I want to show that for the special case of normally distributed random variables - with mean equal to zero and variance equal to one - the strong law of large numbers holds, by integrating the density function twice.
Let \begin{equation} S_n = \frac{1}{n}(X_1+\dots+X_n); \text{ for }X_n \in N(0;1) \iff S_n \in N(0;\frac{1}{n})\text{ with density } f(x;n) \end{equation} Then \begin{equation} P(|S_n|>\epsilon)=1- \int_{-\epsilon}^{\epsilon} f(x,n)dx = \frac{1}{2}\text{erfc}(\epsilon\sqrt{\frac{n}{2}}) \end{equation} If we integrate with respect to n, we get for
\begin{equation} \int_{0}^{\infty}P(|S_n|>\epsilon)dn= \frac{1}{2\epsilon^2} \end{equation}
(I am not sure if I have to integrate from 1 to $\infty$, instead of zero, but the only singularity is at $\epsilon = 0$ no matter where you start integrating)
This implies that for all $\epsilon > 0:\sum_{}^{\infty}$ $P(|S_n|>\epsilon) < \infty$ and there fore $S_n\rightarrow0$ a.s. (using borel-cantelli)
I would be glad if some one can share some advise if this makes sense or not. Thanks a lot!