Theorem. (Gauss’s Summation Formula)For $\text{Re}c>\text{Re}b>0$, $$ {}_2 F_1 (a,b;c;1) = \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)} $$ holds. |

**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.$\square$

Corollary. (Chu–Vandermonde Identity)$$ {}_2 F_1 (-n,b;c;1) = \frac{(c-b)_n}{(c)_n} $$ |

**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.

## 2 Comments

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Dear Leun Kim: I am happy to see that you are doing some calculations related to hypergeoemtric and generalized hypergeometric series. I am working in this area for the last several years. To know more about the results you mentioned, send me your e-mail address so t hat I may send you some of mynelementary results.

All best wishes

Leun KimThanks for the comment! I’m just a beginner to these kind of topics, but you can send me your works if you don’t mind : )