Given $a>1, b>0$, and $N \in \mathbb{N}$, such that $b^N < a$, I'd love your help with proving that the series $\sum_{n=1}^{\infty} \frac{\sin(b^nx)}{a^n}$ converges to a function which is $N$ times differentiable for $x\in (0,1)$.
I tried to bound it and use Dirichlet's M-test so there's a uniform convergence
$\sum_{n=1}^{N}\frac{\sin(b^nx)}{a^n}=\sum_{n=1}^{N}\frac{\sin(b^nx)}{a^n}+\sum_{N}^{\infty}\frac{\sin(b^nx)}{a^n} ,\\\ \sum_{N}^{\infty}\left|\frac{\sin(b^nx)}{a^n}\right|<\sum_{N}^{\infty}\frac{b^nx}{a^n}<\sum_{N}^{\infty}\frac{ax}{a^n}=\sum_{N}^{\infty}\frac{x}{a^{n-1}}.$
How should I prove the aforementioned claim?
Thanks a lot!