For the following series representation of the Bessel function:
$w = J_n = \sum_{k=0}^{\infty} \frac{(-1)^k z^{n+2k}}{k!(n+k)!2^{n+2k}}.$
I want to show that w is indeed the Bessel function, such that $w'' + \frac{1}{z}w' + (1-\frac{n^2}{z^2})w = 0$, with the series definition.
I got the following first and second derivatives:
$w' = \sum_{k=1}^{\infty} \frac{(-1)^k (n+2k) z^{n+2k-1}}{k!(n+k)!2^{n+2k}},$
and
$w'' = \sum_{k=1}^{\infty} \frac{(-1)^k (n+2k) (n+2k-1) z^{n+2k-2}}{k!(n+k)!2^{n+2k}}.$
I tried summing the equations by brute force, which got me as far as
$w'' + \frac{1}{z}w' + (1-\frac{n^2}{z^2})w = \sum_{k=1}^{\infty} \frac{(-1)^k (4nk+4k^2) z^{n+2k-2}+(-1)^k z^{n+2k}}{k!(n+k)!2^{n+2k}} + (1-\frac{n^2}{z^2})(\frac{z^n}{n!2^n}),$ but that doesn't seem to equal zero.
Any ideas much appreciated.