Prove or Disprove $2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin(nx) $ converge uniformly to $x$ on $(-\pi,\pi)$

Question

I want to prove $2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin(nx) $ converges pointwise and uniformly to $x$ on $[-\pi,\pi]$. I know $\sum_{k=1}^{\infty}\frac{(-1)^n}{n}$ converge by alternating series test. And $\sum a_n \sin(nx)$ converge by Dirichlet test if $a_n$ is decreasing sequence. But in this case this does not work. Maybe it just we can just consider the interval without $-\pi$,$\pi$. I get lost. Please help. Thanks a lot
After trying, I think maybe there is no uniform convergence?

Answer 1

Convergence is not uniform on $(-\pi,\pi)$ (although it is on compact subintervals).

To prove non-uniform convergence note that

$$2\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\sin nx }{n} = -2\sum_{n=1}^{\infty} \cos n\pi \frac{\sin nx }{n} = -2 \sum_{n=1}^{\infty} \frac{\sin n(\pi+x) }{n} $$

However, taking $x_n = -\pi + \frac{\pi}{4n} \in (-\pi,\pi)$ we have for $n < k \leqslant 2n$ that $\frac{\pi}{4} < k (\pi+x_n) \leqslant \frac{\pi}{2}$ which implies $\frac{1}{\sqrt{2}} < \sin k (\pi+x_n) \leqslant 1$ and for all $n \in \mathbb{N}$,

$$\sup_{x \in (-\pi,\pi)}\left| \sum_{k = n+1}^{2n}\frac{\sin k(\pi+x) }{k} \right|\geqslant \sum_{k = n+1}^{2n}\frac{\sin k(\pi+x_n) }{k} > \frac{1}{\sqrt{2}}\ \sum_{k=n+1}^{2n} \frac{1}{k} > \frac{1}{\sqrt{2}} \cdot n \cdot \frac{1}{2n} = \frac{1}{2\sqrt{2}}$$

The LHS fails to converge to $0$ as $n \to \infty$ and the Cauchy criterion for uniform convergence is violated.

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