Is there a trick how to find all $z\in \Bbb{C}$ such that the following series converge? I consider the following series
$$\sum_{n=1}^\infty \frac{1}{n^3}(z+i)^{n^2}$$
I want to find all $z\in \Bbb{C}$ such that the series converge. My idea was to use the root test. We have seen the following version:

Let $\sum_{n=0}^\infty a_n(z-z_0)^n$ be a power serie, let $R=\frac{1}{\lim sup_{n\rightarrow \infty} |a_n|^{1/n}}$ then the series

*

*converges absolutely in $|z-z_0|<R$


*converges uniformly in $|z-z_0|\leq r$ where $r<R$


*diverges in $|z-z_0|>R$

But I can't rewrite our sum as needed. So I wanted to ask if someone could give me a hint because substitute $n^2$ does not work.
 A: Alternatively (as well as a similar approach to @JoséCarlosSantos), you can also apply the ratio test:
\begin{align*}
\limsup_{n\to\infty}\left|\frac{a_{n+1}(z + i)^{(n+1)^{2}}}{a_{n}(z + i)^{n^{2}}}\right| & = \limsup_{n\to\infty}\frac{n^{3}}{(n+1)^{3}}\times|z + i|^{2n + 1}\\\\
& = \limsup_{n\to\infty}\left(1 + \frac{1}{n}\right)^{-3}\times|z + i|^{2n+1}\\\\
& = \limsup_{n\to\infty}|z + i|^{2n + 1}
\end{align*}
whence proposed power series converges when $|z + i| < 1$, it diverges when $|z + i| > 1$.
Now it remains to study its behavior when $|z + i| = 1$.
In such case, we obtain the following numerical sequence:
\begin{align*}
\sum_{n=1}^{\infty}\frac{|z + i|^{n^{2}}}{n^{3}} = \sum_{n=1}^{\infty}\frac{1}{n^{3}} 
\end{align*}
which converges absolutely.
Consequently, the proposed power series converges when $|z + i| \leq 1$ and diverges otherwise.
Hopefully this helps !
A: For each $n\in\Bbb N$,$$\sqrt[n]{\left|\frac1{n^3}(z+i)^{n^2}\right|}=\frac1{\sqrt[n]n^3}|z+i|^n$$and therefore$$\limsup_{n\to\infty}\sqrt[n]{\left|\frac1{n^3}(z+i)^{n^2}\right|}=\begin{cases}0&\text{ if }|z+i|<1\\1&\text{ if }|z+i|=1\\\infty&\text{ if }|z+i|>1.\end{cases}$$So, your series converges absolutely if $|z+i|<1$ and diverges if $|z+i|>1$. If $|z+i|=1$, then:

*

*if $z=1-i$, the series converges, since it is equal to $\displaystyle\sum_{n=1}^\infty\frac1{n^3}$;

*otherwise, it converges, by Dirichlet's test: the partial sums $\displaystyle\left(\sum_{n=1}^N(z+i)^n\right)_{N\in\Bbb N}$ are bounded and the sequence $\displaystyle\left(\frac1{n^3}\right)_{n\in\Bbb N}$ is monotonic and converges to $0$.

