I need some help proving that: $$j_2(x)=(\frac{3}{x^3} - \frac{1}{x} )\sin(x) - \frac{3}{x^2} \cos(x)$$ where $$j_m(x)=\sqrt\frac{\pi}{2x} J_{m+\frac{1}{2}}(x)$$
and $$J_n(x)=\sum_{s=0}^\infty\frac{(-1)^{s}}{s!(s+n)!}\left(\frac{x}{2}\right)^{2s+n}$$
After substituting $J_{2+\frac{1}{2}}(x)$ into the formula I get:
$$\sqrt\frac{\pi}{2x}\sum_{s=0}^\infty\frac{(-1)^{s}}{s!(s+\frac{5}{2})!}\left(\frac{x}{2}\right)^{2s+\frac{5}{2}}$$ but I have absolutely no clue as to what I should do next. I know I've pretty much done nothing but I would really appreciate it if someone could tell me where I could find a proof of this or wouldn't mind showing me how to proceed?
You have to analyse closely the denominator $s!(s+5/2)!$.
What does $(s+5/2)!$ mean anyway? It's not a factorial of an integer, so must be interpreted via the gamma function: $(s+5/2)!=\Gamma(s+7/2)$. Using $\Gamma(x+1)=x\Gamma(x)$ and $\Gamma(1/2)=\sqrt\pi$ gives $$(s+5/2)!=(s+5/2)(s+3/2)\cdots(1/2)\sqrt\pi =\frac{1\cdot3\cdot5\cdot(2s+5)}{2^{s+3}}\sqrt\pi =\frac{(2s+5)!}{2^{2s+5}(s+2)!}\sqrt\pi.$$ For $j_{5/2}(x)$ you then get something like $$4\sum_{s=0}^\infty\frac{(-1)^s(s+1)(s+2)}{(2s+5)!}x^{2s+2}$$ which ought to be convertible to a combination of sines/cosines.