Show that $\Sigma_{k=1}^{\infty} \frac{\sin kx}{k}$ converges uniformly on any compact subset of $(0,2\pi)$ This is an old qual problem I'm working on. It asks me to prove that $\Sigma_{k=1}^{\infty} \frac{\sin kx}{k}$ is uniformly convergent on any compact subset of $(0,2\pi)$.
If I'm not wrong, it follows from Dirichlet's test that it's convergent. However, I couldn't make much progress for why it must be uniform. I would appreciate any kind of help.
 A: We have:
$$ S_N(x)=\sum_{n=1}^{N}\frac{\sin(n x)}{n}=\int_{0}^{x}\sum_{n=1}^{N}\cos(ny)\,dy\tag{1} $$
but:
$$ \sin\frac{y}{2}\sum_{n=1}^{N}\cos(ny)=\frac{1}{2}\sum_{n=1}^{N}\left(\sin((n+1/2)y)-\sin((n-1/2)y)\right)=\frac{\sin((N+1/2)y)-\sin(y/2)}{2}\tag{2}$$
so:
$$ S_N(x) = \int_{0}^{x}\left(-\frac{1}{2}+\frac{\sin((N+1/2)y)}{2\sin(y/2)}\right)\,dy=-\frac{x}{2}+\int_{0}^{x/2}\frac{\sin((2N+1)y)}{\sin y}\,dy\tag{3} $$
or:
$$ S_N(x) = -\frac{x}{2}+\frac{\sin(Nx)}{2N}+\int_{0}^{x/2}\sin(2N y)\cot y\,dy. \tag{4}$$
Now: $\cot y-\frac{1}{y}$ is an analytic function over $[0,\pi)^\color{red}{*}$, so the integral $$\int_{0}^{x/2}\sin(2N x)\left(\cot y-\frac{1}{y}\right)\,dy$$ goes to zero very fast when $N\to +\infty$ by any quantitative version of the Riemann-Lebesgue lemma. The problem boils down to estimate:
$$ \int_{0}^{x/2}\frac{\sin(2N y)}{y}\,dy = \int_{0}^{Nx}\frac{\sin y}{y}\,dy\tag{5}$$
that is just a "truncated" Dirichlet integral. Since $\frac{\sin y}{y}$ is improperly Riemann-integrable over $\mathbb{R}^+$ by integration by parts, and the value of the Dirichlet integral is $\frac{\pi}{2}$, the RHS of $(5)$ is $\frac{\pi}{2}$ minus something that goes to zero$^\color{blue}{*}$ as $N\to +\infty$.
This proves that $\color{purple}{S_N(x)}$ converges $\color{purple}{\text{uniformly}}$ to $\color{purple}{\frac{\pi-x}{2}}$ over any compact subset of $(0,2\pi)$.

$\color{red}{*}$: here I am exploiting that $x$ is bounded away from $2\pi$;
$\color{blue}{*}$: here I am exploiting that $x$ is bounded away from $0$.

Since $\sum_{n\geq 1}\frac{\sin(nx)}{n}$ is the Fourier series of $f(x)=\frac{\pi-x}{2}$ over $(0,2\pi)$, it is straightforward to check that Parseval's theorem gives:
$$\|f(x)-S_N(x)\|_2^2 = \sum_{n>N}\frac{\pi}{n^2} < \frac{\pi}{N},\tag{6}$$
so:
$$ S_N(x)\xrightarrow[L^2(0,2\pi)]{} f(x).\tag{7}$$
However, the $L^2$ convergence is weaker than the uniform convergence, so if we follow this path, we have to prove something more on $E_n(x)=\left|f(x)-S_N(x)\right|$ to be able to state uniform convergence. For instance, we may prove that $E_n(x)$ has $2n+1$ zeroes in $(0,2\pi)$ and:
$$ E_n(x) \leq \frac{2}{\pi n \sin(x/2)}, \tag{8}$$
or exploit the Fejer-Jackson inequality, that gives us the non-negativity of $S_N(x)$ over $(0,\pi)$ and the non-positivity of $S_N(x)$ over $(\pi,2\pi)$.
A: If $a_n \ge a_{n+1}$, $a_n \to 0$ and $\left|\sum_{j=1}^N b_j(x)\right| \le M$ for all $N$ and for all $x$ in some set $K$, then summation by parts (which is how Dirichlet's test is proven) shows that $\sum_n a_n b_n(x)$ converges uniformly on $K$.
