proof that $1 = \sum\limits_{k=0}^n (-1)^k { 2n \choose n,k,n-k } \frac{n}{n+k}$

Question

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 $P_n(1) = 1$.

Answer 1

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.

Answer 2

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 $$ \sum_{k=0}^n \binom{n}{k} (-1)^k \frac{1}{n+k}.$$

Answer 3

First, notice that the $n^\text{th}$ forward difference of $1/x$ is $$ \sum_{k=0}^n(-1)^k\binom{n}{k}\frac1{k+x}=\frac{n!}{x(x+1)(x+2)\dots(x+n)}\tag{1} $$ To prove $(1)$, apply the Heaviside Method for Partial Fractions to the right hand side, as is done in this answer with $f(j)=1$.

Next, notice that expanding into factorials, $$ \binom{2n}{n,k,n-k}=\binom{2n}{n}\binom{n}{k}\tag2 $$ Thus, $$ \begin{align} \sum_{k=0}^n(-1)^k\binom{2n}{n,k,n-k}\frac{n}{n+k} &=n\binom{2n}{n}\sum_{k=0}^n\frac{(-1)^k}{n+k}\binom{n}{k}\tag{3a}\\ &=\color{#C00}{n}\binom{2n}{n}\frac{n!}{\color{#C00}{n}\color{#090}{(n+1)(n+2)\cdots(2n)}}\tag{3b}\\ &=\binom{2n}{n}\frac{\color{#090}{n!}n!}{\color{#090}{(2n)!}}\tag{3c}\\[6pt] &=1\tag{3d} \end{align} $$ Explanation:
$\text{(3a):}$ apply $(2)$
$\text{(3b):}$ apply $(1)$
$\text{(3c):}$ cancel the $\color{#C00}{n}$'s and note that
$\phantom{\text{(3c):}}$ $\color{#090}{(n+1)(n+2)\cdots(2n)}=\color{#090}{\frac{(2n)!}{n!}}$
$\text{(3d):}$ $\binom{2n}{n}=\frac{(2n)!}{n!n!}$

Answer 4

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.

Answer 5

Probably the simplest way to obtain the result is: $$\begin{align} \sum_{k=0}^n (-1)^k { 2n \choose n,k,n-k } \frac{n}{n+k}&= \sum_{k=0}^n (-1)^k { 2n \choose n-k }{ n+k \choose k } \frac{n}{n+k}\tag1\\ &=\sum_{k=0}^n (-1)^k { 2n \choose n-k }{ n+k-1 \choose k } \tag2\\ &=\sum_{k=0}^n { 2n \choose n-k }{ -n \choose k } \tag3\\ &=\binom nn\tag4\\ &=1, \end{align} $$ where we used:

$\displaystyle(1): { 2n \choose n,k,n-k }={ 2n \choose n-k }{ n+k \choose k }$.

$\displaystyle(2): { n+k \choose k }\frac n{n+k}={ n+k-1 \choose k }$.

$\displaystyle(3): (-1)^k{ n+k-1 \choose k }={ -n \choose k }$.

$\displaystyle(4):$ Vandermonde's identity.