Convergence of characteristic function

Question

Let $$S_n = \sum_{k=1}^n \frac{X_k}{\sqrt{k}}$$ where $X_1, X_2, \ldots$ are iid symmetric Bernoullis with parameter $\frac{1}{2}$: $$X_k = \begin{cases} 1 &p=\frac{1}{2}\\ -1 &p=\frac{1}{2} \end{cases} $$ I found that the characteristic function for $S_n$ is $$\varphi_n(t)=\prod_{k=1}^n \cos\left(\frac{t}{\sqrt{k}}\right)$$ I want to prove that $\lim_{n \rightarrow \infty}\varphi_n(t)=0, \ t\neq 0$. I first considered $\varphi_n(t) = \exp\{\log(\varphi_n(t))\}$. Then $$\log(\varphi_n(t))=\sum_{k=1}^n \log\left(\cos\left(\frac{t}{\sqrt{k}}\right)\right)=\sum_{k=1}^nf_k(t)$$ The Taylor expansion for $f_k(t), t=0$ is \begin{split} f_k(0) &= -\frac{t^2}{2k} + \int_0^t \frac{f^{'''}(x)}{3!}(t-x)^3 dx\\ &= -\frac{t^2}{2k} - \frac{2}{6k\sqrt{k}} \int_0^t \sec^2\left(\frac{x}{\sqrt{k}}\right)\tan\left(\frac{x}{\sqrt{k}}\right)(t-x)^3dx \end{split} but I'm stuck there. I want to prove that $$\sum_{k=1}^nf_k(t)\xrightarrow[n \rightarrow \infty]{} - \infty$$ Any help is welcome! Thanks

Answer 1

You're almost there! For a fixed $t\neq 0$, $$f_k(t) = \log \cos \frac t {\sqrt{k}} = \log\left ( 1 - 2\sin^2\left ( \frac t {2\sqrt k}\right)\right)$$ and since $\log (1+u)\leq u$, $$f_k(t)\leq -2\sin^2\left ( \frac t {2\sqrt k}\right)$$ Finally, since $\sin (v )\geq \frac 2 \pi v$ for $v\in [0, \frac \pi 2 ]$, $$f_k(t) \leq -\frac 2 {\pi^2} \frac {t^2} {k}$$ Finally, since $$\lim_{n\rightarrow+\infty}\sum_{k=1}^n \frac 1 k = +\infty$$ you get your result.