Proving that $zJ''(z)+J'(z)+zJ(z)=0$ In my complex analysis book there is the following problem that I'm having some trouble solving:

Consider the function:
$$J(z)=\sum_{n\geq0} \frac{(-1)^n}{(n!)^2} \left(\frac{z}{2}\right)^{2n}$$
1 - Determine the radius of convergence of this power series.
2 - Show that:
$$zJ''(z)+J'(z)+zJ(z)=0$$

I already solved number one and got that the radius of convergence is $+\infty$, but I'm having some trouble with the second one

My approach:
Lets rewrite:
$$J(z)=\sum_{n\geq0} \frac{(-1)^n}{(n!)^2} \left(\frac{z}{2}\right)^{2n}=\sum_{n\geq0} \frac{(-1)^n}{2^{2n}(n!)^2} z^{2n}$$
We can define a sequence $\alpha_n$ to be:

*

*$1$, if $n = 0$

*$0$, if $n$ is odd

*$\frac{(-1)^{\frac{n}{2}}}{\left(\left(\frac{n}{2}\right)!\right)^2 2^{n}}$, if $n$ is even

Then we end up with:
$$J(z)=\sum_{n \geq 0}\alpha_n z^n$$
So now we know that:

*

*$$zJ(z)=\sum_{n \geq 0}\alpha_n z^{n + 1}$$

*$$J'(z)=\sum_{n \geq 1}n\alpha_n z^{n - 1}=\sum_{n \geq 2}n\alpha_n z^{n - 1}$$

*$$zJ''(z)=\sum_{n \geq 2}n(n - 1)\alpha_n z^{n - 1}$$
So if we sum all these 3 we end up with:
$$ 
zJ''(z)+J'(z)+zJ(z)= $$
$$\sum_{n \geq 2}n(n - 1)\alpha_n z^{n - 1} + \sum_{n \geq 2}n\alpha_n z^{n - 1} + \sum_{n \geq 0}\alpha_n z^{n + 1}=$$
$$=\sum_{n \geq 2} n^2\alpha_n z^{n - 1} + \sum_{n \geq 0}\alpha_n z^{n + 1}=$$
$$=\sum_{n \geq 1} (n + 1)^2\alpha_{n + 1} z^{n} + \sum_{n \geq 1}\alpha_{n - 1} z^{n}=$$
$$=\sum_{n \geq 1} \left[(n + 1)^2\alpha_{n + 1} + \alpha_{n - 1}\right]z^{n}$$
But I don't know how to proceed now. How can I prove that this is equal to $0$?
 A: You've shown $[z^{2n}]J=\frac{(-1/4)^n}{n!^2}$; that's a good start. Since $J$ is even, $zJ^{\prime\prime}+J^\prime+zJ$ is odd. Its $z^{2n+1}$ coefficient is$$\begin{align}[z^{2n}]J^{\prime\prime}+[z^{2n+1}]J^\prime+[z^{2n}]J&=(2n+1)(2n+2)[z^{2n+2}]J^{\prime\prime}+(2n+2)[z^{2n+2}]J+[z^{2n}]J\\&=(2n+2)^2[z^{2n+2}]J+[z^{2n}]J,\end{align}$$which you can verify is $0$.
A: Introducing the $\alpha_n$ is making a lot of extra work. Just compute directly:
Note that $zJ(z) = \sum_{n \ge 0} 2 {(-1)^n \over (n!)^2} ({z \over 2})^{2n+1}$
$J'(z) = \sum_{n \ge 0} (n+1) {- 1 \over (n+1)^2} {(-1)^n \over (n!)^2} ({z \over 2})^{2n+1}$,
$zJ''(z) = \sum_{n \ge 0} 2(n+1)(n+{1 \over 2}) {- 1 \over (n+1)^2} {(-1)^n \over (n!)^2} ({z \over 2})^{2n+1}$.
Hence the coefficient of $({z \over 2})^{2n+1}$ in $zJ(z)+J'(z)+zJ''(z)$ is
${(-1)^n \over (n!)^2}(2 -{1 \over n+1} - {2 (n+{1 \over 2}) \over n+1}  ) = 0$.
A: You are almost done. If $n\geq 1$, then
$$
(2n)^2 \alpha _{2n}  + \alpha _{2n - 2}  = (2n)^2 \frac{{( - 1)^n }}{{n!^2 2^{2n} }} + \frac{{( - 1)^{n - 1} }}{{(n - 1)!^2 2^{2n - 2} }} \\ = \frac{{( - 1)^n }}{{(n - 1)!^2 2^{2n - 2} }} + \frac{{( - 1)^{n - 1} }}{{(n - 1)!^2 2^{2n - 2} }} = 0
$$
and
$$
(2n + 1)^2 \alpha _{2n + 1}  + \alpha _{2n - 1}  = (2n + 1)^2  \cdot 0 + 0 = 0.
$$
