Complex sequence $\big(\frac{z^n}{n}\big)$ vs $\big(\frac{z^n}{n^k}\big)$ Suppose I want to know for which complex values of $z$ the given sequences converge:
$$\bigg(\frac{z^n}{n}\bigg)\text{ and } \bigg(\frac{z^n}{n^k}\bigg)\text{, for $k\in\mathbb{Z}^+$ }$$
Clearly, one can write $z=r(\cos\theta+i\sin\theta)$ so that $z^n=r^n(\cos n\theta+i\sin n\theta)$.
If $|z|>1$, then $r>1$ so $$\frac{z^n}{n}=\frac{r^n}{n}(\cos n\theta +i\sin n\theta)\to\infty$$ as $n\to\infty$ since $r>1\implies\frac{r^n}{n}\to\infty$ and $|\cos n\theta+i\sin n\theta|=1$ for all $n$.
If $|z|\leq1$, then $z$ can be written as $$z=\frac{\cos\theta + i\sin \theta}{r}, \text{with $r\geq1$}$$
so that $$z^n=\frac{\cos n\theta+i\sin n\theta}{r^n}\implies\frac{z^n}{n}= \frac{\cos n\theta+i\sin n\theta}{nr^n}\to0 $$
since $|\cos n\theta +i\sin n\theta|=1$ and $nr^n\to\infty$ since $r\geq1$.
This shows the first sequence converges for $|z|\leq1$.
What is the point in asking about the second sequence? Isn’t the argument the precise same?
 A: You can apply the ratio test for the first one as follows:
\begin{align*}
\limsup_{n\to\infty}\left|\frac{z^{n+1}}{n+1}\times\frac{n}{z^{n}}\right| & = \limsup_{n\to\infty}\frac{n}{n+1}|z| = |z|
\end{align*}
Hence it converges whenever $|z| < 1$.
As to the second series, you can apply the root test:
\begin{align*}
\limsup_{n\to\infty}\sqrt[n]{\left|\frac{z^{n}}{n^{k}}\right|} & = \limsup_{n\to\infty}\frac{|z|}{n^{k/n}} = |z|
\end{align*}
Hence it converges whenever $|z| < 1$.
At last, we can answer the proposed question.
Whenever $|z| < 1$, both series converges.
Whence we conclude that both sequences converge to zero for such values.
When $|z| = 1$, then $z = \cos(\theta) + i\sin(\theta)$.
Since the $\sin$ and $\cos$ are bounded, we conclude both sequences converge to zero as well.
Finally, as you have noticed, both sequences diverge when $|z| > 1$.
Observation
The underlying concept on which my answer is based is the concept of absolute convergence.
We know that if a series converges absolutely, then the sequence converges conditionally.
Since we are dealing with real numbers (because of the norm), we can apply such result.
The thing is that, even in the context of complex numbers, such results holds.
If you still have any other questions, please let me know.
