Let $(X_j)_{j \in \mathbb{N}}$ be a sequence of independent random variables such that $X_j \sim U(-j,j)$ (uniformly distributed). Show that $$\lim_{n \to \infty}\frac{S_n}{n^{3/2}} \sim N(0,1/9)$$ i.e. normally distributed, where $S_n := \sum_{j = 1}^n X_j$.
Now the hint was to use characteristic functions. So we have (omitting the unnecessary computations) $$\varphi_{S_n/n^{3/2}}(t) = \prod_{j = 1}^n \frac{\sin(tj/n^{3/2})}{tj/n^{3/2}}$$ How do I calculate now $$\boxed{\lim_{n \to \infty}\varphi_{S_n/n^{3/2}}(t)}$$ ?
I never did infinite products so I am a bit lost. I tried using $e^{\log}$ but somehow it does not lead somewhere.
The key is to use big-Oh notation and expand $\sin$. I'd recommend trying to do that on your own; if you can't, then read this solution and give it another shot by yourself.
We know that $$\sin(tj/n^{3/2}) = (tj/n^{3/2}) - \frac{1}{3!} (tj/n^{3/2})^3 + O\left(\left(\frac{tj}{n^{3/2}}\right)^5\right).$$
So this implies $$\sin(tj/n^{3/2}) = 1- \frac{1}{3!} (tj/n^{3/2})^2 + O\left(\left(\frac{tj}{n^{3/2}}\right)^4\right).$$
Using the fact that $j \leq n$ we can change the error term to $O(t^4/n^2)$, giving: $$\prod_{j = 1}^n \left(1 - \frac{1}{3!}\frac{t^2 j^2}{n^3} + O(t^4/n^2)\right).$$ Taking logs gives $$\sum_{j = 1}^n \log\left(1 - \frac{1}{3!}\frac{t^2 j^2}{n^3} + O(t^4/n^2)\right)$$ but $$\log \left(1 - \frac{1}{3!}\frac{t^2 j^2}{n^3} + O(t^4/n^2)\right) = -\frac{1}{3!}\frac{t^2 j^2}{n^3} + O(t^4/n^2).$$ So this gives \begin{align} \sum_{j = 1}^n \log\left(1 - \frac{1}{3!}\frac{t^2 j^2}{n^3} + O(t^4/n^2)\right) &= \sum_{j=1}^n\left(-\frac{1}{3!}\frac{t^2 j^2}{n^3} + O(t^4/n^2). \right) \\ &= -\frac{1}{3!}\frac{t^2}{n^3}\left(\frac{n(n+1)(2n+1)}{6}\right) + O(t^4/n). \end{align}
Taking $n \to \infty$ this approaches $\frac{-1}{18}t^2$, implying the characteristic function is $e^{-t^2 / 18}$ (since we took logs), thereby completing the proof.