If we uniformly choose a lattice path from $(0,0)$ to $(n, n)$ (i.e. at each step we can move from $(x,y)$ to $(x+1,y)$ or $(x,y+1)$), what is the expected number of times that the path crosses the diagonal?

From http://onlinelibrary.wiley.com/doi/10.2307/3314829/pdf we see that the probability that a uniformly chosen path has $r$ crossings is $2\binom{n-1}{r} / \binom{n+r+1}{n}$. And so the expectation is $2 n! (n-1)! \sum_{r=0}^{n-3} \frac{(r+2)(r+1)}{(n+r+2)!(n-r-2)!}$

Is a closed form possible for this expression?