In a colloquium I saw the following result:
In simple and symmetrical random walk $S_n$ d-dimensional starting in $0$ you have:
$P(S_{2n}=0) \sim \frac{1}{n^{d/2}}$
In which book can I see your demonstration? I find it very interesting because it is a easy form to show transience.
$\sim$ means that the quotient of the sequences tends to 1.
I assume that the steps of the rw are $(\pm 1,\ldots, \pm 1)$ with equal probability. Let $T_n$ be a vector of projections of $S_n$ on each axis. Then $S_{2n}=0$ iff $T_{2n}=0$. The coordinates of $T_n$ are independent one-dimensional simple random walks and $$ \mathsf{P}\left(T_n^{(k)}=0\right)=\binom{2n}{n}2^{-2n}\sim\frac{1}{\sqrt{\pi n}}. $$ Then, by independence, $$ \mathsf{P}(T_n=0)=\prod_{k=1}^d \mathsf{P}\left(T_n^{(k)}=0\right)\sim\frac{1}{(\pi n)^{d/2}}. $$
In the standard case (when the steps are the $d$ unit vectors) showing this result is more involved. See, for example, these notes.