Prove that the series converges to a function which is $N$ times differentiable 
Let $a>1$ and $b>0$ such that there's an $N\in\mathbb{N}$ such that $b^N < a$.
  Let the series $$S(x) = \sum_{n=1}^\infty \frac{\sin b^nx}{a^n}$$
  Prove that $S(x)$ converges uniformly on $(-1,1)$ to some function which is $N$ times differentiable.

My Proof
First, we can see that the series converges uniformly on $\mathbb{R}$ by Weierstrass M-test, since for every $x\in\mathbb{R}$: $$\sum_{n=1}^\infty \left| \frac{\sin b^n x}{a^n}\right| \le \sum_{n=1}^\infty \frac{1}{a^n}< \infty$$
Now, since $S(x)$ converges uniformly, $$S'(x) = \sum_{n=1}^\infty\frac{b^n \cos b^nx}{a^n}$$
We may notice that every high-order derivative has this pattern.
Now, for the $k$-th derivative where $k\le N$ we have:  
$$S^{(k)} = \sum_{n=1}^\infty \frac{g(x)b^{kn}}{a^n}$$
where $g(x)$ is a function of the form $\pm \sin(b^nx)$ or $\pm \cos(b^nx)$
And since $k\le N$ and $a>b^N$:
$$\sum_{n=1}^\infty \frac{g(x)b^{kn}}{a^n} = \sum_{n=1}^\infty g(x)q^n $$
where $q\in (0,1)$.
The last series converges uniformly and absolutely by the $M$-test and so, our $N$-th derivative converges to some $S^{(N)}(x)$.
Is my proof okay?
Thanks. 
 A: The theorem (See This and THIS ) that we need is as follows:
If $\sum u_n(x)$ is a series of differentiable functions on $[a,b]$ and converges at one point of $[a,b]$, and if the derived series $\sum u_n'(x)$ converges uniformly on $[a,b]$, then the original series converges uniformly on $[a,b]$ to a differentiable function whose derivative is represented on $[a,b]$ by the derived series.
Here, we proceed recursively and examine the series formed summing the $N$'th derivative of the terms of the original series.  This yields the series
$$\sum_{n=1}^\infty \frac{\phi_N(b^nx)}{(a/b^N)^n}$$   
where $\phi_N(x)=\frac{d^N\sin(x)}{dx^N}$.  Since $|\phi_N(x)|\le 1$ and $a/b^N>1$, then the series $\sum_{n=1}^\infty \frac{\phi_N(b^nx)}{(a/b^N)^n}$ converges uniformly on $(-1,1)$.  
We also have that the series $\sum_{n=1}^\infty \frac{\phi_{N-1}(b^nx)}{(a/b^{N-1})^n}$ converges since if $b>1$, then $a/b^{N-1}>b>1$ and if $b\le 1$, then $a/b^{N-1}\ge a>1$.  
Finally, the aforementioned theorem, applied recursively, yields the desired result. 
