# How can I show $\lim_{n\to\infty}\prod_{j}^n\frac{\sin(jn^{-3/2}u)}{jn^{-3/2}u}=e^{-u^2/18}$?

This is from and exercise of probability theory:

Let $(X_j)_{j\geq 1}$ be independent and let $X_j$ have the uniform distribution on $(-j,j)$. Show that $$\lim_{n\to\infty}\frac{S_n}{n^{3/2}}=Z\sim N(0,\frac{1}{9})$$ in distribution.

In terms of characteristic functions, it suffices to show that $$\lim_{n\to\infty}\prod_{j=1}^n\frac{\sin(jn^{-3/2}u)}{jn^{-3/2}u}=e^{-u^2/18}.$$ Take $\log$ on both side one gets: $$\lim_{n\to\infty}\sum_{j=1}^n\log\frac{\sin(jn^{-3/2}u)}{jn^{-3/2}u}=-\frac{u^2}{18}.$$ How can this limit be proved?

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I think it should be an n^{-3/2} instead of n^{3/2} –  blabler Nov 27 '12 at 2:55
@blabler: Corrected. Thanks! –  Jack Nov 27 '12 at 12:41

For each $u \in \Bbb{R}$, we have $n^{-1/2}|u| < 1$ for every sufficiently large $n$. Then for any $1 \leq j \leq n$ we have $jn^{-3/2}|u| < 1$. Now, from the Taylor expansion
$$\log\left(\frac{\sin x}{x}\right) = -\frac{x^2}{6} + O\left(x^4\right),$$
Taking $n \to \infty$ we obtain the desired result.