Find the maximal region in $\mathbb C$ in which the following series $\sum\limits_{n=1}^\infty \frac{(-1)^{3n}n-n^2}{n+4}z^n$ is uniformly convergentI would like some help with this question.
The series seems to be a power series, so I tried to compute the radius of convergence to use Cauchy-Hadamard, but am stuck on how to take the limit $$\limsup \limits_{n\to \infty }|\sqrt[n]{a_n}|=\limsup \limits_{n\to \infty }\left| \frac{(-1)^{3n}n-n^2}{n+4} \right|^\frac 1n$$ considering the fact that the numerator and denominator are binomials. That is, it's a bit unclear how I can simplify this expression to be able to apply elementary limits.
I also tried the alternative ratio version $$\limsup \limits_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|= \frac{(-1)^{3(n+1)}(n+1)-(n+1)^2}{(n+1)+4}\cdot \frac{n+4}{(-1)^{3n}n-n^2},$$ but this seems to have the same issue with the binomial terms not cancelling easily.