How to show? $${}_2F_1(1,1;\frac{1}{2}, \frac{1}{2}) = 2 + \frac{\pi}{2} $$

It numerically is very close, came up when evaluating:

$$ \frac{1}{1} + \frac{1 \times 2}{1 \times 3} + \frac{1 \times 2 \times 3}{1 \times 3 \times 5} + \frac{1 \times 2 \times 3 \times 4}{1 \times 3 \times 5 \times 7} + \cdots = 1 + \frac{\pi}{2} ~ ?$$

I am missing something trivial, I am sure.


This is equivalent to showing that $\displaystyle\sum_{n=2}^\infty\frac{n!}{(2n-1)!!}=\frac\pi2$ , which, after multiplying both the

numerator and the denominator with $(2n)!!=2^n\,n!$, and taking into account that $\dfrac{(2n)!}{n!^2}=$

$=\displaystyle{2n\choose n}$, can be rewritten as $\displaystyle\sum_{n=2}^\infty\frac{2^n}{\displaystyle{2n\choose n}}=\frac\pi2$ , which can ultimately be deduced from the

wider identity $\displaystyle\sum_{n=1}^\infty\frac{(2x)^{2n}}{\displaystyle{2n\choose n}n^2}=2\arcsin^2x$, itself ultimately a consequence of integrating
the Cauchy product between the binomial series expansion of $\arcsin'x=\dfrac1{\sqrt{1-x^2}}$ and

that of its primitive. Hope this helps.


You may use the general formula from Brychkov, Prudnikov, Marichev, Integral and Series, Vol.3: $$ _{2}F_{1}(1,1;\frac{1}{2};x)=(1-x)^{-1}\left(1+\frac{\sqrt{x}\arcsin{\sqrt{x}}}{\sqrt{1-x}}\right) $$ and put in it $x=\frac{1}{2}$.


Your result can be proved very easily by using a result given in Brychkov, Prudnikov, Marichev, Integral and Series, Vol.3, p-91, $ by taking a = b = 1. For more results of this type, send an e-mail to me for my research paper. All the very best, Arjun K. Rathie, India, e-mail address: akrathie@gmail.com


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