Prove that $f(x) = \sum_{n=1}^{\infty} \frac{sin^3(nx)}{n}$ is discontinuous at point $0$

Prove that $f(x) = \sum_{n=1}^{\infty} \frac{\sin^3(nx)}{n}$ is discontinuous at point $0$.
I rewrote this function using $\sin^3(x) = \frac{1}{4} (3 \sin(x) - \sin(3 x))$.$f(x) = \frac{3}{4}(\sum \frac{\sin(nx)}{n} - \sum \frac{\sin(3nx)}{3n})$So now my problem is that I have to find some cool epsilon that approaches zero so that $|f(\epsilon) - f(2\epsilon)|$ would be GREATER than some constant (or function that is greater than some constant).$f(\epsilon) - f(2\epsilon) = \sum \frac{\sin(n\epsilon)(1 - 2\sin(\pi/2 - n\epsilon))}{n}$ if sum contains only n that are not divisible by $3$. I checked this series in the desmos and indeed it can be quite large but for some particular epsilon. I will be very grateful.