Show that $\pi =2 F\left(\frac{1}{2}, \frac{1}{2}; \frac{3}{2}; 1\right)$

Question

Recently learned about Hypergeometric functions.

\begin{equation} F(a, b; c; z) = \sum_{n=0}^{\infty} \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!} \end{equation}

which has a special case example

\begin{equation} F\left(\frac{1}{2}, \frac{1}{2}; \frac{3}{2}; z^2\right) = \frac{\arcsin(z)}{z} \end{equation}

Rearranging

\begin{equation} \arcsin(z) = z F\left(\frac{1}{2}, \frac{1}{2}; \frac{3}{2}; z^2\right) \end{equation}

and substituting $a=\frac{1}{2}$, $b=\frac{1}{2}$, $c=\frac{3}{2}$, $z=1$

\begin{align} \arcsin(1) &= F\left(\frac{1}{2}, \frac{1}{2}; \frac{3}{2}; 1^2\right)\\ &= \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_n \left(\frac{1}{2}\right)_n}{\left(\frac{3}{2}\right)_n} \frac{1^n}{n!} \end{align}

Given that $\arcsin(1) = \frac{\pi}{2}$, then $\pi$ can be expressed as

\begin{align} \pi &= 2 \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_n \left(\frac{1}{2}\right)_n}{\left(\frac{3}{2}\right)_n} \frac{1^n}{n!}\\ &=\sum_{n=0}^{\infty} \frac{(2n-1)!!}{(\frac{1}{2}+n)n!2^n} \end{align}

and plotted using a Wolfram Hypergeometric Function:

How to prove that result?

Answer 1

By DLMF 15.6.1 $$ 2F\left({1/2,1/2\atop3/2};1\right)=\int_0^1\frac{\mathrm dt}{\sqrt{t(1-t)}}=\operatorname{B}(1/2,1/2)=\Gamma^2(1/2)=\pi. $$

Answer 2

You just need to transform the arcsin series expansion to the hypergeometric form using some properties of the Pocchammer symbol.

Recall that

$$ \arcsin(x) = \sum_{n=0}^{\infty} \binom{2n}{n}\frac{x^{2n+1}}{2^{2n}(2n+1)}$$

Note that

$$\binom{2n}{n} =\frac{(2n)!}{n!^2} \tag{1}$$

The definition of the Pocchhammer symbol is

$$ (x)_{n}=x(x+1)(x+2)\cdots (x+n-1)$$ So

$$(1)_{n}=(1)(2)(3)\cdots(1+n-1)=n!\tag{2}$$

And $$(2n)! =(1)_{2n} \tag{3}$$

The Pochhammer symbol satisfies the following duplication formula

$$ (x)_{2n}=2^{2n}\left(\frac{x}{2}\right)\left(\frac{x+1}{2}\right)$$

To show this note

\begin{align*}2^n\left(\frac{x}{2}\right)=&2^n\left(\frac{x}{2}\right)\left(\frac{x}{2}+1\right)\cdots\left(\frac{x}{2}+n-1\right)\\ =&\left(x\right)\left(x+2\right)\cdots\left(x+2(n-1)\right) \end{align*}

\begin{align*}2^n\left(\frac{x+1}{2}\right)=&2^n\left(\frac{x+1}{2}\right)\left(\frac{x+1}{2}+1\right)\cdots\left(\frac{x+1}{2}+n-1\right)\\ =&\left(x+1\right)\left(x+3\right)\cdots\left(x+1+2(n-1)\right)\\ \end{align*}

Hence

$$ (1)_{2n} = 2^{2n}\left(\frac{1}{2}\right)_{n}\left(1 \right)_{n} \tag{4}$$

Finally

Note that

$$x(x+1)_{n} = x(x+1)\cdots (x+n)=x(x+1)\cdots(x+n-1)(x+n) =(x)_{n}(x+n)$$ So

$$ x+n = \frac{x(x+1)_n}{(x)_n} $$

Then

$$ 2n+1=\frac{\left(\frac{3}{2}\right)_n}{\left(\frac{1}{2}\right)_n} \tag{5}$$

The proof follows from results (1) to (5).