Determining radius of convergence of the series $\displaystyle{\sum_{n=0}^{\infty} \sin(nz^n)}$

Question

I have the following question: for what values of $z\in \mathbb{C}$ does the following series converges:

$\displaystyle{\sum_{n=0}^{\infty} \sin(nz^n)}$

My initial idea was to use the series expansion for $\sin(z)$, but I believe you end up with the double indexed series: $\displaystyle{\sum_{m=0}^{\infty}}\displaystyle{\sum_{n=0}^{\infty}} \displaystyle{\frac{(-1)^m(nz^n)^{2m+1}}{(2m+1)!}}$

I'm unsure of how to deal with this series. I thought in order for the series to converge, we need to start by finding for what values of $z\in \mathbb{C}$ the series $\displaystyle{\sum_{m=0}^{\infty} \frac{(-1)^m(nz^n)^{2m+1}}{(2m+1)!}}=\sin(nz^n)$ converges for every $n\in \mathbb{N}$. Would this be the correct way to approach this problem? If not, could anyone give me a hint as to where I should begin?

Answer 1

Some suggestions:

For $|z| < r < 1$, use $|\sin w | < C |w|$ for $|w| \leq 1$ to show it converges.

For $z \geq 1$, use that $|\sin w|^2 = |\sin Re(w)|^2 + |\sinh Im(w)|^2$ to show the series diverges; $\{nz^n: n \in \mathbb{N}\}$ would have to stay near integer multiples of $\pi$ for the series to converge, which clearly isn't the case.