15
$\begingroup$

I was wandering if someone knows an elementary proof of the following identity:

$$ \frac{(a)_n (b)_n}{(n!)^2} = \sum_{k=0}^n (-1)^k {1-a-b \choose k} \frac{(1-a)_{n-k}(1-b)_{n-k}}{((n-k)!)^2}\ , $$ where $a,b$ are arbitrary real numbers, $$(x)_0=1,\quad (x)_n:=x(x+1)\cdots(x+n-1) \quad \mbox{for $n\geq 1$} $$ is the Pochhammer's symbol, and $$ {x\choose 0}=1,\quad {x\choose k} = \frac{x(x-1)\cdots (x-k+1)}{k!} \quad \mbox{for $k\geq 1$}$$ is a binomial coefficient.

The proof that I know uses the Hypergeometric differential equation. One has to continue analytically the solutions along a path connecting two singular points. This could be done by some well known integral representation of the Hypergeometric function.

I think that there should be a combinatorial proof. Since this is an identity between polynomials in $a$ and $b$, it is enough to prove it for $a$ and $b$ negative integers, i.e., we may assume that $a=-p$ and $b=-q$ where $p,q\in \mathbb{Z}_{\geq 0}$. The identity then turns into $$ {p\choose n}{q\choose n} = \sum_{k=0}^n (-1)^k {1+p+q\choose k}{p+n-k\choose n-k}{q+n-k\choose n-k}. $$ I did not try very hard to proof the above identity and I did not search the literature that much, but since it comes from an interesting subject I think that it is worth finding an alternative proof.

  • 4
    $$\sum_k\binom{m-r+s}k\binom{n+r-s}{n-k}\binom{r+k}{m+n}=\binom rm\binom sn$$, Graham, Knuth, Patashnik: *Concrete Mathematics*, second edition, section 5.1, page 171.2012-07-23
  • 1
    Put another way, we want to prove that $$\binom{a+b+n-2}{n}{}_3 F_2\left({{-n,1-a,1-b}\atop{1,2-a-b-n}}\mid 1\right)=\binom{a+n-1}{n}\binom{b+n-1}{n}$$2012-07-23

2 Answers 2