I have this Ode $zS''+S'+zS=0$ $z\in\mathbb{C}$ and $S$ a 2 times differentiable function. I was looking into the formal power series $S(X)=\sum_{n\geq0}a_nX^n$ that verify this ODE and i'm supposed to show that the convergence radius is infinite.
I did some calculation and got this $$z\sum_{n \geq 2} n(n-1)a_nX^{n-2} +\sum_{n \geq 1} na_nX^{n-1} +z\sum_{n\geq 0} a_nX^n =0 \\ z\sum_{n\geq 0} (n+2)(n+1)a_{n+2}X^n+\sum_{n\geq 0} (n+1)a_{n+1}X^n+z\sum_{n\geq 0} a_nX^n =0 $$
By unicity of the coefficients $$z((n+2)(n+1)a_{n+2}+a_n)+(n+1)a_{n+1}=0$$
And i'm stuck here. I've tried separating the real and imaginary part, i get $$x((n+2)(n+1)a_{n+2}+a_n)+(n+1)a_{n+1}=0\\ y((n+2)(n+1)a_{n+2}+a_n)=0$$
But i don't know what to do to get further. the imaginary oe gets me a recursive relation : $$a_{n+2}=-\dfrac{a_n}{(n+2)(n+1)}$$
I think your solution needs correction and you should try this $$z\sum_{n \geq 2} n(n-1)a_nz^{n-2} +\sum_{n \geq 1} na_nz^{n-1} +z\sum_{n\geq 0} a_nz^n =0 $$ $$\sum_{n \geq 2} n(n-1)a_nz^{n-1} +\sum_{n \geq 1} na_nz^{n-1} +\sum_{n\geq 0} a_nz^{n+1} =0 $$ and after simplification you have $a_1 =0 $ and $a_{n+2}=-\dfrac{a_n}{(n+2)^2}$. Thus odd terms are zero and the convergence radius follows by the ratio test $$\lim_\infty\frac{|a_{n+2}z^{2n+2}|}{|a_nz^n|}\to0$$