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, enter image description here

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?

Thanks in advance~

  • $\begingroup$ The root test applies here without any proble $\endgroup$ Feb 23, 2022 at 3:16
  • $\begingroup$ $\limsup\sqrt[n]{(2+(-\frac{1}{2})^n)^n}=2$. $\endgroup$
    – user912011
    Feb 23, 2022 at 3:37

1 Answer 1



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 $|z| = 1/2$.

Can you take it from here?


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