Is it true that the sum of zero mean i.i.d. random variables oscillates around $0$ infinitely often? Is this true in general? (Without assuming existence of variance etc.). Only thing we know is that the r.v.s are positive with a non-zero probability and that they are integer valued, integrable with mean 0.
Only thing I could come up with was that $S_n/n$ goes to 0 (where $S_n$ is the partial sum). But this can happen even if $S_n$ was always positive. 
My intuition is that this should be true because it happens with Simple Random Walks (which have a unit step size), and a larger step size should only increase the probability of it becoming negative if it was positive before. I am not sure how to formalize this. Can someone please help?
 A: $S_n=X_1+\dots+X_n$. This proof will work without the assumption that $X_i$ is integer valued. Let
$$
\begin{array}{cc}
\alpha=\inf \{n>0:S_n>0\} &&\beta=\inf \{n>0:S_n<0\}\\
\alpha'=\inf \{n>0:S_n\ge 0\} &&\beta'=\inf \{n>0:S_n\le 0\}
\end{array}
$$
Using Wald's equation, you can show that 
$$
E[\alpha]=E[\beta]=\infty.
$$
Otherwise, you would have $E[S_\alpha]=E[X_1]E[\alpha]=0\cdot E[\alpha]=0$, contradicting $S_\alpha>0$.
Next is the real tricky part. For each $n\ge 0$, define $I_n$ to be the index $i\in\{0,1,\dots,n\}$ for which $S_i$ is minimized, with ties going to the latest such index. I claim that

$$
P(I_n=m)=P(\alpha>m)P(\beta'>n-m).
$$

To see this, note that $\{I_n=m\}$ is determined by the first $n$ steps $X_1,\dots,X_n$ of the process, while $\{\alpha>m\}$ is determined by the first $m$ steps and $\{\beta'>n-m\}$ by the first $n-m$ steps. Furthermore, $\{I_n=m\}$ occurs for $(X_1,\dots,X_n)$ if and only if $\{\alpha>m\}$ occurs for $(X_m,X_{m-1},\dots,X_1)$ (note the reversal!) and $\{\beta'>n-m\}$ occurs for $(X_{m+1},X_{m+2},\dots,X_n)$. $\square$
The payoff to that tricky Lemma is we can show

$\alpha'$ and $\beta'$ are almost surely finite.

We start with $$
1=\sum_{m=0}^n P(I_n=m)=\sum_{m=0}^n P(\alpha>m)P(\beta'>n-m)
$$
Now, let $n\to\infty$. Each summand $ P(\alpha>m)P(\beta'>n-m)$ converges to $P(\alpha>m)P(\beta'=\infty)$, so we get
$$
E[\alpha]=\sum_{m=0}^\infty P(\alpha>m)=\frac1{P(\beta'=\infty)}
$$
We already proved $E[\alpha]=\infty$, so this shows that $\beta'$ is almost surely finite. The same goes for $\alpha'$.
Finally, 

$S_n$ is positive and negative infinitely often.

Just like $\alpha'$ is the first time after $0$ the process is nonnegative, we define $\alpha'(k)$ inductively to be the first time after $\alpha'({k-1})$ that $S_n\ge S_{\alpha'({k-1})}$. Now, the sequence of steps
$$
X_{\alpha'+1},X_{\alpha'(2)+1},X_{\alpha'(3)+1},\dots
$$
are iid distributed like $X_1$. Assuming $X_1$ is nontrivial, these have a nonzero probability of being positive, so with probability $1$, infinitely many of them are positive. If $X_{\alpha'(k)+1}$ is positive, then $S_{\alpha'(k)+1}=S_{\alpha'(k)}+X_{\alpha'(k)+1}>0$. Therefore, there are infinitely many times of the form $S_{\alpha'(k)+1}$ which are greater than $0$. The same goes for times where $S_n<0$.
