proof that $1 = \sum\limits_{k=0}^n (-1)^k { 2n \choose n,k,n-k } \frac{n}{n+k}$ 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 $P_n(1) = 1$.
 A: Here's an integral-free computational proof.
$$\begin{align*}
\sum_{k=0}^n (-1)^k { 2n \choose n,k,n-k } \frac{n}{n+k}&=\sum_{k=0}^n(-1)^k\binom{2n}n\binom{n}k\frac{n}{n+k}\\
&=\sum_{k=0}^n(-1)^k\frac{(2n)^{\underline{n+1}}}{k!(n-k)!(n+k)}\\
&=\sum_{k=0}^n(-1)^k\frac{(2n)^{\underline{n-k}}(n+k-1)^{\underline{k}}}{k!(n-k)!}\\
&=\sum_{k=0}^n(-1)^k\binom{2n}{n-k}\binom{n+k-1}k\\
&=\sum_{k=0}^n(-1)^k\binom{2n}{n-k}\binom{n+k-1}{n-1}\;.
\end{align*}$$
Identity (5.24) in Graham, Knuth, & Patashnik, Concrete Mathematics, is
$$\sum_k(-1)^k\binom{\ell}{m+k}\binom{s+k}n=(-1)^{\ell+m}\binom{s-m}{n-\ell}$$ 
for integer $\ell\ge 0$ and integers $m$ and $n$. We almost have the special case
$$\sum_k(-1)^k\binom{\ell}{m+k}\binom{s+k}s=(-1)^{\ell+m}\binom{s-m}{s-\ell}\;,$$
with $\ell=2n$, $m=n$, $s=n-1$: 
$$\sum_k(-1)^k\binom{2n}{n+k}\binom{n-1+k}{n-1}=(-1)^{3n}\binom{-1}{-1-n}=0\;.$$
The summation in the identity is over all integers $k$, so
$$\begin{align*}\sum_{k\ge 0}(-1)^k\binom{2n}{n+k}\binom{n-1+k}{n-1}&=\sum_{k<0}(-1)^{k+1}\binom{2n}{n+k}\binom{n-1+k}{n-1}\\
&=\sum_{k=1}^n(-1)^{k+1}\binom{2n}{n+k}\binom{n-1-k}{n-1}\\
&\stackrel{*}=\sum_{k=1}^n(-1)^{k+1}\binom{2n}{n+k}(-1)^{n-1}\binom{n-1-(n-1-k)-1}{n-1}\\
&=\sum_{k=1}^n(-1)^{n+k}\binom{2n}{n+k}\binom{k-1}{n-1}\\
&=(-1)^{2n}\binom{2n}{2n}\\
&=1\;,
\end{align*}$$
where the starred step is by what GKP calls negating the upper index.
A: 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 $$ \sum_{k=0}^n \binom{n}{k} (-1)^k \frac{1}{n+k}.$$
A: Suppose we seek to evaluate 
$$\sum_{k=0}^n (-1)^k
{2n\choose n,k,n-k}\frac{n}{n+k}.$$
First do the binomial simplification as already noted by Brian M. Scott.
We start with
$$\frac{(2n)!}{n!\times k!\times (n-k)!}
\frac{n}{n+k}
\\ = \frac{(2n)!\times (n+k-1)!}
{(n-1)!\times k!\times (n-k)!\times (n+k)!}
\\ = {2n\choose n-k} {n+k-1\choose n-1}$$
which gives for the sum
$$\sum_{k=0}^n (-1)^k
{2n\choose n-k} {k+n-1\choose n-1}.$$
Introduce the integral representation
$${2n\choose n-k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n-k+1}} \; dz.$$
Note that  this integral is  zero when $k\gt n$  so we may  extend the
range of the sum to infinity, getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}} 
\sum_{k=0}^\infty (-1)^k {k+n-1\choose n-1} z^k
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}} 
\frac{1}{(1+z)^n} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{n+1}} \; dz.$$
This evaluates to one by inspection, QED.
