I'm trying to prove that $J_0(x)+J_2(x)=\frac{2}{x}J_1(x)$, given the following definition: $J_p(x)=\sum_{i=0}^{\infty} \frac{(-1)^n}{\Gamma(n+1)\Gamma(n+p+1)}\frac{x}{2}^{2n+p}$ where $\Gamma(n)$ is the Gamma function. This is my progress so far:
$$(J_0(x)+J_2(x))x/2=\sum_{i=0}^{\infty} \frac{(-1)^n}{\Gamma(n+1)\Gamma(n+2)}\frac{x}{2}^{2n+1}(\frac{x^2}{4(n+3)}+n+2)$$
$$=\sum_{i=0}^{\infty} \frac{(-1)^n}{\Gamma(n+1)\Gamma(n+2)}\frac{x}{2}^{2n+1}\frac{x^2}{4(n+3)}+\sum_{i=0}^{\infty} \frac{(-1)^n}{\Gamma(n+1)\Gamma(n+2)}\frac{x}{2}^{2n+1}n+2J_1(x)$$
What can I do to obtain my original equation?
Thanks a lot