The series is continuously differentiable

Question

I want to show that series $$f(x)=\sum_{k=1}^{\infty}2^k\sin (3^{-k}x)$$ is continuously differentiable.

We have that each $f_k=2^k\sin (3^{-k}x)$ is differentiable and its derivative is equal to $f_k'(x)=2^k\cdot 3^{-k}\cdot \cos (3^{-k}x)$.

It holds that $$|2^k\cdot 3^{-k}\cdot \cos (3^{-k}x)|\leq 2^k\cdot 3^{-k}$$

To show that the series is continuously differentiable do we have to show that it converges uniformly?

Answer 1

$$\sum_{k\geq 1}\frac{2^k}{3^k}\cos(x/3^k) $$ is an absolutely convergent series of continuous functions, hence a continuous function which can be termwise-integrated, leading to a continuously differentiable function, $f(x)$.