Prove $S_n/s_n\rightarrow N(0,1)$ in distribution Given $B_1, B_2,\ldots$ are independent and bounded variables with $E(B_i) = 0$ for all $i=1,2,\ldots$. Define $S_n = B_1+ B_2+\ldots + B_n$ with variance $s_n^2\rightarrow \infty$. Prove that $\frac{S_n}{s_n}$ has a central limit.
My attempt: Due to the given condition, without i.i.d property, I try to prove that this sequence satisfies the Lindeberg condition and then applying Lindeberg-Feller theorem, we're done. So for every $\epsilon>0$ , I need to show for each positive $\varepsilon$, 
$$\lim_{n\to +\infty}   \frac 1{\sum_{j=1}^n\sigma_j^2 }\sum_{j=1}^n\mathbb E\left[B_j^2\mathbf 1\left\{ \left|B_j\right|^2\gt \varepsilon\sum_{i=1}^n   \sigma_i^2\right\}\right]=0 .$$
Since $\sigma_i^{2} = E(B_i^2)$ for all $\ i=1,2,\ldots$ and $B_is$ are bounded variables, $B_j^2\leq M=max(|B_1|,|B_2|,\ldots)$. Thus, $$\lim_{n\to +\infty}   \frac 1{\sum_{j=1}^n\sigma_j^2 }\sum_{j=1}^n\mathbb E\left[B_j^2\mathbf 1\left\{ \left|B_j\right|^2\gt \varepsilon\sum_{i=1}^n   \sigma_i^2\right\}\right]\leq \lim_{n\to +\infty} \frac 1{\sum_{j=1}^n\sigma_j^2 } M^2\sum_{j=1}^n \mathbb E\left[\mathbf 1\left\{ \left|B_j\right|^2\gt \varepsilon\sum_{i=1}^n   \sigma_i^2\right\}\right] .$$ As $n\rightarrow \infty$, $s_n^2\rightarrow \infty$, so all the sets $\mathbf 1\left\{ \left|B_j\right|^2\gt \varepsilon\sum_{i=1}^n \sigma_i^2\right\}\rightarrow 0$. And we would be done if we could show that $\lim_{n\rightarrow \infty} \sum_{j=1}^n \mathbb E\left[\mathbf 1\left\{ \left|B_j\right|^2\gt \varepsilon\sum_{i=1}^n   \sigma_i^2\right\}\right]\rightarrow 0$. But this might not be true (counter example is the harmonic series with $p=1$) unless there is something that I was missing.
My question: Could someone please help me overcome this last step? In case I was on the wrong track, please let me know as well.
 A: You are on the right track. Notice that (given $\varepsilon>0$) there exists $N$ such that 
$$
I\left( |B_j|^2>\varepsilon s_n^2\right) = 0
$$
for all $j$ and all $n\ge N$. (The reason is that $s_n$ is a deterministic sequence tending to infinity, while the $B$'s are bounded, so eventually the inequality is not satisfied.) This means the sum 
$$
\sum_{j=1}^n \mathbb E\left[\mathbf 1\left( \left|B_j\right|^2\gt \varepsilon s_n^2\right)\right]
$$
stops at $j=N$, so it's bounded by $N$.
A: Since each $X_k$ is bounded then for any $\epsilon>0$ there exists $N\in\mathbb N$ such that
$$\mathbf{1}\left(|X_k|>\epsilon s_n\right)=0$$
for all $k$ and all $n\geq N$. Hence
$$\lim_{n\rightarrow\infty}\sum_{k=1}^n\mathbb E\left[\mathbf{1}\left(|X_k|>\epsilon s_n\right)\right]=0$$
Thus taking $M$ as the bound for each $|X_k|$ we have
\begin{align*}
\lim_{n\rightarrow\infty}\frac{1}{s_n^2}\sum_{k=1}^n\int_{|X_k|>\epsilon s_n}X_k^2dP
&=\lim_{n\rightarrow\infty}\frac{1}{s_n^2}\sum_{k=1}^n\mathbb E\left[X_k^2\mathbf{1}\left(|X_k|>\epsilon s_n\right)\right]\\\\
&\leq\lim_{n\rightarrow\infty}\underbrace{\frac{1}{s_n^2}}_{\rightarrow0}M^2\underbrace{\sum_{k=1}^n\mathbb E\left[\mathbf{1}\left(|X_k|>\epsilon s_n\right)\right]}_{\rightarrow0}\\\\
&\rightarrow 0
\end{align*}
Therefore the Lindeberg condition is met and we have that $\frac{S_n}{s_n}$ has a central limit.
