Lindeberg-Feller CLT for weighted sumation. Problem : Let $\{a_n\}_{n\ge 1}$ be an increasing positive sequence of integers $s.t.$ the following condition holds $$\frac{a_n^2}{a_1^2+\cdots+a_n^2} \rightarrow 0.$$
Assume $\{X_n\}_{n \ge 1}$ are $iid$ sequence of random variables with mean $0$ and variance $1$. TO SHOW
$$\frac{a_1X_1 + a_2X_2 + \cdots + a_nX_n}{\sqrt{a_1^2+\cdots+a_n^2}} \rightarrow N(0,1)$$
Direction: Well I feel we need to use the Linderberg's condition for CLT. By the definition of triangular arrows define $X_{n,k} = \frac{a_kX_k}{\sqrt{a_1^2+\cdots+a_n^2}}$, where $k \le n$. The Lindeberg condtion states $\sum_{k=1}^n\mathbf{E}[X_{n,k}^2\mathbf{1}_{X_{n,k} > \delta}] \rightarrow 0$ as $n \rightarrow \infty$ for any arbit $\delta$ implies the result we set out to show.
Thus trying to show the condition holds:
$$\mathbf{E}[X_{n,k}^2\mathbf{1}_{X_{n,k} > \delta}]$$
$$= \frac{a^2_k}{a_1^2+\cdots+a_n^2} \mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_k}}\right] $$
$$\leq \frac{a^2_n}{a_1^2+\cdots+a_n^2}\mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_n}}\right]$$
But now summing over $n$ terms give me
$$\frac{na^2_n}{a_1^2+\cdots+a_n^2}\mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_n}}\right]$$
I can conclude $\mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_n}}\right]$ converges to $0$ because of finite variance and the condtion stated. However there remains a $\frac{na^2_n}{a_1^2+\cdots+a_n^2}$ term which I am pretty sure I cant show it goes to $0$.
Any help is appreciated
Edit : I was wondering if it could be shown that $\frac{na^2_n}{a_1^2+\cdots+a_n^2}$ this could be shown to be a constant as $\frac{na^2_n}{a_1^2+\cdots+a_n^2} \rightarrow 0$ for increasing sequences of $(a_n)$
 A: The bound
$$
\frac{a^2_k}{a_1^2+\cdots+a_n^2} \mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_k}}\right]\leqslant \frac{a^2_n}{a_1^2+\cdots+a_n^2} \mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_n}}\right]
$$
is a bit too crude; instead, use
$$
\frac{a^2_k}{a_1^2+\cdots+a_n^2} \mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_k}}\right]\leqslant \frac{\color{red}{a^2_k}}{a_1^2+\cdots+a_n^2} \mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_n}}\right].
$$
The last term does not dependent of $k$ and the summation over $k$ leaves us only with $ \mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_n}}\right]$. We can then conclude by dominated convergence.
A: Assume $ \{X_n,n\ge 1\} $ are $ iid $ sequence of random variables with $ \mathsf{E}[X_n]=0 $ and $ \mathsf{var}[X_n]=1 $.
Let $ \{a_n, n\ge1\} $ be a sequence of real number and $ s_n^2=\sum_{i=1}^{n}a^2_i $, $ S_n=\sum_{i=1}^{n}a_iX_i $.
1. If
\begin{equation*}
 \lim_{n\to\infty}s_n^2=\lim_{n\to\infty}\sum_{i=1}^{n}a^2_i<+\infty, \tag{1}
\end{equation*}
then
\begin{equation*}
 \lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{i=1}^{n}a_iX_i=S, \quad \text{a.s.}
\end{equation*}
where $ S $ is a nondegenerated random variable, its distribution depends on the $ X_1 $'s. This is a consequence of Kolmogorov theorem(For series of independent r.v. terms with zero mean, the convergence of series of variance implies the convergence a.s.).
2. If
\begin{gather*}
 \lim_{n\to\infty}s_n^2=\lim_{n\to\infty}\sum_{i=1}^{n}a^2_i=+\infty, \tag{2}\\
 \lim_{n\to\infty}\frac{a_n^2}{s_n^2}=0.     \tag{3}
\end{gather*}
Notice that, for $ 1\le k\le n $,
\begin{gather*}
 b_n\triangleq\max_{1\le j\le n}|a_j|\le b_k+\max_{k\le j\le n}\Big(\frac{|a_j|}{s_j}\Big) s_n,\\
 \frac{b_n}{s_n}\le \frac{b_k}{s_n}+\max_{k\le j\le n}\Big(\frac{|a_j|}{s_j}\Big),\\ 
  \limsup_{n\to\infty}\frac{b_n}{s_n}\le \max_{j\ge k}\frac{|a_j|}{s_j},  \\
\limsup_{n\to\infty}\frac{b_n^2}{s_n^2}\le \limsup_{j\to\infty}\frac{a_j^2}{s_j^2}=0.\tag{4}\\
  \mathsf{E}[a_k^2X^2_k1_{|X_k|>\delta s_n/|a_k|}]\le
  a_k^2\mathsf{E}[X^2_k1_{|X_k|>\delta s_n/b_n}]\\
  \sum_{k=1}^{n}\mathsf{E}[a_k^2X^2_k1_{|X_k|>\delta s_n/|a_k|}]
  \le \sum_{k=1}^{n}a_k^2\mathsf{E}[X^2_11_{|X_1|>\delta s_n/b_n}] \tag{5}   
\end{gather*}
hence from (4),
\begin{equation*}
 \lim_{n\to\infty}\frac{\max\limits_{1\le i\le n}a_i^2}{s_n^2}=
   \lim_{n\to\infty}\frac{b_n^2}{s_n^2}=0, \tag{6}
\end{equation*}
and the Lindeberg's condition holds  from (5)(6), i.e.,
\begin{equation*}
 \lim_{n\to\infty}\frac{1}{s_n^2}\sum_{k=1}^{n}\mathsf{E}[a_k^2X^2_k1_{|X_k|>\delta s_n/a_k}]=0,\quad \forall\delta>0. \tag{7}
\end{equation*}
3  If  $ \{a_n, n\ge1\} $ is a sequence of increasing positive real numbers and (3) holds also, then
\begin{equation*}
 s_n^2\ge a_1^2\Big[\frac{a_n^2}{s_n^2}\Big]^{-1}\to+\infty,\quad \text{ as } n\to+\infty, 
\end{equation*}
(6) and Lindeberg's condition (7) holds too.
