficient hypothesis for the Lindeberg's condition to hold and get the CLT for the set $\{ X\_i \}\_{i=1}^{n}$? If yes how could I show it formally, *i.e.* that (**Hp 1**) implies the Lindeberg's condition?


**Follow up question**


Thanks to *Iosif Pinelis* for the nice and clear answer.


Since Lindeberg's condition intuitively guarantees that fluctuations of the random variables from the mean stays "small", I was wondering if assuming finite variances that doesn't grow with $n \rightarrow \infty$, *i.e.* $\textrm{Var} \left( X\_n \right) = \mathcal{O} \left( 1 \right)$ (**Hp 2**) would be a sufficient condition to satisfy the Lindeberg condition.


In the example that you proposed (**Hp 2**) is not satisfied since
$$
\textrm{Var} \left( Y\_n \right) = (n!)^2 2^{-n} \xrightarrow{n\rightarrow\infty} (2 \pi n) \left( \frac{n^2}{e^2 2} \right)^n = \mathcal{O} \left( n^{2n+1} \right)
$$



Hypothesis to guarantee Lindeberg's condition
$\newcommand\ep\varepsilon$No, of course the Lindeberg condition is not necessary for the CLT.


E.g., for each natural $i$ let
\begin{equation\*}
 X\_i:=Z\_i+Y\_i,
\end{equation\*}
where $Z\_i\sim N(0,1)$, $P(Y\_i=i!)=2^{-i-1}=P(Y\_i=-i!)$, $P(Y\_i=0)=1-2^{-i}$, and the random variables (r.v.'s) $Z\_1,Y\_1,Z\_2,Y\_2,\dots$ are independent.


Then
\begin{equation\*}
 \frac1{\sqrt n}\,\sum\_{i=1}^n X\_i=\frac{S\_n}{\sqrt n}+\frac{T\_n}{\sqrt n},
\end{equation\*}
where $S\_n:=\sum\_{i=1}^n Z\_i$ and $T\_n:=\sum\_{i=1}^n Y\_i$.
For any real $\ep>0$, any natural $k$, and any natural $n\ge k$,
\begin{equation\*}
 P\Big(\Big|\frac{T\_n}{\sqrt n}\Big|>2\ep\Big)\le P\_{1n}+P\_{2n},
\end{equation\*}
where
\begin{equation\*}
P\_{1n}:=P\Big(\Big|\frac{T\_k}{\sqrt n}\Big|>\ep\Big),\quad 
P\_{2n}:=P\Big(\Big|\frac{T\_n-T\_k}{\sqrt n}\Big|>\ep\Big). 
\end{equation\*}
Clearly, $P\_{1n}\to0$ for each $k$ as $n\to\infty$. Also,
\begin{equation\*}
 P\_{2n}\le\sum\_{i>k}P(Y\_i\ne0)=\sum\_{i>k}2^{-i}=2^{-k}\to0
\end{equation\*}
as $k\to\infty$. So, $P\big(\big|\frac{T\_n}{\sqrt n}\big|>2\ep\big)\to0$ as $n\to\infty$ for each real $\ep>0$. That is, $\frac{T\_n}{\sqrt n}\to0$ in probability and hence in distribution. Also, $\frac{S\_n}{\sqrt n}\sim N(0,1)$. Thus, the CLT for the $X\_i$'s holds.


On the other hand,
\begin{equation\*}
 s\_n^2=\sum\_{i=1}^n EX\_i^2=\sum\_{i=1}^n EZ\_i^2+\sum\_{i=1}^n EY\_i^2
 =n+\sum\_{i=1}^n (i!)^2 2^{-i}\sim (n!)^2 2^{-n}\to\infty \tag{1}\label{1}
\end{equation\*}
(as $n\to\infty$), whereas for each real $\ep>0$ and all large enough $n$ (such that $n!\ge2\ep s\_n$) we have
\begin{equation\*}
\begin{aligned}
 s\_{n,\ep}^2&:=\sum\_{i=1}^n EX\_i^2\,1(|X\_i|\ge\ep s\_n) \\ 
 &\ge EX\_n^2\,1(|X\_n|\ge\ep s\_n) \\ 
 & \ge E(Z\_n+Y\_n)^2\,1(|Y\_n|\ge2\ep s\_n,|Z\_n|\le\ep s\_n) \\ 
 & = E(Z\_n^2+Y\_n^2)\,1(|Y\_n|\ge2\ep s\_n)\,1(|Z\_n|\le\ep s\_n) \\ 
 & \ge EY\_n^2\,1(|Y\_n|\ge2\ep s\_n)\,1(|Z\_n|\le\ep s\_n) \\ 
 & = EY\_n^2\,1(|Y\_n|\ge2\ep s\_n)\,P(|Z\_n|\le\ep s\_n) \\ 
 & = (n!)^2 2^{-n}\,P(|Z\_n|\le\ep s\_n) \\ 
 & = (n!)^2 2^{-n}\,P(|Z\_1|\le\ep s\_n) \\ 
 & \sim (n!)^2 2^{-n}\sim s\_n^2, 
\end{aligned}
\end{equation\*}
by \eqref{1}. Thus, the Lindeberg condition $s\_{n,\ep}^2/s\_n^2\to0$ fails to hold. $\quad\Box$




