Gauss's Hypergeometric Theorem/Historical Note

Historical Note on Gauss's Hypergeometric Theorem

The proof shown above is a more detailed version of a proof by Srinivasa Ramanujan.

Based on Ramanujan's Notebook, as transcribed in Chapter $10$ of Berndt's book, Ramanujan's proof goes as follows:

\(\text {(8.1)}: \quad\)

\(\ds \map F {-x, -y; n + 1; 1}\)

\(=\)

\(\ds \dfrac {\map \Gamma {x + y + n + 1} \map \Gamma {n + 1} } {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} }\)

Assume that $n$ and $x$ are integers with $n \ge 0$ and $n + x \ge 0$.

Expanding $\paren {1 + u}^{y + n}$ and $\paren {1 + \dfrac 1 u}^x$ in their formal binomial series and taking their product, we find that, if $a_n$ is the coefficient of $u^n$:

\(\ds a_n\)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty \binom {y + n} {k + n} \binom x k\)

\(\text {(8.2)}: \quad\)

\(\ds \)

\(=\)

\(\ds \dfrac {\map \Gamma {y + n + 1} } {\map \Gamma {n + 1} \map \Gamma {y + 1} } \sum_{k \mathop = 0}^\infty \dfrac { \paren {-x}^{\overline k} \paren {-y}^{\overline k} } { k! \paren {n + 1}^{\overline k} }\)

On the other hand, expanding $\paren {1 + u}^{x + y + n}$ in its binomial series and dividing by $u^x$, we find that:

\(\ds a_n\)

\(=\)

\(\ds \binom {x + y + n} {x + n}\)

\(\text {(8.3)}: \quad\)

\(\ds \)

\(=\)

\(\ds \dfrac {\map \Gamma {x + y + n + 1} } {\map \Gamma {x + n + 1} \map \Gamma {y + 1} }\)

Comparing $(8.2)$ and $(8.3)$, we deduce $(8.1)$.

As can be seen, Ramanujan jumped over several intermediate steps in $(8.2)$, but his assertions were all correct.

Sources

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