# [Calculation 4] Saalschütz’s Theorem

Edited by Leun Kim

 Theorem. (Saalschütz’s Theorem) $${}_3 F_2 (a,b,-n;c,1+a+b-c-n;1) = \frac{(c-a)_n (c-b)_n}{(c)_n (c-a-b)_n}$$ holds for $n\in \mathbb N_0$.

Proof. We recall the Euler Transformation Formula:
$$(1-z)^{a+b-c} {}_2 F_1 (a,b;c;z) = {}_2 F_1 (c-a,c-b;c;z).$$
Equating the coefficients of $z^n$ from both sides, we get
\tag{1}
\sum_{k=0}^n \frac{(a)_k (b)_k}{(c)_k k!} \frac{(c-a-b)_{n-k}}{(n-k)!}
=
\frac{(c-a)_n (c-b)_n}{(c)_n n!}

where we used the binomial theorem
$$(1-z)^{a+b-c} = \sum_{k=0}^\infty \frac{(c-a-b)_k}{k!}z^k.$$
In view of the relations
$$\frac{n!}{(n-k)!} = (-1)^k (-n)_k$$
and
$$(c-a-b)_{n-k} = \frac{(c-a-b)_n}{(-1)^k (1+a+b-c-n)_k},$$
we can rewrite (1) as
$$\sum_{k=0}^n \frac{(a)_k (b)_k}{(c)_k k!} \frac{(c-a-b)_n (-n)_k}{(1+a+b-c-n)_k n!} = \frac{(c-a)_n (c-b)_n}{(c)_n n!}.$$
Therefore we finally obtain
\begin{eqnarray*}
{}_3 F_2 (a,b,-n;c,1+a+b-c-n;1)
&=&
\sum_{k=0}^\infty \frac{(a)_k (b)_k (-n)_k}{(c)_k (1+a+b-c-n)_k k!}\\
&=&
\sum_{k=0}^n \frac{(a)_k (b)_k (-n)_k}{(c)_k (1+a+b-c-n)_k k!}\\
&=&
\frac{(c-a)_n (c-b)_n}{(c)_n (c-a-b)_n}.
\end{eqnarray*}
This proves the theorem.$\square$

References.
[1] W. N. Bailey, Generalized Hypergeometric Series, Cambridge University Press, 1964, p.9.
[2] Leun Kim, Euler’s Transformation Formula

#### Leun Kim

Ph.D Candidate at The University of Tokyo
I was born and raised in Daegu, S. Korea. I majored in electronics and math in Seoul from 2007 to 2012. I've had a great interest in math since freshman year, and I studied PDE in Osaka, Japan from 2012-2014. I worked at a science museum and HUFS from 2014 in Seoul. Now I'm studying PDE in Tokyo, Japan. I also developed an interest in music, as I met a great piano teacher Oh in 2001, and joined an indie metal band in 2008. In my spare time, I enjoy various things, such as listening music, blogging, traveling, taking photos, and playing Go and Holdem. Please do not hesitate to contact me with comments, email, guestbook, and social medias.

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