Series Question: $\sum_{n=0}^\infty\frac{2n!}{(2n+1)!!}=\pi$ How to prove:
$$\sum_{n=0}^\infty\frac{2n!}{(2n+1)!!}=\pi$$
I tried to use property of double factorial
$$(2n+1)!!=\frac{(2n+1)!}{2^nn!}$$
then
$$\frac{2n!}{(2n+1)!!}=\frac{2^{n+1}(n!)^2}{(2n+1)!}$$
The next step I am stuck. Any help would be appreciated. Thanks in advance.

Consider the series
\begin{align}
S = \sum_{n=0}^{\infty} \frac{ 2 \ n!}{(2n+1)!!} = 2 \sum_{n=0}^{\infty} \frac{ 2^{n} (n!)^{2} }{(2n+1)!} 
\end{align}
which can quickly be cast into the form
\begin{align}
S = 2 \sum_{n=0}^{\infty} \frac{ (1)_{n} (1)_{n} }{ n! (3/2)_{n} \ 2^{n} }
\end{align}
which yields the hypergeometric form
\begin{align}
S = 2 {}_{2}F_{1} ( 1, 1; 3/2; 1/2).
\end{align}
Using the formulas
\begin{align}
{}_{2}F_{1}(1,1; c; 1/2) &= (c-1) \left[ \psi\left(\frac{c}{2}\right) - \psi\left( \frac{c-1}{2}\right) \right] \\
\psi\left(\frac{3}{4} \right) &= \frac{\pi}{2} - 3 \ln 2 - \gamma \\
\psi\left(\frac{1}{4} \right) &= - \frac{\pi}{2} - 3 \ln 2 - \gamma
\end{align}
then the series becomes
\begin{align}
S &= \psi\left(\frac{3}{4}\right) - \psi\left( \frac{1}{4}\right) = \pi.
\end{align}
Hence
\begin{align}
\sum_{n=0}^{\infty} \frac{ n!}{(2n+1)!!} = \frac{\pi}{2}.
\end{align}
