TITLE: How to sum this series for $\pi/2$ directly?
QUESTION [110 upvotes]: The sum of the series
$$
\frac{\pi}{2}=\sum_{k=0}^\infty\frac{k!}{(2k+1)!!}\tag{1}
$$
can be derived by accelerating the Gregory Series
$$
\frac{\pi}{4}=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\tag{2}
$$
using Euler's Series Transformation. Mathematica is able to sum $(1)$, so I assume there must be some method to sum the series in $(1)$ directly; what might that method be?

REPLY [4 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\sum_{k = 0}^{\infty}{k! \over
\pars{2k + 1}!!}} =
\sum_{k = 0}^{\infty}{k! \over \prod_{j = 0}^{k}\pars{2j + 1}}
\\[5mm] = &\
\sum_{k = 0}^{\infty}{k! \over 2^{k + 1}
\prod_{j = 0}^{k}\pars{j + 1/2}} =
{1 \over 2}\sum_{k = 0}^{\infty}{k! \over
\pars{1/2}^{\overline{k + 1}}}\,\pars{1 \over 2}^{k}
\\[5mm] = &\
{1 \over 2}\sum_{k = 0}^{\infty}{k! \over
\Gamma\pars{k + 3/2}/\Gamma\pars{1/2}}\,\pars{1 \over 2}^{k}
\\[5mm] = &\
{1 \over 2}\sum_{k = 0}^{\infty}{\Gamma\pars{1/2}\Gamma\pars{k + 1} \over
\Gamma\pars{k + 3/2}}\,\pars{1 \over 2}^{k}
\\[5mm] = &\
{1 \over 2}\sum_{k = 0}^{\infty}\pars{1 \over 2}^{k}
\int_{0}^{1}t^{-1/2}\pars{1 - t}^{k}\,\dd t
\\[5mm] = &\
{1 \over 2}\int_{0}^{1}t^{-1/2}\sum_{k = 0}^{\infty}
\pars{1 - t \over 2}^{k}\,\dd t
\\[5mm] = &\
{1 \over 2}\int_{0}^{1}t^{-1/2}\,
{1 \over 1 - \pars{1 - t}/2}\,\dd t
=
\int_{0}^{1}{t^{-1/2} \over 1 + t}\,\dd t
\\[5mm] \stackrel{t\ \mapsto\ t^{2}}{=}\,\,\,&
2\int_{0}^{1}{\dd t \over 1 + t^{2}}\,\dd t
=
2\,{\pi \over 4} = \bbx{\pi \over 2} \\ &
\end{align}