TITLE: Showing that Lindeberg condition does not hold
QUESTION [5 upvotes]: Let $X_1, X_2, \dots$ be independent random variables and $$X_n = Y_n + Z_n$$ where
$Y_n$ takes values $1$ and $-1$ with chance $1/2$ each, and $$P(Z_n = \pm n) = 1/(2n^2) = (1 - P(Z_n = 0))/2$$ and $S_n:=X_1+ \dots X_n$. Show that Lindeberg condition does not hold, yet $$S_n/\sqrt{n} \rightarrow N(0,1).$$

REPLY [2 votes]: It seems that the goal of the example is to give an example where the classical central limit theorem holds  but the Lindeberg condition is not satisfied. Hence the later is sufficient for the central limit theorem but not necessary. 
I will assume that the collection of random variables $\left\{Y_i,Z_j,i,j\geqslant 1\right\}$ is independent. 
By the Borel-Cantelli lemma, there exists $\Omega'$ of probability one such that for all $\omega\in \Omega'$, there exists an integer $N(\omega)$ for which $X_n(\omega)=0$ for all $n\geqslant N(\omega)$ hence $n^{-1/2}\sum_{i=1}^nZ_i\to 0$ almost surely. From the classical central limit theorem applied to the i.i.d. sequence $\left(Y_i\right)_{i\geqslant 1}$ and Slutsky's theorem, it follows that $n^{-1/2}\sum_{i=1}^nX_i\to N(0,1)$.
Observe that $\left(X_i\right)_{i\geqslant 1}$ is independent and that $\mathbb E\left[X_i^2\right]= \mathbb E\left[Y_i^2\right]+\mathbb E\left[Z_i^2\right]=2$
hence Lindeberg's condition reads 
$$
\forall \varepsilon\gt 0,\frac 1n\sum_{i=1}^n
\mathbb E\left[X_i^2\mathbb 1\left\{\left\lvert X_i\right\rvert \gt \varepsilon \sqrt n\right\}\right]\to 0.
$$
Observe that for all $i$, 
$$X_i^2\mathbb 1\left\{\left\lvert X_i\right\rvert \gt \varepsilon \sqrt n\right\}\geqslant \left(i+1\right)^2\left[i+1>\varepsilon\sqrt n\right] \mathbf 1\{Y_i=1\}\mathbf 1\{Z_i=i\}$$
where $[P]$ equal $i$ if assertion $P$ holds and $0$ otherwise. Therefore, 
$$\mathbb E\left[X_i^2\mathbb 1\left\{\left\lvert X_i\right\rvert \gt \varepsilon \sqrt n\right\}\right]\geqslant \left(i+1\right)^2\left[i+1>\varepsilon\sqrt n\right] \Pr\{Y_i=1\}\Pr 1\{Z_i=i\}\geqslant \frac 14\left[i+1>\varepsilon\sqrt n\right].$$
It follows that 
$$
\frac 1n\sum_{i=1}^n
\mathbb E\left[X_i^2\mathbb 1\left\{\left\lvert X_i\right\rvert \gt \varepsilon \sqrt n\right\}\right]\geqslant \frac 1n\sum_{\varepsilon\sqrt n\leqslant i\leqslant n}\frac 14=\frac{n-\varepsilon\sqrt n}{4n}.
$$