getting sine from Bessel function of order 1/2

Question

I am trying derive the following relationship $$ J_{1/2}(x) = \sqrt{\frac{2}{\pi x}} \sin x, $$ starting with $$ J_{\nu} = \left(\frac{x}{2}\right)^\nu \sum_{k=0}^\infty \frac{(-1)^k}{k!\,\Gamma\left(k+\nu+1\right)}\left(\frac{x}{2}\right)^{2k}, $$ where $J_{1/2}(x)$ is the Bessel function of the first kind of order $1/2$. Letting $\nu = 1/2$, this becomes $$ J_{1/2}(x) = \sqrt{\frac{x}{2}} \sum_{k=0}^\infty \frac{(-1)^k}{k!\,\Gamma\left(k+\frac{3}{2}\right)}\left(\frac{x}{2}\right)^{2k}. $$ Since $\Gamma(k+3/2) = \Gamma(k+1/2)/2$, this becomes $$ J_{1/2}(x) = 2\sqrt{\frac{x}{2}} \sum_{k=0}^\infty \frac{(-1)^k}{k!\,\Gamma\left(k+\frac{1}{2}\right)}\left(\frac{x}{2}\right)^{2k}. $$ Now use the following property of gamma functions $$ \Gamma\left(k + \frac{1}{2}\right) = \frac{(2k)!}{4^k k!} \sqrt{\pi}, $$ we get $$ J_{1/2}(x) = \frac{2}{\sqrt{\pi}}\sqrt{\frac{x}{2}} \sum_{k=0}^\infty \frac{(-1)^k 4^k k!}{k!(2k)!}\left(\frac{x}{2}\right)^{2k}, $$ or $$ J_{1/2}(x) = \frac{2}{\sqrt{\pi}}\sqrt{\frac{x}{2}} \sum_{k=0}^\infty \frac{(-1)^k}{(2k)!}x^{2k}. $$ Now multiply the numerator and denominator by $x$ and you get $$ J_{1/2}(x) = \sqrt{\frac{2}{\pi x}} \sum_{k=0}^\infty \frac{(-1)^k}{(2k)!}x^{2k+1}. $$ So, I get almost the Taylor series for sine, except the denominator in the sum needs to be $(2k+1)!$, not $(2k)!$. What am I getting wrong here?