Expand the function into a Maclaurin series and find the radius of convergence

Question

Well, I've solved similar problems before, but this one isn't giving me anything to work with as every higher level derivative keeps getting more unpleasant.

Expand the function into a Maclaurin series and find the radius of convergence:

$f(x)=\ln\sqrt[5]{3+x-6x^2-2x^3}$

Could someone help out?

Answer 1

Try to factorize a polynomial, it is easy. Then you arrive at the sum of three logarithms. If you know the expansion of $\ln(1+x)$ (easy), you could easily find what you are looking for.

Answer 2

Just use the fact that\begin{align}\log\left((3+x-6x^2-2x^3)^{1/5}\right)&=\frac15\log\bigl((x+3)(1-2x^2)\bigr)\\&=\frac15\left(\log(3)+\log\left(1+\frac x3\right)+\log\left(1-2x^2\right)\right).\end{align}Now, you only have to expand $\log\left(1+\frac x3\right)$ and $\log\left(1-2x^2\right)$