Gauss's Hypergeometric Theorem/Examples/2F1(1,2;4;1)

Example of Use of Gauss's Hypergeometric Theorem

$1 + \dfrac 2 4 + \paren {\dfrac {2 \times 3} {4 \times 5} } + \paren {\dfrac {2 \times 3 \times 4} {4 \times 5 \times 6} } + \cdots = 3$

Proof

From Gauss's Hypergeometric Theorem:

$\map F {a, b; c; 1} = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }$

where:

$\map F {a, b; c; 1}$ is the Gaussian hypergeometric function: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {1^k} {k!}$

$x^{\overline k}$ denotes the $k$th rising factorial power of $x$

$\map \Gamma {n + 1} = n!$ is the Gamma function.

We have:

\(\ds \map F {1, 2; 4; 1}\)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty \dfrac { 1^{\overline k} 2^{\overline k} } { 4^{\overline k} } \dfrac {1^k} {k!}\)

Definition of Gaussian Hypergeometric Function

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty \dfrac { 2^{\overline k} } { 4^{\overline k} }\)

One to Integer Rising is Integer Factorial, $1^k = 1$

\(\ds \)

\(=\)

\(\ds 1 + \dfrac 2 4 + \paren {\dfrac {2 \times 3} {4 \times 5} } + \paren {\dfrac {2 \times 3 \times 4} {4 \times 5 \times 6} } + \cdots\)

Number to Power of Zero Rising is One

and:

\(\ds \map F {1, 2; 4; 1}\)

\(=\)

\(\ds \dfrac {\map \Gamma 4 \map \Gamma {4 - 1 - 2} } {\map \Gamma {4 - 1} \map \Gamma {4 - 2} }\)

Gauss's Hypergeometric Theorem

\(\ds \)

\(=\)

\(\ds \dfrac {\map \Gamma 4 \map \Gamma 1 } {\map \Gamma 3 \map \Gamma 2 }\)

\(\ds \)

\(=\)

\(\ds \dfrac {3! \times 0!} {2! \times 1!}\)

$\map \Gamma {n + 1} = n!$ is the Gamma function

\(\ds \)

\(=\)

\(\ds 3\)

Therefore:

$1 + \dfrac 2 4 + \paren {\dfrac {2 \times 3} {4 \times 5} } + \paren {\dfrac {2 \times 3 \times 4} {4 \times 5 \times 6} } + \cdots = 3$

$\blacksquare$