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.
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).