Why does the series $\sum_{n=1}^{\infty} \frac{\sin nx}{n}$ not converge uniformly at a neighbourhood containing $2k\pi$?

Question

I have known that the series $$\sum_{n=1}^{\infty} \frac{\sin nx}{n}$$ is convergent at every point on $(-\infty,+\infty)$ but as I suggest in the title, how to prove that this series isn't uniformly convergent in a neighbourhood containing the point $2k\pi$,where k is an integer?

Answer 1

Consider the interval $I = (2k\pi, (2k+1)\pi)$.

The Cauchy criterion for uniform convergence of a series $\sum_j f_j(x)$ on $I$ is given any $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $m > n \geqslant N$ and for all $x \in I$ we have $\left|\sum_{j=n+1}^m f_j(x) \right| < \epsilon$.

For any $n \in \mathbb{N}$ choose $x_n = 2k\pi + \frac{\pi}{4n}$.

For all $j \in \mathbb{N}$ such that $n < j \leqslant 2n$ we have $2kn\pi + \frac{\pi}{4} < jx_n \leqslant 2kn\pi + \frac{\pi}{2}$ and, hence,

$$\frac{1}{\sqrt{2}} = \sin \left(2kn\pi + \frac{\pi}{4}\right) < \sin jx_n \leqslant \sin \left(2kn\pi + \frac{\pi}{2}\right) = 1$$

Thus,

$$\left|\sum_{j=n+1}^{2n} \frac{\sin jx_n}{j}\right| = \sum_{j=n+1}^{2n} \frac{\sin jx_n}{j}> \frac{1}{\sqrt{2}} \sum_{j = n+1}^{2n} \frac{1}{j} > \frac{1}{\sqrt{2}} \frac{n}{2n} = \frac{1}{2\sqrt{2}}$$

From this we see that the Cauchy criterion for uniform convergence is violated.