Now, $\mathrm{J_n(x)=\displaystyle\sum_{m=0}^\infty\dfrac{(-1)^m~(x/2)^{2m+n}}{m!~|\overline {n+m+1}}}$

Putting n=1/2, we get,

$\mathrm{J_{1/2}(x)=\displaystyle\sum_{m=0}^\infty\dfrac{(-1)^m~(x/2)^{2m+(1/2)}}{m!~|\overline {m+(3/2)}}}$

But $\mathrm{\Bigg|\overline{m+\dfrac{3}{2}}=\bigg(m+\dfrac{1}{2}\bigg)\bigg(m-\dfrac{1}{2}\bigg)\bigg(m-\dfrac{3}{2}\bigg)\cdots \cdots\bigg(\dfrac{3}{2}\bigg)\bigg(\dfrac{1}{2}\bigg)\Bigg|\overline{\dfrac{1}{2}}}$

$\mathrm{\Bigg|\overline{m+\dfrac{3}{2}}=\dfrac{(2m+1)(2m-1)(2m-3) \cdots \cdots 3\cdot1\cdot \sqrt\pi}{2^{m+1}}}$

Multiplying and dividing by $\mathrm{2\cdot 4\cdot 6\cdots \cdots (2m-2)(2m)}$in the numerator and denominator on r.h.s.

$\mathrm{\Bigg|\overline{m+\dfrac{3}{2}}=\dfrac{(2m+1)(2m)(2m-1) \cdots \cdots 3\cdot 2\cdot 1}{2^{m+1}\cdot 2 \cdot 4\cdot 6\cdots \cdots (2m-2)(2m)}\cdot \sqrt\pi}$

$\mathrm{=\dfrac{1\cdot 2\cdot 3 \cdots \cdots(2m-1)(2m)(2m+1)}{2^{m+1}\cdot 2^m\cdot 1 \cdot2\cdot 3\cdots \cdots (m-1)m}\cdot \sqrt\pi}$

$\mathrm{=\dfrac{(2m+1)!}{2^{2m+1}\cdot m!}\cdot \sqrt\pi}$

$\mathrm{J_{1/2}(x)=\displaystyle\sum_{m=0}^\infty\dfrac{(-1)^m~2^{2m+1}\cdot m!}{m!~(2m+1)!}\cdot \dfrac{1}{\sqrt\pi}\cdot \dfrac{x^{2m+(1/2)}}{2^{2m+(1/2)}}}$

$\mathrm{\therefore J_{1/2}(x)=\displaystyle\sum_{m=0}^\infty\dfrac{(-1)^m\cdot x^{2m+1}}{(2m+1)!}\cdot \dfrac{x^{-1/2}}{\sqrt\pi}\cdot 2^{1/2}}$

$\mathrm{\therefore J_{1/2}(x)=\sqrt{\dfrac{2}{\pi x}}\cdot \Bigg(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}- \cdots\cdots\Bigg)=\sqrt{\dfrac{2}{\pi x}}\cdot sinx}$