Find all complex $z$ such that $z=\sum_{n=2}^{\infty}\sum_{k=1}^{n}e^{\frac{2\pi ikz}{n}}$

Question

Here is what I tried: The inner sum is a geometric series so $$\sum_{k=1}^{n}e^{\frac{2\pi ikz}{n}} =\frac{e^{c}\left(e^{nc}-1\right)}{e^{c}-1}$$ where $c=\frac{2\pi iz}{n}$ and $nc=2\pi i z$. So the sum becomes $$z=\sum_{n=2}^{\infty}\sum_{k=1}^{n}e^{\frac{2\pi ikz}{n}} = \sum_{n=2}^{\infty}\frac{e^{c}\left(e^{nc}-1\right)}{e^{c}-1}$$ $$\frac{z}{e^{2\pi iz}-1}= \sum_{n=2}^{\infty}\frac{e^{c}}{e^{c}-1}$$ But as $n$ approaches infinity $c$ approaches $0$ and $ \left|\frac{e^{c}}{e^{c}-1}\right|\ $ approaches infinity and the series diverges for all $z$ with $\left|z\right|\ $finite. What did I miss?