let $(\epsilon_n)_{n \geq 1}$ be a sequence of i.i.d. random variables such that $\mathbb{P}(\epsilon_1 = \pm 1) = \frac12$.
let $S_0 = 0, \,\,\,\, S_n = \sum_{k = 1}^{n} \epsilon_k$
show that $S_n$ diverges a.s. and in $L^2$.
since we have either $Var(|\epsilon_n| \mathbb{1}_{|\epsilon_n| \leq A}) = 1$ or $\mathbb{P}(|\epsilon_{n}|\geq A) \geq \frac12$ for all possible values of $A$
then from Kolmogorov's three-series theorem we conclude that it can't converge therefore it diverges.
but what about the case in $L^2$ ?
I thought of that this way, because the $(\epsilon_n)_{n \geq 1}$ are i.i.d and $S_n^2$ is a sum of products of $\epsilon^{m}_{i}$'s and $\mathbb{E}\epsilon^{m}_{i} = 0, 1$.
then $\mathbb{S_n^2} = c(n)$ with $ n < c(n) < n^n$
therefore using the squeeze theorem it can't converge in $L^2$.
is this correct ? if so, can it be formulated in a better way ?
Re pointwise convergence: Kolmogorov three series theorem is somewhat overkill here. Just note that $$|S_n(\omega)-S_{n-1}(\omega)| = 1, \qquad n \geq 1,$$ for almost all $\omega \in \Omega$. In particular, for any $\epsilon \in (0,1)$ there does not exist $N \in \mathbb{N}$ such that $|S_n(\omega)-S_{n-1}(\omega)| \leq \epsilon$ for all $n \geq N$; this means that $(S_n(\omega))_{n \geq 1}$ is not a Cauchy sequence and, hence, not convergent.
Re $L^2$-convergence: If a sequence $(X_n)_{n \in \mathbb{N}}$ converges in $L^2$, then $(X_n)_{n \in \mathbb{N}}$ is $L^2$-bounded, i.e. $\sup_{n \in \mathbb{N}} \mathbb{E}(X_n^2)< \infty$. Consequently, convergence in $L^2$ fails to hold for sequences which are unbounded in $L^2$. As $$\mathbb{E}(S_n^2) = \sum_{k=1}^n \mathbb{E}(\epsilon_k^2) = n$$ we have $$\sup_{n \in \mathbb{N}} \mathbb{E}(S_n^2) = \infty,$$ i.e. $(S_n)_{n \geq 1}$ is unbounded in $L^2$; hence, it cannot converge in $L^2$.