Why does $\sum_{n=0}^\infty \frac{n!}{(2n+1)!!}$ converge to $\frac{\pi}{2}$ Why does:

$$\sum_{n=0}^\infty \frac{n!}{(2n+1)!!}$$

converge to $\frac{\pi}{2}$?

I'm honestly not sure where to start.
$$\sum_{n\geq 0}\frac{n!}{(2n+1)!!}=\sum_{n\geq 0}\frac{2^n n!^2}{(2n+1)\cdot (2n)!}=\sum_{n\geq 0}\frac{2^n}{(2n+1)\binom{2n}{n}}=\sum_{n\geq 0}\frac{2^n \Gamma(n+1)^2}{\Gamma(2n+2)}$$
can be written, through Euler's Beta function, also as
$$ \sum_{n\geq 0}2^n\int_{0}^{1}x^n(1-x)^n\,dx = \int_{0}^{1}\frac{dx}{1-2x(1-x)}\stackrel{\text{symmetry}}{=}2\int_{0}^{1/2}\frac{2\,dx}{1+(2x-1)^2} $$
or, through the substitution $x=\frac{1-z}{2}$, as
$$ \int_{0}^{1}\frac{2\,dz}{1+z^2}=2\arctan(1)=\color{red}{\frac{\pi}{2}}.$$



As an alternative, you may notice your series (related to the Taylor series of the arcsine function) is half the Euler transform of Gregory series. See pages 20-21 of my notes, for instance.
What about finding a function $f$ for which there exists an $x$ such that $f(x)=\frac{\pi}{2}$, calculating its Taylor series and using the Euler Transform for series?

Start with $a\tan(1)=\frac{\pi}{4}$ and search for the Taylor series of $a\tan(x)$:
$$a\tan(x)=f(x) = \sum _{n=0}^{\infty } \frac{(-1)^n}{(2n+1)}x^n.$$

Looking at the Euler transform  gives us $$ \left(\frac{1}{1-x}\right)f\left(\frac{x}{1-x}\right) = \sum _{n=0}^{\infty } \left(\sum _{k=0}^n {n \choose k}\frac{(-1)^k}{(2k+1)}\right)x^n.$$

Using that  $$\sum _{k=0}^n {n \choose k}\frac{(-1)^k}{(2k+1)} = \frac{(2n)!!}{(2n+1)!!}$$

We see that the series $$ \sum_{n=0}^{\infty} \frac{n!}{(2n+1)!!}  =\sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} \left(\frac{1}{2}\right)^n = 
 2f(1)=\frac{\pi}{2}$$