TITLE: Series Question: $\sum_{n=0}^\infty\frac{2n!}{(2n+1)!!}=\pi$
QUESTION [6 upvotes]: 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.

REPLY [8 votes]: 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.