Does the following necessarily converge to a normal random variable in distribution? Suppose $X_1, \dots, X_n$ are i.i.d. random variables with $E(X_1) = 0$ and $\operatorname{Var}(X_1) = 1$. Let
$$
S_n = \frac{1}{n}\sum_{i=1}^n\sqrt{i}X_i.
$$
Does $S_n$ converge to a normal random variable?
Originally, I attempted to use Lindeberg CLT to prove this. However, I ran into a wall because I can't figure out a way to check the Lindeberg condition that for $\forall \varepsilon > 0$
$$
\sum_{i=1}^nE(|Y_i|^2\mathbf{1}(|Y_i| > \varepsilon)) \to 0
$$
where $Y_i = \sqrt{i}X_i/s_n$ and $s_n^2 = \sum_{i=1}^ni\operatorname{Var}(X_i) = \frac{n(n+1)}{2}$. If I can prove this, then I can use Slutsky, then we are done. But I have no idea what $X_i$ actually is so I don't know how to verify the condition.
Then I tried using characteristic functions and try to do expansion and approximation. However, I also hit a wall due to the changing index of $i$.
I also tried finding counterexample, but nothing came up.
Can anyone provide some hint? Thank you!
 A: For $c_n=\sqrt n$, observe that $\frac{\max_{1\le k\le n}c_k^2}{\sum_{k=1}^n c_k^2}=\frac{n}{n(n+1)/2}=\frac{2}{n+1}\to 0$ as $n\to \infty%$.
As $X_i$'s are i.i.d, by Hajek-Sidak's CLT,
$$\frac{\sum_{k=1}^n c_k X_k}{\sqrt{\sum_{k=1}^n c_k^2}}=\sqrt{\frac{2}{n(n+1)}}\sum_{k=1}^n \sqrt k X_k \stackrel{d}\longrightarrow N(0,1)$$
That is, $$\sqrt{\frac{2n}{(n+1)}}S_n\stackrel{d}\longrightarrow N(0,1)$$
Hajek-Sidak's CLT can be shown using Lyapounov's condition (which implies Lindeberg's condition) under the additional assumption $E|X_1|^3<\infty$. But I am not aware of the general proof.
A: Following up on @StubbornAtom's answer, here is a proof of the Hájek - Šidák CLT.
Claim: Let $(Y_n)$ be IID with zero mean and unit variance. For each $n$ let $c_{n1},\ldots,c_{nn}$ be constants. If
$$\frac{\max_{1\le k\le n}c_{nk}^2}{\sum_{k=1}^n c_{nk}^2}\to 0\qquad\text{as $n\to\infty$,}$$ then $$\frac{\sum_{k=1}^n c_{nk}Y_k}{\left(\sum_{k=1}^n c_{nk}^2\right)^{1/2}}\stackrel d\to N(0,1).$$
Proof: WLOG we may take $\sum_{k=1}^n c_{nk}^2=1$. Put $X_{nk}:=c_{nk}Y_k$. Then each $X_{nk}$ has mean zero. Put $s_n^2:=\operatorname{Var}(\sum_{k=1}^n X_{nk})=\sum_{k=1}^n c_{nk}^2=1$.
Let $\epsilon>0$. For each $n$ and $k$ we have
$$
\begin{aligned}
E(X_{nk}^2 ; |X_{nk}|>\epsilon)&=E(c_{nk}^2 Y_k^2 ; |c_{nk}Y_k|>\epsilon)\\
&=c_{nk}^2 E( Y^2 ; c_{nk}^2Y^2>\epsilon^2)\\
&\le c_{nk}^2 E( Y^2 ; m_nY^2>\epsilon^2),\\
\end{aligned}
$$ where for the sake of brevity we write $m_n:=\max_{1\le k\le n}c_{nk}^2$. Summing over $k$ gives for each $n$
$$\frac1{s_n^2}\sum_{k=1}^nE(X_{nk}^2 ; |X_{nk}|>\epsilon)\le\sum_{k=1}^nc_{nk}^2 E( Y^2 ; m_nY^2>\epsilon^2)= E( Y^2 ; m_n Y^2>\epsilon^2).
$$
Since $m_n\to0$ as $n\to\infty$, this last expectation tends to zero by dominated convergence. This verifies the Lindeberg condition, hence $s_n^{-1}\sum_{k=1}^n X_{nk}$ converges in distribution to a standard normal, and the claim follows.
