Centered random walk on $\mathbf{Z}^2$ is recurrent? Is the random walk on $\mathbf{Z}^2$ started at $\left\lbrace \left(0,0\right) \right\rbrace $, in which the probability of advance to right is $p$, the probability of retreat to left is $p$, the probability of advance to up is $q$ and the probability of retreat to down is $q$, with  $p \not = q$, recurrent?
Let $\left( S_n\right) $ the random walk, then 
the random walk is recurrent if and only if $\sum_{n=1}^{\infty}P(S_{2n}=\left(0,0 \right) )=+\infty$
$P(S_{2n}=\left(0,0 \right) )=\sum_{j=0}^{n}\frac{\left( 2n\right)! }{j!j!\left(n-j \right)! \left(n-j \right)! }p^{2j}q^{2n-2j}=q^{2n} {2n \choose n}\sum_{j=0}^{n}{n\choose j}^2 \left(\frac{p^{2}}{q^{2}} \right)^{j} $
is 
$\sum_{n=1}^{\infty}{q^{2n} {2n \choose n}\sum_{j=0}^{n}{n\choose j}^2 \left(\frac{p^{2}}{q^{2}} \right)^{j}}=+\infty$ ?
 A: In $\mathbb Z$ and $\mathbb Z^2$, a finitely supported probability measure $\mu$ yields a recurrent random walk if and only if it is centered, meaning that
$$\sum_{x\in \mathbb Z^d}x\mu(x)=0.$$
This is usually proved using Fourier transforms.
Indeed, you can prove that the $n$th power convolution of $\mu$ satisfies the local following local limit theorem:
$$\mu^{*n}(e)\sim CR^{-n}n^{d/2}$$
and $R=1$ if and only if $\mu$ is centered.
Actually, $R$ is the inverse of the minimum of the function
$$\phi(u)=\sum_{x\in \mathbb Z^d}\mu(x)\mathrm{e}^{u\cdot x}.$$
This function is strictly convex and it reaches its minimum where its gradient vanishes.
You can check that the gradient is given by
$$\nabla \phi(u)=\sum_{x\in \mathbb Z^d}x\mu(x)\mathrm{e}^{u\cdot x}.$$
Thus, the minimum is reached at 0 if and only if
$$\sum_{x\in \mathbb Z^d}x\mu(x)=0$$
that is $\mu$ is centered. In other words, the minimum is equal to $\phi(0)=1$ if and only if $\mu$ is centered.
Now, if $d\geq 3$, you see that $\mu^{*n}(e)$ is always summable, whether $R=1$ or not, but if $d\leq 2$, then $\mu^{*n}(e)$ is summable if and only if $R>1$, if and only if $\mu$ is non centered.
Finally, summability of $\mu^{*n}$ is by definition equivalent to finiteness of the Green function, which is equivalent to the random walk being transient.
For all the details I have not proved,
see for example Woess's book random walks on infinite graphs and groups, in particular Chapter III.13 for local limit theorems on $\mathbb Z^d$.
