[Calculation 5] Gauss’s Summation Formula

 
Proof. We remember that the Euler Integral Representation for the hypergeometric function is
$$
_2 F_1 (a,b;c;z)
=
\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}
\int_0^1 \frac{t^{b-1} (1-t)^{c-b-1}}{(1-tz)^a}\,dt.
$$
Taking the limit $z\to 1$ both sides, we obtain
\begin{eqnarray*}
_2 F_1 (a,b;c;1)
&=&
\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}
\int_0^1 t^{b-1} (1-t)^{c-a-b-1}\,dt\\
&=&
\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}
\frac{\Gamma(b)\Gamma(c-b-a)}{\Gamma(c-a)}\\
&=&
\frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)},
\end{eqnarray*}
which proves the theorem.$\square$
 

 
Proof. Taking $a = -n$ in Gauss’s Summation Formula, we have
$$
{}_2 F_1 (-n,b;c;1) = \frac{\Gamma(c)\Gamma(c-b+n)}{\Gamma(c+n)\Gamma(c-b)} = \frac{(c-b)_n}{(c)_n}.
$$
 
References.
[1] en.wikipedia.org/wiki/Hypergeometric_function
[2] Leun Kim, Fundamentals of Hypergeometric Functions.