A: As per answer by @StubbornAtom, I found the original proof of the Hajek-Sidak's CLT which is by deducing the Lindeberg condition. It turns out one can also use characteristic function to prove the claim which is almost in the same spirit as proving the Lindeberg CLT. I will provide the proof for both approaches below:
Proof using Lindeberg CLT:
Let $s_n^2 = \sum_{i=1}^niVar(X_i) = n(n+1)/2$, we will prove this by deducing the Lindeberg condtion. Let $\epsilon > 0$, we want to show that
$$
\frac{1}{s_n^2}\sum_{i=1}^n\int_{|x| > \epsilon s_n}x^2dP(\sqrt{i}X_i \leq x) \to 0.
$$
Now for each $i$,
$$
\int_{|x| > \epsilon s_n}x^2dP(\sqrt{i}X_i \leq x) = i\int_{|\sqrt{i}y|>\epsilon s_n}y^2dP(X_i \leq y) \leq i\int_{|y| > \epsilon v_n}y^2dP(X_i \leq y),
$$
where
$$
v_n^2 = \frac{n+1}{2}.
$$
Then
$$
\frac{1}{s_n^2}\sum_{i=1}^n\int_{|x| > \epsilon s_n}x^2dP(\sqrt{i}X_i \leq x) \leq \int_{|y| > \epsilon v_n}y^2dP(X_i \leq y).
$$
Since $E(X_i^2) < \infty$, we have that the RHS of above goes to zero as $n \to \infty$ by DCT. Therefore, the Lindeberg condition holds. Hence, after some simple calculation and application of Slutsky's theorem, we will have that
$$
S_n \to N(0, 1/2).
$$
Proof using characteristic functions:
Let $\phi(t)$ be the c.f. of $X_1$, and $z_{k,n}(t) = \phi(\frac{\sqrt{k}t}{n})$. Then the c.f. of $S_n$ is then
$$
\prod_{k=1}^nz_{k,n}(t).
$$
Moreover, let $w_{k,n}(t) = 1 - \frac{kt^2}{2n^2}$. Note that for any $t \in \mathbb{R}$,
$$
\sup_{k}|w_{k,n}(t)| \leq 1
$$
for sufficiently large $n$. Now since $E(X_1^2) = 1 < \infty$, we can consider the standard inequality regarding taylor expansion of complex exponential, and we have
$$
|z_{k,n}(t) - w_{k,n}(t)| \leq \frac{kt^2}{n^2}E\left[\min\left(\frac{\sqrt{k}|t|}{n}|X_1|^3, X_1^2\right)\right].
$$
Now since $|z_{k,n}(t)| \leq 1$ and $\sup_{k}|w_{k,n}(t)| \leq 1$, by a standard inequality of complex numbers, we have
\begin{align*}
\bigg|\prod_{k=1}^nz_{k,n}(t) - \prod_{k=1}^nw_{k,n}(t)\bigg| &\leq \sum_{k=1}^n|z_{k,n}(t) - w_{k,n}(t)| \\
&\leq \sum_{k=1}^n\frac{kt^2}{n^2}E\left[\min\left(\frac{\sqrt{k}|t|}{n}|X_1|^3, X_1^2\right)\right] \\
&\leq \frac{t^2(n+1)}{2n}E\left[\min\left(\frac{|t|}{\sqrt{n}}|X_1|^3, X_1^2\right)\right].
\end{align*}
Now the last term of the above converge to $0$ by DCT. Hence, we have that
$$
\lim_{n\to \infty}\prod_{k=1}^nz_{k,n}(t) = \lim_{n\to \infty}\prod_{k=1}^nw_{k,n}(t).
$$
Now
\begin{align*}
\lim_{n\to \infty}\prod_{k=1}^nw_{k,n}(t) &= \lim_{n\to \infty}\prod_{k=1}^n\left(1 - \frac{kt^2}{2n^2}\right)\\
 &= \lim_{n\to \infty}\exp\left(\sum_{k=1}^n\log\left(1 - \frac{kt^2}{2n^2}\right)\right) \\
&= \exp\left(\lim_{n\to \infty}\sum_{k=1}^n\log\left(1 - \frac{kt^2}{2n^2}\right)\right)\\
&= \exp\left(\lim_{n\to\infty}\sum_{k=1}^n- \frac{kt^2}{2n^2} + o\left(\frac{kt^2}{2n^2}\right)\right) \\
&= \exp\left(-\frac{t^2}{4}\right).
\end{align*}
The above is the c.f. of $N(0,1/2)$. Therefore,
$$
S_n \to N(0,1/2).
$$
