Gauss's Hypergeometric Theorem/Corollary 1

Corollary to Gauss's Hypergeometric Theorem

Let $\map \Re {1 - a} > 0$.

Let $c \notin \Z_{\le 0}$ and $c \ne 1$.

Then:

$\ds \sum_{k \mathop = 0}^\infty \dfrac {a^{\overline k} } {\paren {c - 1 + k} k!} = \dfrac {\map \Gamma {c - 1} \map \Gamma {1 - a} } {\map \Gamma {c - a} }$

Proof

Set $b = c - 1$ in Gauss's Hypergeometric Theorem

Before substitution:

\(\ds \sum_{k \mathop = 0}^\infty \dfrac {a^{\overline k} b^{\overline k} } {c^{\overline k} } \dfrac {1^k} {k!}\)

\(=\)

\(\ds \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }\)

Gauss's Hypergeometric Theorem

After substitution:

\(\ds \sum_{k \mathop = 0}^\infty \dfrac {a^{\overline k} \paren {c - 1}^{\overline k} } {c^{\overline k} } \dfrac {1^k} {k!}\)

\(=\)

\(\ds \dfrac {\map \Gamma c \map \Gamma {c - a - \paren {c - 1} } } {\map \Gamma {c - a} \map \Gamma {c - \paren {c - 1} } }\)

\(\ds \leadsto \ \ \)

\(\ds \sum_{k \mathop = 0}^\infty \dfrac {a^{\overline k} \paren {c - 1}^{\overline k} } {c^{\overline k} } \dfrac {1^k} {k!}\)

\(=\)

\(\ds \dfrac {\paren {c - 1} \map \Gamma {c - 1} \map \Gamma {1 - a} } {\map \Gamma {c - a} }\)

Definition of Gamma Function and $\map \Gamma 1 = 1$

\(\ds \leadsto \ \ \)

\(\ds \sum_{k \mathop = 0}^\infty \dfrac {a^{\overline k} \paren {c - 1} c^{\overline {k - 1} } } {\paren {c - 1 + k} c^{\overline {k - 1} } } \dfrac {1^k} {k!}\)

\(=\)

\(\ds \dfrac {\paren {c - 1} \map \Gamma {c - 1} \map \Gamma {1 - a} } {\map \Gamma {c - a} }\)

Sum of Indices of Rising Factorial

\(\ds \leadsto \ \ \)

\(\ds \sum_{k \mathop = 0}^\infty \dfrac {\paren a^{\overline k} \paren {c - 1} } { \paren {c - 1 + k} k!}\)

\(=\)

\(\ds \dfrac {\paren {c - 1} \map \Gamma {c - 1} \map \Gamma {1 - a} } {\map \Gamma {c - a} }\)

$c^{\overline {k - 1} }$ cancels

\(\ds \leadsto \ \ \)

\(\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} } {\paren {c - 1 + k} k!}\)

\(=\)

\(\ds \dfrac {\map \Gamma {c - 1} \map \Gamma {1 - a} } {\map \Gamma {c - a} }\)

$\paren {c - 1}$ cancels

$\blacksquare$

Sources

• 1989: Bruce C. Berndt: Ramanujan's Notebooks: Part II: Chapter $\text {10}$. Hypergeometric Series: $\text I$