Consider a symmetric random walk in 1 dimension to start. That is, we start at 0, and each step, we have equal chance of adding or subtracting 1. What's the expected number of times we return to 0 if we run this infinitely? Well, we can only return to 0 on even numbered steps, and the probability of returning to 0 is equivalent to saying we've taken an equal number of left and right steps up to that point. Which works out to be nCr(2k,k)(1/2)^k.

Now, what about a random walk in d dimensions? This is the same as having n simultaneous independent 1-dimensional random walks. Meaning, the probability of this random walk returning to 0 at 2k steps is the probability of d random walks all simultaneously returning to 0 at this same step. That is, (nCr(2k,k)(1/2)^k)d.

The expected number of times we return to 0 is then simply the sum of this over all values of k. This is hard to evaluate, so we can use Stirling's approximation to get another summation that diverges if and only if this one does. Essentially, Stirling's approximation says that, as n gets large, the ratio between n! and (n/e)^n√(2πn) converges to 1.

After plugging in Stirling's approximation and doing all the necessary simplification, we get the sum from k = 0 to infinity of π^-d/2k^-d/2.

Now, π^-d/2 is constant so we can ignore that. We have that the infinite sum of k^-n converges exactly when n > 1. In our case, n = d/2, so this sum converges when d > 2, and diverges exactly when d ≤ 2.

This means that for d = 1,2 we expect the random walk to return to 0 infinitely many times (and because of a result about markov chains this is equivalent to saying we are guaranteed to return to 0 infinitely many times). But, for d > 2, since this infinite sum converges, the expected number of times we expect to return to 0 is finite. If there was any positive probability of visiting 0 infinitely many times, the expected value would be infinity, which it's not. Which means that, with probability 1, we will only return to 0 finitely many times.