Showing Bessel function of first kind (order 1/2) is $J_{1/2}(x)=\sqrt{\frac{2}{\pi x}}\sin(x)$ The bessel function of the first kind with order $p$ is 
$$J_p(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+p+1)}\Big(\frac{x}{2}\Big)^{2n+p}$$
What I'm thinking about is finding a way to get $\sqrt{\frac{2}{\pi x}}$, factor it out, and then reduce the remaining term as the Taylor series representation of the sine function. But I haven't been able to achieve this.
With $p=1/2$ we have that $$\Gamma(n+p+1)=\frac{(2n+1)!!}{(2)^n}\sqrt{\pi}$$ and my overall remaining expression has been so far: $$\frac{(-1)^n (2)^n}{n!(n+1/2)(2n+1)!!\sqrt{\pi}} \left( \frac{x^{2n+1}}{2^{2n+1/2}x^{1/2}}\right)$$
Which looks like a big mess and is not going to reduce to the Taylor series representation of the sine function.
Any ideas? Anyone have a source were they carry out this demonstration?
Thanks a lot.
 A: From the representation you have
$$J_{1/2}(x) = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{n! \Gamma(n+3/2)} \left(\dfrac{x}2\right)^{2n+1/2} \implies \sqrt{\dfrac{x}2}J_{1/2}(x) = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{n! \Gamma(n+3/2)} \left(\dfrac{x}2\right)^{2n+1} \, (\spadesuit)$$
We have
$$\Gamma(n+3/2) = \left(n+\dfrac12\right)\left(n-\dfrac12\right) \cdots \dfrac12 \sqrt{\pi} = \dfrac{(2n+1)(2n-1)\cdots 3 \cdot 1}{2^{n+1}} \sqrt{\pi} = \dfrac{(2n+1)!}{2^{2n+1}n!} \sqrt{\pi}$$
Hence,
$$2^{2n+1} n! \Gamma(n+3/2) = (2n+1)! \sqrt{\pi}$$
Plugging this in $(\spadesuit)$, we obtain
$$\sqrt{\dfrac{x}2}J_{1/2}(x) = \dfrac1{\sqrt{\pi}}\sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n+1}}{(2n+1)!} = \dfrac1{\sqrt{\pi}}\sin(x)$$

Also, avoid the double factorial notation. It is a very very poor notation.
A: Your main tool can be the duplication formula for the Gamma function,
$$ \Gamma(z)\Gamma(z+1/2) = 2^{1-2z}\sqrt{\pi}\Gamma(2z), $$
which you can use on the denominator of your expression. It should then fall almost straight out.
