Why is $\sum_{k=0}^{\infty} \frac{k!}{\prod_{j=0}^{k} \left(2j+3\right)} = 2-\frac{\pi}{2}$ and how is this solution derived? I recently came across the problem:
$$\sum_{k=0}^{\infty} \frac{k!}{\prod_{j=0}^{k} \left(2j+3\right)}$$ and decided to try finding its solution. I started off by writing a program that gave me the answer $0.4292036732051...$ which I found probably meant the solution was $2-\frac{\pi}{2}$, but I wanted to see if I could prove it. I started by manipulating the denominator into a double factorial resulting in:
$$\sum_{k=0}^{\infty} \frac{k!}{\left(2k+3\right)!!}$$
Then, I thought things would be easier to work with if I only had regular factorials. This gave me:
$$\sum_{k=0}^{\infty} \frac{4\cdot2^k k! \left(k+2\right)!}{\left(2k+4\right)!}$$
I unfortunately had little clue to go from here considering I haven't dealt with factorials often, much less in infinite series such as this one. I am curious how it is possible to go forth from here and also how to solve infinite series that are similarly structured. Any help would be appreciated.
 A: Starting from
$$4\sum_{k=0}^{\infty} \frac{2^k k! \left(k+2\right)!}{\left(2k+4\right)!} $$Consider
$$4\sum_{k=0}^{\infty}\frac{k!  (k+2)!}{ (2 k+4)!}(2t)^{2k}$$ and, now, the trick is to recognize (not so obvious) that this is
$$\frac{1}{t^2}-\frac{\sqrt{1-t^2} }{t^3}\sin ^{-1}(t)$$ Make $t=\frac 1 {\sqrt 2}$ and get the result.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\sum_{k = 0}^{\infty}{k! \over \prod_{j = 0}^{k}\pars{2j + 3}} & =
\sum_{k = 0}^{\infty}{k! \over 2^{k + 1}\prod_{j = 0}^{k}\pars{j + 3/2}} =
\sum_{k = 0}^{\infty}{k! \over 2^{k + 1}\pars{3/2}^{\overline{k + 1}}}
\\[5mm] & =
\sum_{k = 0}^{\infty}{1 \over 2^{k + 1}}\,{k! \over 
\Gamma\pars{3/2 + k + 1}/\Gamma\pars{3/2}}
\\[5mm] & =
\sum_{k = 0}^{\infty}\,{1 \over 2^{k + 1}}\,
{\Gamma\pars{k + 1}\Gamma\pars{3/2} \over \Gamma\pars{k + 5/2}}
\\[5mm] & =
\sum_{k = 0}^{\infty}{1 \over 2^{k + 1}}\,
\int_{0}^{1}t^{k}\pars{1 - t}^{1/2}\,\dd t
\\[5mm] & =
{1 \over 2}\int_{0}^{1}\root{1 - t}
\sum_{k = 0}^{\infty}\pars{t \over 2}^{k}\,\dd t
\\[5mm] & =
\int_{0}^{1}{\root{1 - t} \over 2 - t}\,\dd t
\,\,\,\stackrel{t\ =\ 1 - x^{2}}{=}\,\,\,
2\int_{0}^{1}\pars{1 - {1 \over 1 + x^{2}}}\,\dd x
\\[5mm] & =
\bbx{2 - {\pi \over 2}}\ \approx\ 0.4292
\end{align}
