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!

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.
