Behavior of a Complex-Valued Series (3.6(d))

Walter Rudin Principles of Mathematical Analysis, chapter 3, exercise 6(d):

Investigate the behavior of $∑a_n$ when $a_n=\dfrac{1}{1+z^n}$ for complex values of $z$.

Proof: Assume $|z| > 1$. Then $$||\dfrac{a_{n+1}}{a_n}|-\dfrac{1}{|z|}|=||\dfrac{\dfrac{1}{1+z^{n+1}}}{\dfrac{1}{1+z^n}}|-|\dfrac{1}{z}||=||\dfrac{1+z^n}{1+z^{n+1}}|-|\dfrac{1}{z}||$$ $$≤|\dfrac{1+z^n}{1+z^{n+1}}-\dfrac{1}{z}|=|\dfrac{z-1}{z+z^{n+2}}|=|\dfrac{z-1}{z}|·\dfrac{1}{|1+z^n|}$$ $$≤|\dfrac{z-1}{z}|·\dfrac{1}{|z|^n-1}→0$$ as $n→∞$, so $\text{lim }|\dfrac{a_{n+1}}{a_n}|=\dfrac{1}{|z|} < 1$ and $∑a_n$ converges by the ratio test. Assume $|z| ≤ 1$. Then $|\dfrac{1}{1+z^n}|≥\dfrac{1}{1+|z|^n}≥\dfrac{1}{2}$ and $∑a_n$ diverges seeing as $a_n$ doesn't converge to $0$.$~\square$