Let $\{X_n\}_{n\ge1}$ be a sequence of independent and identically distributed random variables. Let $S_n = \sum_{n=1}^NX_n$ for $N\ge1$. Now assume that $X_n$ takes values $1, 0 \text{ or} -1$ with probability $\frac{1}{3}$.
The question now is to prove that $S_N$ is unbounded as $N \rightarrow \infty$.
Edit: I think I need to show that $\mathbb{P}(\sup S_N>M)=1$ for any $M$. So if we have the maximum value that $S_N$ can obtain, then $X_{N+1}\le0$ otherwise $M$ is not an upper bound and $S_N$ must be unbounded as $N \rightarrow \infty$. So it works out that if we can prove that $\sum_{n=N+1}^\infty X_n\le0 \ \forall \ n$ then M is an upper bound. However I think I can show that $\mathbb{P}(\sum_{n=N+1}^\infty X_n\le0, \ \forall \ n) = 0$.
I've been considering this from the perspective of a trinomial triangle. I know that after N steps, there are $3^N$ possible paths. I think if I can get an expression for the number of steps that never go above $0$, i.e. $S_j \le 0, \ \forall \ 0\le j \le N $, then the fraction would tend to $0$ as $N \rightarrow 0$ since the denominator grows faster than the numerator.
$$lim_{n\rightarrow\infty}\frac{\text{# Paths that never go above 0}}{3^N} = 0$$
I've done some by hand so far: $$N = 1: \frac{2}{3}$$ $$N = 2: \frac{5}{9}$$ $$N = 3: \frac{13}{27}$$ $$N = 4: \frac{34}{81}$$ $$N = 5: \frac{93}{243}$$