I'm trying to get the radius of convergence of the series:
$\sum_{j=0}^{\infty} \frac{(-ax)^j}{j!} \frac{\partial^{2+j}}{\partial x^{2+j}} \Big[ x^{2j} f(x) \Big]$
where $a$ is a constant and $f(x)$ is just a complicated function of, only, $x$.
I tried to do: $\lim_{j \rightarrow \infty} \Bigg| \Bigg( \frac{(-ax)^{j+1}}{(j+1)!} \frac{\partial^{3+j}}{\partial x^{3+j}} \Big[ x^{2(j+1)} f(x) \Big] \Bigg)\Bigg/ \Bigg( \frac{(-ax)^{j}}{j!} \frac{\partial^{2+j}}{\partial x^{2+j}} \Big[ x^{2j} f(x) \Big] \Bigg) \Bigg|$
but I have no idea how deal with the derivatives, then I found this relation, that could help:
$\frac{1}{n!} \frac{\partial^n f(z')}{\partial z'^n} = \frac{1}{2 \pi i} \oint \frac{f(z)}{(z-z')^{n+1}} dz$
but I don't know how apply this to find the radius of convergence.
Anyone have a idea?
Thanks a lot!
