[Calculation 6] Kummer’s Theorem

 
To prove Kummer’s theorem, we introduce the following lemma, which is called Kummer’s quadratic transformation:
 

 
Proof of Lemma. Note that the right hand side of (1) is analytic inside the loop of the curve $|4z|=|1-z|^2$ which surrounds the origin as below.

Therefore the right hand side of (1) can be expanded in powers of $z$ when $|z|<3-2\sqrt 2$. Thus
\begin{eqnarray*}
\text{(rhs)}
&=&
(1-z)^{-a} \sum_{k=0}^\infty \frac{(\frac{1}{2}a)_k (\frac{1}{2} + \frac{1}{2}a-b)_k}{(1+a-b)_k k!}\left( -\frac{4z}{(1-z)^2}\right)^k\\
&=&
\sum_{k=0}^\infty \frac{(\frac{1}{2}a)_k (\frac{1}{2} + \frac{1}{2}a-b)_k (-4)^k}{(1+a-b)_k k!}z^k (1-z)^{-a-2k}\\
&=&
\sum_{k=0}^\infty \sum_{r=0}^\infty \frac{(\frac{1}{2}a)_k (\frac{1}{2} + \frac{1}{2}a-b)_k (-4)^k}{(1+a-b)_k k!} \frac{(a+2k)_r}{r!}z^{k+r}.
\end{eqnarray*}
Calculating the coefficient of $z^n$, we get
\begin{equation}\tag{2}
\sum_{k=0}^n \frac{(\frac{1}{2}a)_k (\frac{1}{2} + \frac{1}{2}a-b)_k (-4)^k}{(1+a-b)_k k!} \frac{(a+2k)_{n-k}}{(n-k)!}
=
\frac{(a)_n}{n!}\sum_{k=0}^n \frac{(\frac{1}{2}+\frac{1}{2}a-b)_k (a+n)_k (-n)_k}{(1+a-b)_k (\frac{1}{2} + \frac{1}{2}a)_k k!},
\end{equation}
where we applied the relations
$$
\frac{1}{(n-k)!} = \frac{(-1)^k (-n)_k}{n!}
$$
and
$$
(a+2k)_{n-k} = \frac{(a)_n(a+n)_k}{4^k (\frac{1}{2}a)_k (\frac{1}{2} + \frac{1}{2}a)_k}.
$$
Therefore using Saalschütz’s Theorem to the right hand side of (2), we obtain
\begin{eqnarray*}
\frac{(a)_n}{n!}\sum_{k=0}^n \frac{(\frac{1}{2}+\frac{a}{2}-b)_k (a+n)_k (-n)_k}{(1+a-b)_k (\frac{1}{2} + \frac{a}{2})_k k!}
&=&
\frac{(a)_n}{n!} {}_3 F_2 \left( \frac{1}{2} + \frac{a}{2} -b, a+n, -n; 1+a-b, \frac{1}{2} + \frac{a}{2};1 \right)\\
&=&
\frac{(a)_n}{n!} \frac{(\frac{1}{2}+\frac{1}{2}a)_n (a-b-n)_n}{(1+a-b)_n(\frac{1}{2}-\frac{1}{2}a-n)_n}\\
&=&
\frac{(a)_n (b)_n}{n!(1+a-b)_n}
\end{eqnarray*}
so that
$$
\text{(rhs)} = \sum_{n=0}^\infty \frac{(a)_n (b)_n}{n!(1+a-b)_n} z^n = {}_2 F_1 (a,b;1+a-b;z),
$$
which proves Lemma for $|z|<3-2\sqrt 2$, and the complete result follows by analytic continuation.$\square$
 
Proof of Theorem. Limiting $z\to -1$ in Lemma, we obtain
\begin{eqnarray*}
{}_2 F_1 (a,b;1+a-b;-1)
&=&
2^{-a} {}_2 F_1 \left( \frac{1}{2}a, \frac{1}{2}+\frac{1}{2}a-b; 1+a-b; 1\right)\\
&=&
2^{-a} \frac{\Gamma(1+a-b)\Gamma(\frac{1}{2})}{\Gamma(1+\frac{1}{2}a-b)\Gamma(\frac{1}{2}+\frac{1}{2}a)}\\
&=&
\frac{\Gamma(1+a-b)\Gamma\left(1+\frac{1}{2}a\right)}{\Gamma(1+a)\Gamma\left(1+\frac{1}{2}a-b\right)},
\end{eqnarray*}
where we used the Gauss Summation Formula for the second equality, and the Duplication Formula for the Gamma function:
$$
\Gamma\left(\frac{1}{2} + \frac{1}{2}a\right) = \frac{2^{-a}\Gamma(\frac{1}{2})\Gamma(1+a)}{\Gamma(1+\frac{1}{2}a)}
$$
for the last equality.$\square$
 
Errata : $(0,3-2\sqrt 2)$ should be replaced by $(3-2\sqrt 2,0)$.
 
[1] W. N. Bailey, Generalized Hypergeometric Series, Cambridge University Press, 1964, pp.9-10.
[2] Bruno Gauthier, A Proof of Kummer’s Theorem, 2008. (arXiv:math/9904061v2)
[3] Leun Kim, Saalschütz’s Theorem.
[4] Leun Kim, Gauss’s Summation Formula.