<s>[INST] Hypergeometric sum specific value 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.
 [/INST] 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.
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