Show that $S_n:=\sum\limits_{k=1}^n\sin(kx)\frac{1}{k}$ doesn't converge uniformly.
My approach:
First we set $x_n:=\frac{\pi}{2n}$ and observe that for all $k$ with $\lceil\frac{n}{2}\rceil\leq k\leq n$ we have
$$\sin(kx_n)=\sin(k\frac{\pi}{2n})\geq \frac{1}{\sqrt{2}}.$$
Now we apply the Cauchy criterion, $\Vert S_m-S_n\Vert_{\infty}<\epsilon$, and see that no matter how large we choose any index $n_0$, we can always find another index $n_1:= \lceil\frac{n}{2}\rceil\geq n_0$ by increasing $n$ such that
$$\Vert S_m-S_n\Vert_{\infty}\geq \left|\sum\limits_{k=\lceil\frac{n}{2}\rceil}^n\sin(kx_n)\frac{1}{k}\right|\geq \sum\limits_{k=\lceil\frac{n}{2}\rceil}^n\frac{1}{\sqrt{2}}\frac{1}{k}.$$
Since the lower bound is the tail of the series $\sum\limits_{k=1}^n\frac{1}{\sqrt{2}}\frac{1}{k}$ which diverges, the original series $S_n:=\sum\limits_{k=1}^n\sin(kx)\frac{1}{k}$ can't converge uniformly.
Is this correct?
PS: I know there already exists a similar question (see Does $\sum{\frac{\sin{(nx)}}{n}}$ converge uniformly for all $x$ in $[0,2\pi]$) but it didn't really help me to figure out the single steps to come up with a proof.
Best Answer
The approach is correct. To provide more justification in finishing note that
$$\sum\limits_{k=\lceil\frac{n}{2}\rceil}^n\frac{1}{\sqrt{2}}\frac{1}{k} > \frac{1}{\sqrt{2}}\left(n - \frac{n}{2}\right)\frac{1}{n}= \frac{1}{2\sqrt{2}} \underset{n \to \infty}\nrightarrow 0$$