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.