= Multinomial collision proof of three-dimensional walk transience
{title2=$P(S_{2n}=0)=\frac{\binom{2n}{n}}{4^n}\sum_{i+j+k=n}\left(\frac{n!}{3^ni!j!k!}\right)^2=O(n^{-3/2})$}
For the nearest-neighbor symmetric walk on $\mathbb Z^3$, a return at time $2n$ uses $i,j,k$ positive and the same numbers of negative steps along each axis, with $i+j+k=n$. Counting arrangements gives the displayed identity. The sum of squares of the three-category <multinomial distribution> probabilities is at most their maximum. This maximum occurs at balanced counts and is $O(n^{-1})$ by <Stirling's formula>; the central binomial factor is $O(n^{-1/2})$. The total return probability is therefore summable. The expected number of visits is finite, so the probability of returning after time zero is strictly below one, proving that the origin is a <transient state>.
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