Where $f(z)=\sum_{n=1}^{\infty}\frac{(2i)^n}{n}z^n$ is convergent?
I checked that the radius of convergence is equal to $\frac{1}{2}$.
Now, since we know that the series $\sum_{n=0}^{\infty}\frac{1}{n}z^n$ is convergent on a unit ball except for point $1$ (it is a well-known fact), we check when $(2i)^n \cdot z^n=1$. The only point lying on $|z|=1/2$ that satisfies that equation is $z=-\frac{i}{2}$.
So we conclude that $f(z)$ is convergent in $\overline{B}(0,\frac{1}{2})\setminus\{-\frac{i}{2}\}$
Is that the correct answer?
If $\;z=-\frac i2\;$ then sequence's general term is
$$\frac{(2i)^n}{n}\left(-\frac i2\right)^n=(-1)^n\frac{i^{2n}}n=(-1)^n\frac{(-1)^n}n=\frac1n$$
and the series indeed diverges, so yes: it looks your work is fine.