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\bf
A 21st-Century Proof Of Dougall's Hypergeometric 
Sum Identity\rm

\it
Shalosh B. Ekhad
and Doron Zeilberger*
\footnote{ *}{
Supported, in part, by NSF grant DMS8800663.  
}

Department of Mathematics, Drexel University, Philadelphia, PA19104
\rm

{\bf Appeared in: J. Math. Anal. Appl. 147 (1990), 610-611.}

\it
   My guess is that, within fifty or hundred years (or it might be one 
   hundred and fifty) computers will successfully compete with the human 
   brain in doing mathematics, and that their mathematical style will be 
   rather different from ours.  Fairly long computational verifications 
   (numerical or combinatorical) will not bother them at all, and this 
   should lead not just to different sorts of proofs, but more importantly
   to different sorts of  theorems being proved.
\rm

\qquad\qquad\qquad ---David Ruelle[3].
          

Dougall's identity[2] is one of the most beautiful and general
results in the theory of hypergeometric series, and Dougall's
original proof([1], 5.1) is short and elegant.
In this note we present a new proof that is even
shorter, modulo
purely routine verifications (that are easily carried out with
computer algebra. ) It was generated by the first author using an
algorithm written by the second author[4].

{\bf  Theorem } (Dougall[2]; [1], p. 26 and section 5.1):

For any complex number $z$, and any integer $k$, let
$(z)_k = z(z+1) ... (z+k-1)$.  The following identity is true
for any positive integer $n$.

$$
\sum_{k} (-1)^k  (  { n \atop k }  )  \cdot
$$
$$
{{ (1+2a-b-c-d)_k (3/2+a-b/2-c/2-d/2)_k 
(1+a-c-d)_k  (1+a-b-d)_k (1+a-b-c)_k (a+n)_k }\over
{ (1/2+a-b/2-c/2-d/2)_k (1+a-b)_k (1+a-c)_k 
(1+a-d)_k (2+a-b-c-d-n)_k (2+2a-b-c-d+n)_k }}
$$

$$
   = {{ (2+2a-b-c-d)_n (b)_n (c)_n (d)_n }\over{ 
(-1-a-b-c-d)_n (1+a-b)_n (1+a-c)_n (1+a-d)_n }} \, \,. 
$$

{\bf Proof}: Let $R(n)$ and $F(n,k)$ be the sum and the summand
respectively on the left.  Since the theorem is obviously true for $n=0$,
it would follow by induction once we show that

$$
 (-1-a-b-c-d+n)(1+a-b+n)(1+a-c+n)(1+a-d+n)R(n+1) 
$$
$$
     - (2 + 2 a - b - c - d + n) (b + n) (c + n) (d + n)R(n)=0 
\eqno(1)
$$          

Let $G(n,k):= - (k + 1 + a - b - c) (k + 1 + a - b - d) \cdot$
$$
  {{  
(k + 1 + 2 a - b - c - d) (a + n + k)  (k + 1 + a - c - d)(2 n + a + 1)(2 + 2 a - b - c - d + n)}\over{((2 k + 2 a - b - c - d + 1) (2 + 2 a + n + k - b - c - d) (a + n))}} F(n,k)    
$$

It is readily verified (by machine, if possible) that

$$
 (-1-a-b-c-d+n)(1+a-b+n)(1+a-c+n)(1+a-d+n)F(n+1,k) 
$$
$$
     - (2 + 2 a - b - c - d + n) (b + n) (c + n) (d + n)F(n,k)= G(n,k)-G(n,k-1) \, \, \,. 
\eqno(2)
$$         

[ Dividing (2) by $F(n,k)$ yields a routinely verifiable identity
among rational functions].

and we get (1) by summing (2) with respect to $k$.  QED

{\bf References}

1. W. N. Bailey,"Generalized Hypergeometric Series", Cambridge Math.
Tracts {\bf 32}, Cambridge University Press, London, 1935. (Reprinted:
Hafner, New York, 1964.)

2. J. Dougall, \it
On Vandermonde's theorem and some more general
expansions\rm
, Proc. Edinburgh Math. Soc., {\bf 25}(1907), 114-132.

3. Ruelle (David), \it
Is our mathematics natural? the case of
equilibrium statistical mechanics\rm
, Bull. (New Series) Amer. Math.
Soc. {\bf 19} (1988), 259-268.

4. D.Zeilberger, \it
A fast algorithm for proving terminating
hypergeometric identities\rm
, submitted.
\end
