Almost sure convergence of a martingale

Question

Let $X_1,X_2,\ldots$ be i.i.d random variables with standard normal distribution $\mathcal{N}(0,1)$. We define $$N_n = e^{n/2}\sin{(X_1+X_2+\ldots X_n)}$$
and after some calculations, it's easy to show that it is a martingale with respect to the filtration $\mathcal{F}_n = \sigma(X_1,X_2,\ldots ,X_n)$. My goal is to check whether the sequence $(N_n^2)_{n \ge 0}$ is uniformly integrable, so – knowing $N_n$ is a martingale – it is enough to check the $L^2$-convegence. So I wanted to check the almost sure convergence first to have the candidate for $L^2$ limit. But here is where I'm stuck.

Answer 1

Since $(N_n)_{n \in \mathbb{N}}$ is a martingale, $(N^2_n)_{n \in \mathbb{N}}$ is a submartingale. If $(N^2_n)_{n \in \mathbb{N}}$ is uniformly integrable, then $N_n^2\to Z$ to some $Z$ a.s., $Z$ is integrable and $E[N_n^2]\to E[Z]$. Now note ($S_n:=X_1+...+X_n\sim \mathcal{N}(0,n))$ and $$N_n^2=\frac{e^n}{2}(1-\cos(2S_n))$$ So $E[N_n^2]=(e^n/2)(1-e^{-2n})=(e^n-e^{-n})/2\to \infty$. Suppose $(N_n^2)_{n \in \mathbb{N}}$ is uniformly integrable. This would mean $E[Z]=\infty$ so $Z$ is not integrable, a contradiction. So $(N_n^2)_{n \in \mathbb{N}}$ is not uniformly integrable.