# 2D simple symmetric random walks

I have a theorem which says that 2D symmetric random walks are recurrent. I understand this, the way my lecturer shows it is as follows;

$\ p_{(0,0),(0,0)}^{(2n)} = (p_{0,0}^{(2n)})^2 = ((2nCn)(1/4)^n)^2$ He then uses stirlings formula to say that $\Sigma p_{(0,0),(0,0)}^{(2n)} = \Sigma [(pq)^n 2^{2n+1}/\sqrt{2\pi n}]^2$ With p=q=1/2 I can show that the sum is infinite and thus the SSRW must be recurrent. The case I am struggling with is $\ p\neq q$ I know this is transient but I dont know how to show that the above sum is finite for such p and q.

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use the law of large numbers to show transience –  mike Jun 2 '12 at 14:50

If $p\ne q$, $4pq\lt1$. Thus $[(pq)^n2^{2n+1}/\sqrt{2\pi n}]^2\leqslant r^n$ with $r=(4pq)^2\lt1$, hence $_______$.