---


Then OP additionally asked if there is a counterexample with $s\_n^2$ bounded. Here is a simple one: Let $X\_1\sim N(0,1)$ and $X\_2=\cdots=X\_n=0$. Then $\sum\_{i=1}^n X\_i\sim N(0,1)$, whereas $s\_n^2=1$
\begin{equation\*}
 s\_{n,\ep}^2=EX\_1^2\,1(|X\_1|\ge\ep s\_n)\not\to0
\end{equation\*}
for any real $\ep>0$. So, the Lindeberg condition $s\_{n,\ep}^2/s\_n^2\to0$ fails to hold. $\quad\Box$




---


Now, the positive news. Suppose that we have a triangular array $(X\_{nk}\colon n=1,2,\dots,\ k=1,\dots,k\_n)$ of independent zero-mean r.v.'s satisfying the normalization condition
\begin{equation}
 s\_n^2:=\sum\_k EX\_{nk}^2\to1, \tag{$\*$}\label{0}
\end{equation}
the infinite smallness condition
\begin{equation\*}
 \max\_{1\le k\le k\_n}P(|X\_{nk}|\ge\ep)\to0
\end{equation\*}
for each real $\ep>0$, and the CLT condition
\begin{equation\*}
 S\_n:=\sum\_k X\_{nk}\to Z \tag{2}\label{2}
\end{equation\*}
in distribution, where $Z\sim N(0,1)$.


Then, by Theorem 15 of Chapter IV by [Petrov](https://books.google.com/books?id=zSDqCAAAQBAJ&printsec=frontcover#v=onepage&q&f=false), for every real $\ep>0$
\begin{equation\*}
 \sum\_k P(|X\_{nk}|\ge\ep)\to0 \tag{3}\label{3}
\end{equation\*}
and
\begin{equation\*}
 \sum\_k(EY\_{nk}^2-E^2Y\_{nk})\to1, \tag{4}\label{4}
\end{equation\*}
where $Y\_{nk}:=X\_{nk}\,1(|X\_{nk}|<\ep)$.


Since the $X\_{nk}$'s are zero-mean, for any real $\ep>0$ and any real $M>\ep$ we have
\begin{equation\*}
 -EY\_{nk}=EX\_{nk}\,1(|X\_{nk}|\ge\ep)=EX\_{nk}\,1(|X\_{nk}|\in[\ep,M))+EX\_{nk}\,1(|X\_{nk}|\ge M),
\end{equation\*}
\begin{equation\*}
\sum\_k|EX\_{nk}\,1(|X\_{nk}|\in[\ep,M))|\le \sum\_k MP(|X\_{nk}|\ge\ep)\to0
\end{equation\*}
by \eqref{3},
\begin{equation\*}
\sum\_k|EX\_{nk}\,1(|X\_{nk}|\ge M)|\le\frac1M\,\sum\_k EX\_{nk}^2\to0
\end{equation\*}
by \eqref{0} as $M\to\infty$. So,
\begin{equation\*}
 \sum\_k|EY\_{nk}|\to0. 
\end{equation\*}
Also, $|Y\_{nk}|\le\ep$ and hence $E^2Y\_{nk}\le\ep|EY\_{nk}|$. So,
\begin{equation\*}
 \sum\_k E^2Y\_{nk}\to0. 
\end{equation\*}
So, by \eqref{4},
\begin{equation\*}
 \sum\_k EY\_{nk}^2\to1. \tag{5}\label{5}
\end{equation\*}


So, for
\begin{equation\*}
 s\_{n,\ep}^2:=\sum\_k EX\_{nk}^2\,1(|X\_{nk}|\ge\ep)
\end{equation\*}
we get
\begin{equation\*}
 \frac{s\_{n,\ep}^2}{s\_n^2}\sim s\_{n,\ep}^2=\sum\_k EX\_{nk}^2-\sum\_k EX\_{nk}^2\,1(|X\_{nk}|<\ep)
 =\sum\_k EX\_{nk}^2-\sum\_k EY\_{nk}^2\to1-1=0,
\end{equation\*}
so that the Lindeberg condition $s\_{n,\ep}^2/s\_n^2\to0$ holds.


