# radius of convergence with a series that contain derivatives

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!

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In general, this series is not a power series. So it's set of convergence is not necessarily the disk, moreover this set may be complicated. – M. Strochyk Sep 17 '12 at 16:41