Using Lindeberg’s Condition together with the Central Limit Theorem I have the following problem:

Problem. Let $ (X_{n})_{n \in \mathbb{N}} $ be a sequence of independent random variables such that
  $$
  \mathbf{Pr} \! \left( X_{n} = \sqrt{n} + 1 \right)
= \mathbf{Pr} \! \left( X_{n} = - \sqrt{n} - 1 \right)
= \frac{1}{2 (\sqrt{n} + 1)^{2}}
$$
  and
  $$
\mathbf{Pr} \! \left( X_{n} = 0 \right) = 1 - \frac{1}{(\sqrt{n} + 1)^{2}}.
$$
  Then prove that the random variable $ \dfrac{\sum_{i = 1}^{n} X_{i}}{\sqrt{n}} $ converges in distribution to one with a standard normal distribution.

I have to use Lindeberg’s Condition together with the Central Limit Theorem in order to solve this, but I’m stuck. Any assistance would be greatly appreciated. Thanks!
 A: We want to check that for a fixed positive $\varepsilon$, the convergence 
$$\lim_{n\to\infty}\frac 1{s_n^2}\sum_{j=1}^n\mathbb E[X_j^2\mathbf 1\{ |X_j|\gt \varepsilon s_n \}] =0     $$
takes place, with $s_n^2=\sum_{j=1}^n \operatorname{Var}(X_j)=n$ in order to apply Lindeberg-Lévy-Feller theorem. 
We thus have to prove that 
$$\tag{*} \lim_{n\to\infty}\frac 1n \sum_{j=1}^n\mathbb E[X_j^2\mathbf 1\{ |X_j|\gt \varepsilon \sqrt n \}]=0.$$
If $j$ is such that $\sqrt j +1\leqslant \varepsilon\sqrt n$, then $\mathbb E[X_j^2\mathbf 1\{ |X_j|\gt \varepsilon \sqrt n \}]=0$ and if $j$ is such that $\sqrt j +1\gt \varepsilon\sqrt n$, then $\mathbb E[X_j^2\mathbf 1\{ |X_j|\gt \varepsilon \sqrt n \}]=1$, hence
$$\frac 1n \sum_{j=1}^n\mathbb E[X_j^2\mathbf 1\{ |X_j|\gt \varepsilon \sqrt n \}]=\frac 1n\sum_{j\gt(\varepsilon\sqrt n-1)^2}^n1, $$
which does not converge to $0$. 
By Theorem 2.1. page 331 of the book Probability: A Graduate Course by Allan Gut, since $\max_{1\leqslant j\leqslant n}\frac{\operatorname{Var}(X_j)}{s_n^2}\to 0$, the sequence $\left(n^{-1/2}\sum_{j=1}^nX_j    \right)_{n\geqslant 1}  $ cannot converge to a standard normal distribution.      
