TITLE: proof that $1 = \sum\limits_{k=0}^n (-1)^k { 2n \choose n,k,n-k } \frac{n}{n+k}$
QUESTION [11 upvotes]: 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 $P_n(1) = 1$.

REPLY [2 votes]: You asked for a combinatorial proof.  Given the fractions in the identity, perhaps a probabilistic proof makes more sense.  In any case, here's a probabilistic proof of the reformulation of the identity 
$$ \sum_{k=0}^n \binom{n}{k} (-1)^k \frac{1}{n+k} = \frac{1}{n \binom{2n}{n}} = \frac{(n-1)! n!}{(2n)!}.$$
Question: Suppose you have $n-1$ numbered red balls, $n$ numbered blue balls, and 1 black ball in a jar.  Suppose you draw the balls, one-by-one, from the jar without replacing them.  What is the probability that you draw all the red balls, followed by the black ball, followed by the blue balls?
Answer 1: There are $(n-1)!$ ways to draw the red balls, times  1 way to draw the black ball, times $n!$ ways to draw the blue balls, out of $(2n)!$ total ways to draw the balls, for a probability of $$\frac{(n-1)! n!}{(2n)!}.$$
Answer 2: Use inclusion/exclusion.  Let $B$ be the event that the black ball is drawn after all the red balls.  Let $A_i$ be the event that the black ball is drawn before the blue ball numbered $i$.  We want $P(B \cap \left( \cap_{i=1}^n A_i \right)$.  The event consisting of $B$ and any particular $k$ of the $\bar{A}_i$'s is the event that the black ball comes after all the red balls and after $k$ specific blue balls.  This event has probability $1/(n-1 + k + 1) = 1/(n+k)$.  By the principle of inclusion/exclusion, then, the answer to the question is also $$ \sum_{k=0}^n \binom{n}{k} (-1)^k \frac{1}{n+k}.$$