How to sum this series for $\pi/2$ directly? 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?
 A: First, $$(2k+1)!! = (2k+1)(2k-1) \cdots (1) = \frac{(2k+1)!}{(2k)(2(k-1)) \cdots 2(1)} = \frac{(2k+1)!}{2^k k!}.$$
So your sum can be rewritten as
$$\sum_{k=0}^\infty\frac{k! \, k! \, 2^k }{(2k+1)!} = \sum_{k=0}^\infty\frac{2^k}{(2k+1)\binom{2k}{k}}.$$
Variations of the sum of reciprocals of the central binomial coefficients have been well-studied.  For example, this paper by Sprugnoli (see Theorem 2.4) gives the ordinary generating function of $a_k =  \frac{4^k}{(2k+1)}\binom{2k}{k}^{-1}$ to be 
$$A(t) = \frac{1}{t} \sqrt{\frac{t}{1-t}} \arctan \sqrt{\frac{t}{1-t}}.$$
Subbing in $t = 1/2$ says that $$\sum_{k=0}^\infty\frac{2^k}{(2k+1)\binom{2k}{k}} = 2 \arctan(1) = \frac{\pi}{2}.$$
A: We can prove this identity, as well as the corresponding power series identities by using a relation with the Beta function.  Rearranging as done in Mike Spivey's answer we are looking at $$ \sum_{k=0}^\infty\frac{k! k! 2^k}{(2k+1)!}$$  Using induction or a Beta Function identity, we can show that $$\int_0^1 x^{k}(1-x)^k=\frac{k!k!}{(2k+1)!}.$$  Hence your sum becomes 
$$ \sum_{k=0}^\infty 2^k \int_0^1 x^{k}(1-x)^k=\int_0^1 \left(\sum_{k=0}^\infty 2^k x^k (1-x)^k\right)dx.$$ 
Notice that since $0\leq x\leq 1$, $x(1-x)\leq \frac{1}{4}$ and the series converges absolutely.  Summing gives
$$=\int_0^1 \frac{1}{1-2x(1-x)}dx=\int_0^1 \frac{1}{x^2+(1-x)^2}dx$$  Substituting $u=\frac{1}{x}$, and then $v=u-1$, we see that this integral is equal to $$\int_1^\infty \frac{1}{1+(u-1)^2}du=\int_0^\infty \frac{1}{1+v^2}dv=\frac{\pi}{2},$$ as desired.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\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}
A: I had intended for this to be a comment to Mike Spivey's answer, but it is too long.
One of the answers to the related question mentions a result equivalent to
$$
\int_0^\frac{\pi}{2}\sin^{2k+1}(x)\;\mathrm{d}x=\frac{2k}{2k+1}\frac{2k-2}{2k-1}\cdots\frac{2}{3}=\frac{1}{2k+1}\frac{4^k}{\binom{2k}{k}}\tag{1}
$$
Using $(1)$, my sum becomes
$$
\begin{align}
\sum_{k=0}^\infty\frac{k!}{(2k+1)!!}
&=\sum_{k=0}^\infty\frac{2^k}{(2k+1)\binom{2k}{k}}\\
&=\sum_{k=0}^\infty\int_0^\frac{\pi}{2}\sqrt{2}\left(\frac{\sin(x)}{\sqrt{2}}\right)^{2k+1}\mathrm{d}x\\
&=\sqrt{2}\int_0^\frac{\pi}{2}\frac{\left(\frac{\sin(x)}{\sqrt{2}}\right)}{1-\left(\frac{\sin(x)}{\sqrt{2}}\right)^2}\;\mathrm{d}x\\
&=\int_0^\frac{\pi}{2}\frac{2\,\sin(x)}{2-\sin^2(x)}\;\mathrm{d}x\\
&=\int_\frac{\pi}{2}^0\frac{2\;\mathrm{d}\cos(x)}{1+\cos^2(x)}\\
&=\frac{\pi}{2}
\end{align}
$$
A: Notice that for $c_k = \frac{k!}{(2k+1)!!}$ the ratio of successive terms $\frac{c_{k+1}}{c_k} = \frac{k+1}{2k +3} = \frac{1}{2} \frac{k+1}{k+3/2}$.
This means that the series is hypergeometric with the value ${}_2 F_1(1, 1, \frac{3}{2}, \frac{1}{2})$.
This particular Gaussian hypergeometric is elementary:
$$
   {}_2 F_1(1, 1, \frac{3}{2}, x) = \frac{\arcsin\left(\sqrt{x}\right)}{\sqrt{1-x} \sqrt{x}}
$$
Upon substitution of $x=\frac{1}{2}$ we recover the result $ 2 \arcsin(\frac{1}{\sqrt{2}}) = \frac{\pi}{2}$.
