Calculation of a characteristic function

Question

Suppose $X_1, X_2, \ldots X_n \ldots$ are independent random variables with $$P(X_n = 1) = \frac{1}{2}$$$$P(X_n = -1) = \frac{1}{2}$$ Then $$\sqrt{\frac{3}{n^3}}\sum_{k=1}^n kX_k$$ converges to $N(0,1)$ in distribution.

What I have done: trying to use Levy continuity theorem I obtain for the characteristic function of $Y_n = \sqrt{\frac{3}{n^3}}\sum_{k=1}^n kX_k$ $$\phi_{Y_n}(t) = \prod_{k=1}^n \biggl( \frac{e^{ik\sqrt{\frac{3}{n^3}} t} + e^{-ik\sqrt{\frac{3}{n^3}} t}}{2}\biggr) = \prod_{k=1}^n \cos(k\sqrt{\frac{3}{n^3}} t)$$

How to proceed ?

Answer 1

For a fixed $t$, taking logarithm gives $\log \phi_{Y_n}(t) = \sum_{k=1}^n \log \cos(kt \sqrt{\dfrac{3}{n^3}})$

Using $$1-\cos(x) = \dfrac{x^2}{2} - O(x^4), x\to 0$$ $$\log(1-x) = -x + O(x^2), x\to 0$$ we have $$1- \cos(kt \sqrt{\dfrac{3}{n^3}}) = \dfrac{3k^2t^2}{2n^3} + O(\dfrac{9k^4t^4}{n^6}) = \dfrac{3k^2t^2}{2n^3} + O(\dfrac{k^4}{n^6})$$ and \begin{align} \log \cos(kt \sqrt{\dfrac{3}{n^3}}) &= \log\left(1-\left(1 - \cos(kt \sqrt{\dfrac{3}{n^3}})\right)\right) \\ &= -(1 - \cos(kt \sqrt{\dfrac{3}{n^3}})) + O((1 - \cos(kt \sqrt{\dfrac{3}{n^3}}))^2) \\ &= -\dfrac{3k^2t^2}{2n^3} + O(\dfrac{k^4}{n^6}) \end{align}

thus we have \begin{align} \sum_{k=1}^n \log \cos(kt \sqrt{\dfrac{3}{n^3}}) &= \sum_{k=1}^n -\dfrac{3k^2t^2}{2n^3} + \sum_{k=1}^n O(\dfrac{k^4}{n^6}) \\ &=-\dfrac{3t^2n(2n^2 + 3n + 1)}{12n^3} + O(\frac{1}{n}) \end{align}

Sending $n$ to infinity gives $\lim_{n\to+\infty}\log \phi_{Y_n}(t) = -\dfrac{t^2}{2}$, as desired

Answer 2

Hint: now you want to look at how close $$\cos\left(k t \sqrt{\frac{3}{n^3}}\right) \exp\left(\frac{3 k^2 t^2}{2 n^3}\right)$$ is to $1$.