I'm looking for a proof of this identity:
$ 1 = \sum_{k=0}^n (-1)^k { 2n \choose n,k,n-k } \frac{n}{n+k} $
I'll take anything, but a combinatorial proof would be nice - all of the terms in the sum appear to be integers.
Update: Given J.M.'s reformulation, if we start with $ x^{n-1} (1-x)^n = \sum_{k=0}^n { n \choose k } (-1)^k x^{n+k-1} $ and integrate both sides from 0 to 1 wrt $x$ we get: $ \int_0^1 x^{n-1} (1-x)^n dx = \sum_{k=0}^n { n \choose k } \frac{(-1)^k}{n+k} $ and so it is sufficient to prove that the integral is $1/( n { 2n \choose n } )$.
My instinct tells me to try a trigonometric substitution ($x = \cos^2 u$?) to evaluate the integral - haven't worked out all the details, though. (Update: see leslie townes comment below.)
In any case, I would really like to find a combinatorial proof.
Update 2: Found this paper: Walking into an absolute sum and the sum I'm interested in is $P_n(1)$ where $P_n(x)$ is the polynomial defined by: $ P_0(x) = 1 \\ P_{n+1}(x) = x^2 [ P_n(x) - P_n(x-1) ] + x P_n(x-1) $ From this definition it is clear that $P_n(0) = 0$ for $n > 0$ and so $P_n(1) = 1$.