Distribution converges weakly to $N(0,\alpha^{-2})$

Question

Let $(X_n)_{n\in\mathbb N}$ be i.i.d. Laplace distributed, i.e. with pdf $\frac{\alpha}{2}e^{-\alpha|x|}$, and let $Z_n=\frac{1}{n}\sum_{k=1}^{n}\sqrt{k}X_k$, I need to show that the distribution of $Z_n$ converges weakly to $N(0,\alpha^{-2})$. I know weak convergence is equivalent to pointwise convergence of characteristic functions, so I tried to calculate the characteristic function of $Z_n$. I managed to calculate the characteristic function of $X$, namely $\frac{\alpha^2}{\alpha^2+x^2}$, but the characteristic function of $Z_n$ doesn't look like anything that's familiar to me $$\prod_{k=1}^{n}\frac{\alpha^2}{\alpha^2+k\big(\frac{x}{n}\big)^2}.$$ How can I show this converges to $e^{-\frac{1}{2}\big(\frac{x}{\alpha}\big)^2}$ for $n\to\infty$?

Answer 1

Hint. Pass to the logarithm: \begin{align*} \log{\left(\prod_{k=1}^n\frac{\alpha^2}{\alpha^2+k{\left(\frac xn\right)}^2}\right)}&=\log{\left(\prod_{k=1}^n\frac1{1+k{\left(\frac x{\alpha n}\right)}^2}\right)}\\[.4em] &=-\sum_{k=1}^n\log{\!\left(1+k{\left(\frac x{\alpha n}\right)}^2\right)}\\[.4em] &=-\sum_{k=1}^n\left[k{\left(\frac x{\alpha n}\right)}^2 +O\!\left(\frac{k^2}{n^4}\right)\right]\\[.4em] &=-\frac{n(n+1)x^2}{2\alpha^2n^2}+O\!\left(\frac1n\right)\\[.4em] &\xrightarrow[n\to\infty]{}-\frac{x^2}{2\alpha^2}. \end{align*} (There I used that $\log(1+x)=x+O(x^2)$ for $|x|<\frac12$, say.)

Answer 2

Let $t=x/\alpha$. We are reduced to show that $$ \prod_{k=1}^{n}\frac{1}{1+k\big(\frac{t}{n}\big)^2}\to e^{-t^2/2} $$ and letting $s=t^2$ and taking the inverse then the logarithm, we actually have to prove the convergence $$ \sum_{k=1}^n\log\left(1+\frac{k}{n^2}s\right)\to s/2 $$ or in order words, $$ \sum_{k=1}^n\left(\log\left(1+\frac{k}{n^2}s\right)-\frac{k}{n^2}s\right)\to 0. $$ This follows from a Taylor expansion at $0$ of $t\mapsto \log(1+t)$.