Use the series definition $$\sum_{k=0}^{\infty} \frac{(-1)^k(z/2)^{2k+v}}{k!\Gamma (1+k+v)}$$ to show that $$\left(\frac{1}{z}\frac{d}{dz}\right)(z^{-v}J_v(z))=-z^{-v-1}J_{v+1}(z)$$
I have shown that $$\left(\frac{1}{z}\frac{d}{dz}\right)(z^{-v}J_v(z))=\sum_{k=0}^{\infty} \frac{(-1)^k}{(k-1)!\Gamma (1+k+v)} \frac{z^{2k-2}}{2^{2k+v-1}}$$ But when I calculated the RHS, $$-z^{-v-1}J_{v+1}(z)=-z^{-v-1}\sum_{k=0}^{\infty} \frac{(-1)^k}{k!\Gamma (2+k+v)} \frac{z^{2k+v+1}}{2^{2k+v+1}}=\sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{k!\Gamma (2+k+v)} \frac{z^{2k}}{2^{2k+v+1}}$$ this does not seem to equal the LHS. Any advice would be greatly appreciated.
Notice that in your LHS-result the sum actually starts at $k=1$, since under the derivative the $k=0$ term is constant and thus vanishes. After applying the derivative you have to shift $k \to k + 1$ in order to compare to RHS.