Using the notation for the Bessel function as $J_n(z)=\sum \limits_{k=0}^{\infty}\frac{(-1)^kz^{n+2k}}{k!(n+k)!2^{n+2k}}$, I want to show that $w=J_n(z)$ satisfies $w''+\frac{1}{z}w'+\left(1-\frac{n^2}{z^2}\right)w=0$ by first showing it is analytic everywhere and then that it satisfies the equation.
To show it is analytic everywhere, I used the ratio test to show that the ratio of convergence is $R=\infty$. Then I want to show that $$\lim_{k \to \infty}\left|\frac{(-1)}{4(k+1)(n+k+1)}\right|<1$$ for all $n \geq0$, which is clear. Is this enough?
To show it satisfies the Bessel equation, I differentiated $w=J_n(z)$ to find $$w'=\sum \limits_{k=1}^{\infty}\frac{(n+2k)(-1)^kz^{n+2k-1}}{k!(n+k)!2^{n+2k}}\qquad w''=\sum \limits_{k=2}^{\infty}\frac{(n+2k)(n+2k-1)(-1)^kz^{n+2k-2}}{k!(n+k)!2^{n+2k}}$$ which makes the Bessel equation $$\sum \limits_{k=2}^{\infty}\frac{(n+2k)(n+2k-1)(-1)^kz^{n+2k-2}}{k!(n+k)!2^{n+2k}}+\sum \limits_{k=1}^{\infty}\frac{(n+2k)(-1)^kz^{n+2k-2}}{k!(n+k)!2^{n+2k}}+\sum \limits_{k=1}^{\infty}\frac{(-1)^{k-1}z^{n+2k-2}}{(k-1)!(n+k-1)!2^{n+2k-2}}-\sum \limits_{k=0}^{\infty}\frac{(-1)^kn^2z^{n+2k-2}}{k!(n+k)!2^{n+2k}}$$ If I can put all the coefficients together, they become $(n+2k)(n+2k-1)+(n+2k)-4k(n+k)-n^2=0$, but I'm not sure how to make the limits of the sums allow me to do that. If I put all the summands with limits $k=1$, I end up with $$\sum\limits_{k=1}^{\infty}(0)z^{n+2k-2}+\frac{(n+2)(n+1)z^n}{(n+1)!2^{n+2}}-\frac{n^2z^{n-2}}{n!2^n}=\frac{(n+2)z^n}{n!2^{n+2}}-\frac{n^2z^{n-2}}{n!2^n}$$ (where the first term comes from subtracting the term in the first summand when $k=1$ and the second comes from adding the term in the last summand when $k=0$), which needs to equal $0$ for all $n\geq0$ if the equations are satisfied. However, it doesn't seem to. What have I done wrong?