TITLE: How to prove this $\pi$ formula?
QUESTION [24 upvotes]: I am hoping to find out where the formula 
$$\frac{\pi}{2}=\sum_{k=0}^{\infty}\frac{k!}{\left(2k+1\right)!!}$$
comes from. I can't see how one could begin to prove it.

REPLY [25 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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It's well known that
$\ds{\pars{2k + 1}!! = {\pars{2k + 2}! \over 2^{k + 1}\pars{k + 1}!}}$ such that

\begin{align}
\sum_{k = 0}^{\infty}{k! \over \pars{2k + 1}!!} & =
\sum_{k = 0}^{\infty}{k!\pars{k + 1}! \over \pars{2k + 2}!}\,2^{k + 1} =
\sum_{k = 0}^{\infty}{\Gamma\pars{k + 1}\Gamma\pars{k + 2} \over
\Gamma\pars{2k + 3}}\,2^{k + 1}
\\[5mm] & =
\sum_{k = 0}^{\infty}2^{k + 1}\int_{0}^{1}x^{k}\pars{1 - x}^{k + 1}\,\dd x =
2\int_{0}^{1}\pars{1 - x}\sum_{k = 0}^{\infty}\bracks{2x\pars{1 - x}}^{\,k}
\,\dd x
\\[5mm] & =
2\int_{0}^{1}\pars{1 - x}{1 \over 1 - 2x\pars{1 - x}}\,\dd x =
\int_{0}^{1}{1 - x \over x^{2} - x + 1/2}\,\dd x =
\int_{-1/2}^{1/2}{1/2 - x \over x^{2} + 1/4}\,\dd x
\\[5mm] & =
2\int_{0}^{1}{\dd x \over x^{2} + 1} = 2\arctan\pars{1} = 2\,{\pi \over 4} = \bbx{\ds{\pi \over 2}} \\ &
\end{align}