Prove that $3\sum\limits_{k=1}^n \frac{k X_k}{n^{3/2}}$ converges in distribution, for i.i.d. $X_k$ uniform in $[-1,1]$ 
Let $(X_n)$ denote a sequence of independent random variables, uniformly distributed on $[-1,1]$. We want to prove that $S_n=\frac{3}{n^{3/2}}\sum\limits_{k=1}^n k X_k$ converges in distribution as $n$ tends to infinity.

I was trying to compute $F_{S_n}(x)=P(S_n\le x)=P\left(\sum\limits_{k=1}^n k X_k\le \frac13n^{3/2}x\right)$ (1). 
Denote $Y_k=k X_k$, and it is easy to know that $Y_k$ is uniformly distributed on $[-k,k]$. 
Now, we want to find the distribution of $\sum\limits_{k=1}^n Y_k$, and then find the limit of (1). 
I computed $f_{Y_k+Y_{k+1}}(z)$, which is the convolution of $f_{Y_k}$ and $f_{Y_{k+1}}$, and it equals to $\frac{z+2k+1}{2k(2k+2)}$ when $-2k-1\le z<-1$, $\frac{1}{(2k+2)}$ when $-1\le z<1$, and $\frac{-z+2k+1}{2k(2k+2)}$ when $1\le z\le 2k+1$. 
Then, I thought this way can be tedious, and I am not sure if I can continue this way to solve the problem. 
I am wondering if there is any cleverer way to do this. In general, when I deal with problems about the convergence in distribution of some specific random variables, I usually try two perspectives:


*

*Prove a stronger convergence, such as convergence in probability or in $L^p$.

*Compute the exact expression of corresponding distribution function, and take the limit.
However, I am feeling that I missed some tools to prove convergence in distribution. Is there any other theorem, lemma or method used frequently to prove convergence in distribution?
 A: My own solution:
Consider characteristic function of $\eta_n=\frac{3\sum\limits_{k=1}^n kX_k}{n^{3/2}}$: $\phi_n(t)=E(e^{it\eta_n})$. It suffices to show that $\phi_n$ converges to some characteristic function $\phi$.
Since $X_k$ are independent, $\phi_n(t)=E(e^{it\eta_n})=\prod\limits_{k=1}^n E(e^{it\frac{3kX_k}{n^{3/2}}})$. Note that $E(e^{it\frac{3kX_k}{n^{3/2}}})=\int_{-1}^1 e^{it\frac{3kx}{n^{3/2}}}\frac{1}{2}dx=\frac{\sin(t\frac{3k}{n^{3/2}})}{t\frac{3k}{n^{3/2}}}:=1-y_k(t)$.
Now we want to find $\prod\limits_{k=1}^{\infty}(1-y_k(t))$ where $|y_k|\le1$. $\log\prod\limits_{k=1}^{n}(1-y_k(t))=\sum\limits_{k=1}^n(-y_k(t)-\frac{y_k(t)^2}{2}-\cdots)$.
Note that 
$\sum\limits_{k=1}^ny_k(t)=\sum\limits_{k=1}^n(1-\frac{\sin(t\frac{3k}{n^{3/2}})}{t\frac{3k}{n^{3/2}}})$, by taylor expansion, $\sum\limits_{k=1}^ny_k(t)=\sum\limits_{k=1}^n(\frac{1}{6}(t\frac{3k}{n^{3/2}})^2-\frac{1}{120}(t\frac{3k}{n^{3/2}})^4\cdots)$
The first term becomes $\frac{3}{2}\frac{t^2}{n^3}\sum\limits_{k=1}^nk^2=\frac{1}{2}t^2+O(\frac{1}{n})$. All the other terms are $O(n^s)$ for $s<0$, and same for $\sum\limits_{k=1}^ny_k(t)^2,\cdots$.
Hence, $\log\prod\limits_{k=1}^{\infty}(1-y_k(t))=-\frac{t^2}{2}$, and $\prod\limits_{k=1}^{\infty}(1-y_k(t))=e^{-t^2/2}$, which is the characteristic function of Gaussian random variable with mean $0$ and standard deviation $1$. Then we're done.
