Finding Radius of Convergence, when ratio/root tests do not apply

How would you find the radius of convergence of a complex power series when ratio/root tests don't work? For example, in the below,

How would we do this? My gut feeling is that the radius of convergence is when $$|z|<1/2$$, as we can see that in general, $$(-1/2)^n$$ converges as it is an alternating sequence. Then, we would need $$2z$$ to converge which would mean $$|z|<1/2$$. Is my logic flawed? Is there a more standard way of solving when you can't apply root/ratio?

• $\limsup\sqrt[n]{(2+(-\frac{1}{2})^n)^n}=2$. Feb 23, 2022 at 3:37
Let us try to prove absolute convergence according to the root test: \begin{align*} f(z) = \sum_{n=0}^{\infty}\left[2 + \left(-\frac{1}{2}\right)^{n}\right]^{n}z^{n} & \Rightarrow |a_{n}z^{n}| = \left|\left[2 + \left(-\frac{1}{2}\right)^{n}\right]^{n}z^{n}\right|\\\\ & \Rightarrow \lim_{n\to\infty}\sqrt[n]{|a_{n}z^{n}|} = \lim_{n\to\infty}\left|2 + \left(-\frac{1}{2}\right)^{n}\right||z| = 2|z| \end{align*}
Consequently, we conclude the proposed power series converges whenever $$|z| < 1/2$$, as you have guessed. Now it remains to study the behavior of the power series when $$|z| = 1/2$$.