Zero/First order Bessel function (of first kind) identity proof

Question

Here's my attempt to establish that $\dfrac{d}{dx}J_0(x)=-J_1(x)$: $$\begin{align*} J_0(x)&=\sum_{k=0}^{\infty}\frac{(-1)^k}{k! \Gamma(k+1)}\left(\frac{x}{2}\right)^{2k}\\\\ \frac{d}{dx}J_0(x)&=\frac{d}{dx}\left[\sum\cdots\right]\\\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{k!\Gamma(k+1)}\frac{d}{dx}\left[\left(\frac{x}{2}\right)^{2k}\right]\\\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{k!\Gamma(k+1)}2k\left(\frac{x}{2}\right)^{2k-1}\frac{d}{dx}\left[\frac{x}{2}\right]\\\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{(k-1)!\Gamma(k+1)}\left(\frac{x}{2}\right)^{2k-1}\\\\ &=\sum_{p+1=0}^\infty\frac{(-1)^{p+1}}{((p+1)-1)!\Gamma((p+1)+1)}\left(\frac{x}{2}\right)^{2(p+1)-1}\\\\ &=-\sum_{p=-1}^\infty\frac{(-1)^{p}}{p!\Gamma(p+2)}\left(\frac{x}{2}\right)^{2p+1} \end{align*}$$ Regarding the last step, how can I show that the $-1$-th term of the series is $0$? Considering the plot of the gamma function, I figured I could say something along the lines of $p!=\Gamma(p+1)\to+\infty$ as $p\to-1^+$, but I have this gut feeling that this sort of reasoning wouldn't hold up.

Answer 1

The first term of the $J_0$ series is a constant, so its derivative is zero, and hence your sum actually starts at $k=1$ before you reindex it.