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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.

Philipp
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1 Answers1

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The approach is correct. To provide more justification in finishing note that

$$\sum\limits_{k=\lceil\frac{n}{2}\rceil}^n\frac{1}{\sqrt{2}}\frac{1}{k} > \frac{1}{\sqrt{2}}\left(n - \frac{n}{2}\right)\frac{1}{n}= \frac{1}{2\sqrt{2}} \underset{n \to \infty}\nrightarrow 0$$

RRL
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  • Your answer provides another nice proof of why the harmonic series diverges: No matter how large the index $n_0$ is chosen I always find an index $m>n_0$ such that: $|\sum\limits_{k=m}^{2m}\frac{1}{k}|>(2m-\frac{2m}{2})\frac{1}{2m}=\frac{1}{2}$. Maybe in some courses this is the standard proof of divergence? – Philipp Oct 20 '21 at 19:31
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    @Philipp: Thanks. Even if it is not the standard way, it is certainly one of the easiest ways. – RRL Oct 20 '21 at 20:31