The Bessel function of the first kind and order $p$ is given by: $$ J_{p}(x)= \sum_{n=0}^{\infty}\frac{(-1)^n}{n!\, \Gamma(n+p+1)}\left(\frac{x}{2}\right)^{2n+p} $$
I want to show that $J_2(x) = (2/x)J_1(x)-J_0(x)$. Here is what I have tried:
$$ \left(\frac{x}{2}\right)^{-1}J_1(x)-J_0(x) = \left(\frac{x}{2}\right)^{-1}\sum_{n=0}^{\infty}\frac{(-1)^n}{n!\, \Gamma(n+2)}\left(\frac{x}{2}\right)^{2n+1} - \sum_{n=0}^{\infty}\frac{(-1)^n}{n!\, \Gamma(n+1)}\left(\frac{x}{2}\right)^{2n} = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\left[\frac{1}{\Gamma(n+2)}-\frac{1}{\Gamma(n+1)} \right]\left(\frac{x}{2}\right)^{2n} = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\left[\frac{-n}{\Gamma(n+2)} \right]\left(\frac{x}{2}\right)^{2n} $$
But I don't see how I could manipulate this expression to get: $$ J_{2}(x)= \sum_{n=0}^{\infty}\frac{(-1)^n}{n!\, \Gamma(n+3)}\left(\frac{x}{2}\right)^{2n+2} $$
I believe you are correct in arriving at $$\frac{2}{x}J_1(x)-J_0(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\left[\frac{-n}{\Gamma(n+2)} \right]\left(\frac{x}{2}\right)^{2n}$$ The next thing to observe is that the entire sum is equal to zero when $n=0$ because of the $-n$ that appears in the numerator. Thus $$\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\left[\frac{-n}{\Gamma(n+2)} \right]\left(\frac{x}{2}\right)^{2n} = \sum_{n=1}^{\infty}\frac{(-1)^n}{n!}\left[\frac{-n}{\Gamma(n+2)} \right]\left(\frac{x}{2}\right)^{2n} \\ = \sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{(n+1)!}\left[\frac{-(n+1)}{\Gamma(n+3)} \right]\left(\frac{x}{2}\right)^{2(n+1)} \\ = \sum_{n=0}^{\infty}(-1)^{n+1}(-1)\frac{(n+1)}{(n+1)!}\left[\frac{1}{\Gamma(n+3)} \right]\left(\frac{x}{2}\right)^{2n+2} \\ =\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\left[\frac{1}{\Gamma(n+3)} \right]\left(\frac{x}{2}\right)^{2n+2}$$ And you get your result.