Kummer's Hypergeometric Theorem/Examples/2F1(0.33,0.33;1;-1)

Example of Use of Kummer's Hypergeometric Theorem

 * $1 - \paren {\dfrac 1 3}^2 + \paren {\dfrac {1 \times 4} {3 \times 6} }^2 - \paren {\dfrac {1 \times 4 \times 7} {3 \times 6 \times 9} }^2 + \cdots = \dfrac {\pi} {\paren {\map \Gamma {\dfrac 5 6} }^2 \map \Gamma {\dfrac 1 3} }$

Proof
From Kummer's Hypergeometric Theorem:


 * $\ds \map F {n, -x; x + n + 1; -1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } { \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {n + 1} } $

where:
 * $\ds \map F {n, -x; x + n + 1; -1}$ is the Gaussian hypergeometric function of $-1$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren {-x}^{\overline k} } { \paren {x + n + 1}^{\overline k} } \dfrac {\paren {-1}^k} {k!}$
 * $x^{\overline k}$ denotes the $k$th rising factorial power of $x$
 * $\map \Gamma {n + 1} = n!$ is the Gamma function.

We have:

and:

Recall from the Euler Reflection Formula: $\map \Gamma z \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$

Therefore:

Substituting this result back into our equation above:

Therefore:
 * $1 - \paren {\dfrac 1 3}^2 + \paren {\dfrac {1 \times 4} {3 \times 6} }^2 - \paren {\dfrac {1 \times 4 \times 7} {3 \times 6 \times 9} }^2 + \cdots = \dfrac {\pi} {\paren {\map \Gamma {\dfrac 5 6} }^2 \map \Gamma {\dfrac 1 3} }$