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!

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.
