Does $\sum _{ n=1 }^\infty \frac {(-1) ^n} n \sin (nx) $ converge uniformly on $[0, \pi )$? i tried Werierstrass M test, but >$$\left| \frac{(-1)^n}n \sin(nx)\right| \leq \frac 1 n.$$
is there any other way to do it?
 A: It does not converge uniformly on the given interval: loosely speaking, the coefficients $\frac{(-1)^{n+1}}{n}$ do not decay fast enough to zero to ensure uniform convergence. The fact that $\{(-1)^n\sin(nx)\}_{n\geq 1}$ has bounded partial sums ensures that the given Fourier (sine) series is pointwise convergent on $(-\pi,\pi)$, so on $\mathbb{R}\setminus\pi\mathbb{Z}$ we have
$$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\sin(nx) = W(x) \tag{1}$$
with $W(x)$ being a $2\pi$-periodic function equal to $\frac{x}{2}$ on $(-\pi,\pi)$, i.e. a sawtooth wave.
This function is not continuous on $\mathbb{R}$, and that leads to Gibbs phenomenon, contradicting uniform convergence. In a quantitative form,
$$\lim_{N\to +\infty}\sup_{x\in(-\pi,\pi)}\left|\frac{x}{2}-\sum_{n=1}^{N}\frac{\sin(nx)}{n}(-1)^{n+1}\right| = C > 0\tag{2}$$
and that can be proved, for instance, by showing that $\frac{x}{2}-\sum_{n=1}^{N}\frac{\sin(nx)}{n}(-1)^{n+1}$ has constant sign in a left neighbourhood of $\pi$ with size $\approx\frac{\pi}{N}$ and 
$$ \int_{-\pi}^{\pi}\left|\frac{x}{2}-\sum_{n=1}^{N}\frac{\sin(nx)}{n}(-1)^{n+1}\right|^2\,dx =\pi\sum_{n>N}\frac{1}{n^2}\approx\frac{\pi}{N},\tag{3}$$
$$ \frac{\pi}{2}=\lim_{x\to \pi^-}\frac{x}{2}\neq \lim_{N\to +\infty}\lim_{x\to \pi^-}\sum_{n=1}^{N}\frac{\sin(nx)}{n}(-1)^{n+1}=0.\tag{4}$$
A: The series converges to $0$ for $x=\pi$ so if we assume the convergence is uniform on $[0,\pi)$ then the convergence is uniform on $[0,\pi]$ and in particular $f$ has to be continuous at $x=\pi$. Take $x = \pi$ and $x_N = \pi - \frac{c}{N}$ for some $c>0$ and consider the series of partial sums
$$f_N(x) - f_N(x_N) = \sum_{n=1}^N\frac{(-1)^n}{n}2\sin\left(\frac{nx-nx_N}{2}\right)\cos\left(\frac{nx+nx_N}{2}\right) \\= \frac{1}{N}\sum_{n=1}^N\frac{2}{\frac{n}{N}}\sin\left(\frac{cn}{2N}\right)\cos\left(\frac{cn}{2N}\right) = \frac{c}{N}\sum_{n=1}^N\text{sinc}\left(\frac{cn}{N}\right)$$
This is a Riemann sum so we have
$$\lim_{N\to\infty}[f_N(x) - f_N(x_N)] = \int_0^c\text{sinc}(t){\rm d}t \not= 0$$
even though $x_N\to x$ contradicting uniform convergence.
