Let $Y_1, \ldots , Y_n$ be independent with $Y_k \sim U(0,1).$ If $S_n=\Sigma_k kY_k$, show that $4S_n/n^2$ converges in distribution to $1.$ a question on Convergence in distribution:

Let $Y_1, \ldots , Y_n$ be independent random variables with $Y_k \sim U(0,1).$
If $X_k=kY_k$ and $S_n=X_1 + \cdots + X_n$, show that
$$\frac{S_n}{\frac{n^2}{4}}$$ converges in distribution to $1.$.

I have tried to use the Borel-Cantelli Lemma to test the convergence a.c. but I have not been able to estimate anything, another attempt I made was to calculate directly the distribution function but the calculations do not lead me to anything.
Any advice on how to solve this problem will be greatly appreciated.
 A: Convergence in distribution to a constant is equivalent with convergence in probability (to the same constant), which might be easier to establish. Indeed, note that by linearity of expectation $$\mathbb E[S_n] = \sum_{k=1}^n k \mathbb E[Y_k] = \frac{1}{2}\sum_{k=1}^n k = \frac{n(n+1)}{4}$$ Moreover, since variables $Y_k$ are independent, we get $$ Var(S_n) = \sum_{k=1}^n Var(k Y_k) = \sum_{k=1}^n k^2 Var(Y_k) = \frac{1}{12}\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{72}$$
Hence by Chebyshev's inequality, for any $\varepsilon > 0$ we have $$ \mathbb P\left ( \left | \frac{S_n}{\mathbb E[S_n]} - 1 \right| > \varepsilon\right) = \mathbb P( | S_n - \mathbb E[S_n] | > \varepsilon \mathbb E[S_n]) \le \frac{Var(S_n)}{\varepsilon^2 (\mathbb E[S_n])^2} =  \frac{16n(n+1)(2n+1)}{72\varepsilon^2 n^2(n+1)^2}$$
and the last bound on the right converges to $0$ (the numerator is of order $n^3$, while the denominator is of order $n^4$). In other words, $\frac{S_n}{\mathbb E[S_n]} \to 1$ in probability (since $\varepsilon > 0$ was arbitrary). But $\frac{4\mathbb E[S_n]}{ n^2} \to 1$, hence $$\frac{4S_n}{n^2} = \frac{S_n}{\mathbb E[S_n]} \cdot \frac{4\mathbb E[S_n]}{n^2} \to 1 \cdot 1 = 1$$ in probability.
Edit In fact, letting $Z_k \sim \mathcal U((-\frac{1}{2},\frac{1}{2}))$ and by noting that $\mathbb E[ |S_n - \mathbb E[S_n]|^4 ] = \mathbb E[ \big(\sum_{k=1}^n k Z_k\big)^4] \sim Cn^5$ (via expanding), we can get better bound, that is $$ \mathbb P\left ( \left | \frac{S_n}{\mathbb E[S_n]} - 1 \right| > \varepsilon\right) = \mathbb P( | S_n - \mathbb E[S_n] |^4 > \varepsilon^4 \mathbb E[S_n]^4) \le \frac{\mathbb E[|S_n-\mathbb E[S_n]|^4}{\varepsilon^4 \mathbb E[S_n]^4} \sim C \frac{n^5}{\varepsilon^4 n^8} \sim \frac{C}{\varepsilon^4 n^3}$$ Hence for any $\varepsilon > 0$ the series $$\sum_{n=1}^\infty \mathbb P\left ( \left | \frac{S_n}{\mathbb E[S_n]} - 1 \right| > \varepsilon\right) $$ is convergent, meaning that by Borel Cantelli lemma, we have even convergence $\frac{S_n}{\mathbb E[S_n]} \to 1$ almost surely ( so by similar argument as above, $\frac{4S_n}{n^2} \to 1$ almost surely).
