Given $a>1, b>0$,  such that $b^N < a$. Prove that $\sum_{n=1}^{\infty} \frac{\sin(b^nx)}{a^n}$ converges to an $N$ times differentiable function. 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!
 A: You're interested in the (real coefficient of the) imaginary part of
$$
D^{(k)} \sum_{n=1}^{\infty} \frac{e^{ib^nx}}{a^n}
=
\sum_{n=1}^{\infty} D^{(k)} \frac{e^{ib^nx}}{a^n}
=
\sum_{n=1}^{\infty} i^k b^{kn} \frac{e^{ib^nx}}{a^n}
$$
which converges for $\left|\frac{b^k}{a}\right|<1$
by the root test, since:
$$
\left|i^k b^{kn} \frac{e^{ib^nx}}{a^n}\right|
=
\left|\frac{b^{kn}}{a^n}\right|
=
\left|\frac{b^{k}}{a}\right|^n.
$$
The convenience of looking at the augmented complex series
$$
\sum_{n=1}^{\infty} \frac{e^{ib^nx}}{a^n}
=
\sum_{n=1}^{\infty} a^{-n}\left(\cos{b^nx}+i\sin{b^nx}\right)
$$
is that we don't need to bother with the fact that
the sine and cosine functions, and the sign,
keep swapping/interchanging (rotating quadrants)
with each higher derivative.
Without this "convenience", we would proceed as follows.
The original function
$$
f(x)=\sum_{n=1}^{\infty} a^{-n} \sin{b^nx}
$$
converges because
$|\frac{\sin{b^nx}}{a^n}|^\frac1n\leq\frac1a<1$ by the root test.
Likewise, the first derivative
$$
f'(x)=\sum_{n=1}^{\infty} \left(\frac{b}a\right)^n \cos{b^nx}
$$
converges because
$|\frac{b^n\sin{b^nx}}{a^n}|^\frac1n\leq\frac{b}a<\frac{a^{1/N}}a<1$ by the root test. Continuing, we find that
$$
f^{(k)}(x)=
\left\{
\begin{matrix}
\sum_{n=1}^{\infty} (-1)^\frac{k}{2} \left(\frac{b^k}a\right)^n \sin{b^nx}
&\quad& 0\leq k~\text{even}\\
\sum_{n=1}^{\infty} (-1)^\frac{k-1}{2} \left(\frac{b^k}a\right)^n \cos{b^nx}
&\quad& 1\leq k~\text{odd}
\end{matrix}
\right.
$$
converge for $0\leq k\leq N$ because
$$
\left|\frac{b^{kn}}{a^n}\right|^\frac1n\leq\frac{b^k}a<\frac{a^{k/N}}a<1
$$
(since $0 < b^N < a$ and $a>1$, and the sine and cosine functions
are bounded in magnitude by one).
