This is the text of the problem: Let $\left(X_{j}\right)_{j\ \geq\ 1}$ be independent and let $X_{j}$ have the uniform distribution on $\left(-j,j\right)$. Show that $\lim_{n \to \infty}{S_{n} \over n^{3/2}}=Z$ In distribution, where $\mathcal{L}\left(Z\right) = N\left(0,\frac{1}{9}\right)$.
We are given the following hint: Show that the characteristic function of $X_{j}$ is $\varphi_{X_{j}}\left(u\right)=\sin\left(uj\right)/\left(uj\right)$; compute $\varphi_{S_{n}}\left(u\right)$, then $\varphi_{S_{n}/n^{3/2}}\left(u\right)$, and prove that the limit is $exp\left(-u^{2}/18\right)$ by using $\sum_{j=1}^{n}j^{2}=n\left(n + 1\right)\left(2n + 1\right)/6$.
Now, my difficulty is the following: I have shown that $\varphi_{S_{n}/n^{3/2}}\left(u\right)=\prod_{j=1}^{n}\frac{n^{3/2}}{uj}\sin\left(\frac{uj}{n^{3/2}}\right)$. My next step was that I took $\ln$ of both sides, then used a Maclaurin expansion for $\sin\left(\frac{uj}{n^{3/2}}\right)$, and simplified to obtain that $\ln \varphi_{S_{n}/n^{3/2}}(u)=\sum_{j=1}^{n}\ln\left(1-\frac{1}{3!}\frac{u^{2}j^{2}}{n^{3}}+o\left(\frac{j^{4}}{n^{6}}\right)\right)$ (my little oh notation might be slightly off). Now, eventually, I want to be left just with a $j^{2}$ term, so that I can use the remaining part of the hint, simplify, and end up with $-\frac{u^{2}}{18}$. Getting there, though is the problem. I seem to be missing something technical. If somebody could fill in the steps between $\ln \varphi_{S_{n}/n^{3/2}}(u)=\sum_{j=1}^{n}\ln\left(1-\frac{1}{3!}\frac{u^{2}j^{2}}{n^{3}}+o\left(\frac{j^{4}}{n^{6}}\right)\right)$ and $-\frac{u^{2}}{18}$ in detail, that would be great. Or if what I did is the wrong approach, tell me what the right approach would be.
Thank you very much in advance!
We can avoid a further use of Taylor expansion using the inequality $$-x-\frac{x^2}2\leqslant \ln(1-x)\leqslant -x,\quad 0\leqslant x\lt 1.$$
The notation "o" can be replaced by an integral remainder, thought it's not a problem.