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
$$
\textrm{Var} \left( Y_n \right) = (n!)^2 2^{-n} \xrightarrow{n\rightarrow\infty} (2 \pi n) \left( \frac{n^2}{e^2 2} \right)^n = \mathcal{O} \left( n^{2n+1} \right)
$$
Best Answer
$\newcommand\ep\varepsilon$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, 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.