# Show $f(x) = \sum_{n = 1}^{\infty} \frac{\sin(nx)}{2^{n}}$ is infinitely differentiable

I want to show that the function $$f(x) = \sum_{n = 1}^{\infty} \frac{\sin(nx)}{2^{n}}$$ is infinitely differentiable on $$\mathbb{R}$$. I have no idea how to do this. I think it's pretty obvious that the function converges since $$2^{n}$$ increases really quickly. I tried doing something with analyticity which implies infinitely differentiable; however, I got nowhere.

• This series is uniformly convergent on $\Bbb R$, and the series of the derivatives converges uniformly on $\Bbb R$. This is enough to conclude that the derivative of $f(x)$ is simply the series of the derivatives. – Crostul Dec 14 '18 at 16:38
• how do we know the series is uniformly convergent? – user614735 Dec 14 '18 at 17:08
• Do you know Weierstrass' M-test? It's so simple to use here. – Crostul Dec 14 '18 at 17:14
• i am allowed to use weierstrass m test, yes – user614735 Dec 14 '18 at 17:17

Write $$\sin(nx)={1\over 2i}\bigl(e^{inx}-e^{-inx}\bigr)$$, so that your $$f$$ appears as $$f(x)={1\over 2i}\left(\sum_{n\geq1} p^n-\sum_{n\geq1} q^n\right)\ .$$ Simplifying in the end will give you a simple expression for $$f$$, namely $$f(x)={2\sin x\over5-4\cos x}\ .$$