Show that $\sum_{1}^\infty\frac{\sin(nx)}{n^3}$ is differentiable everywhere

Question

I have recently been trying out some questions on series of functions.In one of the questions, I was given a series $$\sum_{1}^\infty\frac{\sin(nx)}{n^3}$$
and now I am supposed to show that the above series is differentiable at every real number and I need to find its derivative.

I was wondering when a series of the form $\sum_{1}^\infty f_n(x)$ said to be differentiable at x?..is it when each $f_n(x)$ is differentiable?.. and if that is so,then I assume that I am done with first half of the question.

In the second half,should I show that the given series is uniformly convergent by using the Weierstrass' M-Test and hence differentiate the series term by term?

Help please!

Answer 1

The main theorem regarding differentiation of series of functions is provided at this link.

Here $f_n(x) = \frac{\sin nx }{n^3}$ is such that $\sum f_n^\prime(x) = \sum \frac{ \cos nx }{n^2}$ is normally convergent, therefore uniformly convergent by Weierstrass M-test as $\sum 1/n^2$ converges. As the series is also convergent at $0$, the theorem states that the series of functions is differentiable and has for derivative $\sum \frac{ \cos nx }{n^2}$.