so let's assume we have independent random variables $X_1,X_2, X_3, \ldots$ with $$\mathbb{E}[X_k]=0 \mbox{ and } \mathbb{Var}[X_k]=\sigma_k^2=1 \quad \forall k\in\mathbb{N}. $$
We define $$s_n^2:= \sum_{k=1}^n \mathbb{E}[(X_k-\mathbb{E}[X_k])^2]=\sum_{k=1}^n\sigma_k^2=n.$$
Now we check Lindenberg's condition: So for $\varepsilon>0$ we must check $$\lim\limits_{n\rightarrow \infty} \frac{1}{s_n^2}\sum_{k=1}^n \mathbb{E}[(X_k-\mathbb{E}[X_k])^2 \cdot \mathbf{1}_{|X_k| > \varepsilon s_n^2} ] =\lim\limits_{n\rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \mathbb{E}[X_k^2 \cdot \mathbf{1}_{|X_k| > \varepsilon n} ].$$
We know,
$$\mathbb{E}[X_k^2 ]=1,$$
so
$$\mathbb{E}[X_k^2 \cdot \mathbf{1}_{|X_k| > \varepsilon n} ]$$
converges quicker than $$\frac{1}{n},$$
whch means we find an $N\in\mathbb{N}$ such that for every $n\geq N$ we have
$$\mathbb{E}[X_k^2 \cdot \mathbf{1}_{|X_k| > \varepsilon n} ]< \frac{1}{n}, $$
which means for every $\varepsilon>0$ we have
$$\lim\limits_{n\rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \mathbb{E}[X_k^2 \cdot \mathbf{1}_{|X_k| > \varepsilon n} ] \leq \lim\limits_{n\rightarrow \infty} \frac{N}{n} + \frac{1}{n} \sum_{k=N+1}^n \mathbb{E}[X_k^2 \cdot \mathbf{1}_{|X_k| > \varepsilon n} ]< \lim\limits_{n\rightarrow \infty} \frac{N}{n} + + \frac{1}{n^2 }=0,$$
so we know that the Central Limit Theorem holds.
Is that right?
Edit: What we really need is uniform integrability of ${ X_k^2: k\in\mathbb{N}} $$ and that gives us that Lindeberg holds.
Otherwise we can construct $X_j$ with $$P(X_j =-j^2 )=P(X_j =j^2 ) = j^{-4}/2$$ and Lindeberg does not hold!