Let $X_1,X_2,\ldots\in L^2$ be i.i.d. random variables with $E(X_i)=0$. Show that $$\mathbb{P}\left(\exp\left(\sum_{i=1}^n X_i\right) >1\right)\to\frac{1}{2}$$ as $n$ tends to infinity.
My approach:
$$\mathbb{P}\left(\exp\left(\sum_{i=1}^n X_i\right) >1\right)=\mathbb{P}\left(\sum_{i=1}^n X_i > 0\right)=\mathbb{P}\left(\frac{1}{n}\sum_{i=1}^n X_i > 0\right)$$
From the law of large numbers, we know that $\frac{1}{n}\sum\limits_{i=1}^n X_i\to 0$ in probability, i.e.
$$\forall\epsilon>0: \mathbb{P}\left(\left|\frac{1}{n}\sum_{i=1}^n X_i\right|>\epsilon\right)\to 0$$
I now want to make some kind of statement about $P\left(\frac{1}{n}\sum\limits_{i=1}^n X_i<0\right)$ in order to proceed with the proof. How do I continue?
Apply CLT instead of SLLN. $P(\frac {S_n} {\sqrt n \sigma } >0) \to 1-\Phi (0)=\frac 1 2$ where $\Phi$ is the standard normal distribution. Hence $P(S_n >0) \to \frac 1 2$.