Dixon's Hypergeometric Theorem/Examples/3F2(0.5,0.5,0.25;1,1.25;1)

Example of Use of Dixon's Theorem

 * $1 + \dfrac 1 5 \paren {\dfrac 1 2}^2 + \dfrac 1 9 \paren {\dfrac {1 \times 3} {2 \times 4} }^2 + \dfrac 1 {13} \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} }^2 + \cdots = \dfrac {\pi^2} {4 \paren {\map \Gamma {\dfrac 3 4} }^4 }$

Proof
From Dixon's Hypergeometric Theorem:


 * $\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {\dfrac n 2 + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {y + \dfrac n 2 + 1} } $

where:
 * $\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1}$ is the generalized hypergeometric function of $1$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren {-x}^{\overline k} \paren {-y}^{\overline k} } { \paren {x + n + 1}^{\overline k} \paren {y + n+ 1}^{\overline k} } \dfrac {1^k} {k!}$
 * $x^{\overline k}$ 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:
 * $1 + \dfrac 1 5 \paren {\dfrac 1 2}^2 + \dfrac 1 9 \paren {\dfrac {1 \times 3} {2 \times 4} }^2 + \dfrac 1 {13} \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} }^2 + \cdots = \dfrac {\pi^2} {4 \paren {\map \Gamma {\dfrac 3 4} }^4 }$