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.
 A: First, a small diversion: consider the general alternating series $S = \sum_{n=0}^\infty (-1)^na_n$.  Then this can formally be written as $\displaystyle S = \sum_{n=0}^\infty (-1)^n E^n a_0 = \left(\frac1{1+E}\right)a_0$, where $E$ is the successor operator, $Ea_n = a_{n+1}$.  Now, we have the formal identity $E=1+\Delta$, where $\Delta$ is the difference operator, $\Delta a_n = (a_{n+1}-a_n)$; thus, $S$ can be expressed as $\displaystyle S=\left(\frac1{1+E}\right)a_0$ $\displaystyle= \left(\frac1{2+\Delta}\right)a_0$ $\displaystyle =\sum_{n=0}^\infty\frac{(-1)^n\Delta^na_0}{2^{n+1}}$  This is known as Euler's Transform; these are all formal manipulations, of course, but they can be proven using what's known as operator or Umbral calculus.
Now, consider the classic series for pi, $\displaystyle\frac\pi4 = \arctan(1) = \sum_{n=0}^\infty\frac{(-1)^n}{2n+1}$.  Then we have $a_k = \frac1{2k+1}$, so $\Delta a_k = \dfrac1{2k+3}-\dfrac1{2k+1} = \dfrac{-2}{(2k+1)(2k+3)}$, $\Delta^2 a_k = \dfrac{-2}{(2k+3)(2k+5)}-\dfrac{-2}{(2k+1)(2k+3)} = \dfrac{(-2)\cdot(-4)}{(2k+1)(2k+3)(2k+5)}$, and generically $\Delta^na_k = \dfrac{(-1)^n2^nn!}{(2k+1)(2k+3)\cdots(2k+2n+1)}$, so $\Delta^na_0 = \dfrac{(-1)^n2^nn!}{(2n+1)!!}$.  Plugging this in to the Euler transform formula, we get $\displaystyle\frac\pi4=\sum_{n=0}^\infty\frac{(-1)^n}{2^{n+1}}\cdot\frac{(-1)^n2^nn!}{(2n+1)!!}$ $\displaystyle=\sum_{n=0}^\infty \frac{n!}{2\cdot(2n+1)!!}$.  Multiplying this by 4 yields the original sum.
A: 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}
A: Notice the summands can be expressed in terms of Beta function. We can use this to convert the series into an integral.
$$\sum_{n=0}^\infty\frac{2\cdot n!}{(2n+1)!!}
= \sum_{n=0}^\infty\frac{2^{n+1}(n!)^2}{(2n+1)!}
= \sum_{n=0}^\infty 2^{n+1}\frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)}\\
= \sum_{n=0}^\infty 2^{n+1} \int_0^1 t^n(1-t)^n dt
= 2 \int_0^1 \frac{dt}{1-2t(1-t)}
$$
Now let $x = 2t-1$, we can evaluate the integral as
$$\int_{-1}^1 \frac{dx}{1-(1+x)\left(\frac{1-x}{2}\right)}
= 2\int_{-1}^{1}\frac{dx}{1+x^2} = 2\left[\tan^{-1}(x)\right]_{-1}^1 = 4\tan^{-1}(1) = \pi$$
