TITLE: Asymptotic behavior for the return to zero of a simple random walk
QUESTION [6 upvotes]: I got stuck today trying to understand an argument of the Frank den Hollander Book's. The problem is described below. 
Let $S_n=\sum_{i=1}^n X_i$ be the simple random walk in $\mathbb{Z}^d$, that is 
$$
\mathbb{P}(X_i=x)=
\left\{
\begin{array}{ll} 
\frac{1}{2d}&\text{if}\ \|x\|=1;\\
&&\\
 0&\text{otherwise.} 
\end{array}\right.
$$
I would like to know how to prove that 
$$
\mathbb{P}(S_{2n}=0)\sim 2\left(\frac{d}{4\pi n}\right)^{\frac{d}{2}},  \qquad n\to\infty.
$$
I learn from the Gregory Lawler book's that this is a consequence of the Local Central Limit Theorem. But I would like to know if one can prove this fact without use this result. I tried to Taylor Expand 
$$
\hat{p}(k)=\frac{1}{d}\sum_{j=1}^d \cos k_j
$$
$k=(k_1,\dots,k_d)\in [-\pi,\pi)^d$
and use that 
$$
\mathbb{P}(S_{2n}=0)=\left(\frac{1}{2\pi}\right)^d\int_{[-\pi,\pi)^d} [\hat{p}(k)]^{2n} dk.
$$ 
But It is not working. Any help or reference is welcome.
Thanks.

REPLY [5 votes]: The approach is taken  on pages 78 and 79 of Principles of Random Walk (2nd edition) by Frank Spitzer. I was able to see these pages using Google Books. 
Spitzer first translates $[-\pi,\pi)^d$ by the vector $(\pi/2,\pi/2,\dots,\pi/2)$
 which doesn't change the value of the integral. 
Then he argues that the bulk of the integral is concentrated at two points, 
the origin and  $(\pi,\pi,\dots,\pi)$ both contributing the same value asymptotically.
The Taylor's series expansion  ${1\over d}\sum_{j=1}^d \cos k_j\approx \exp(-|k|^2/2d)$
near the origin finishes the result.