calculus - Show that $S_n(x)=sumlimits^{n}_{k=1}frac{sin(kx)}{k},$ is uniformly bounded.

Show that

$$S_n(x)=\sum^{n}_{k=1}\frac{\sin(kx)}{k},$$
is uniformly bounded in $\mathbb{R}.$

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I have
$$|S_n(x)|=\Bigg|\sum^{n}_{k=1}\frac{\sin(kx)}{k}\Bigg|\leq\sum^{n}_{k=1}\frac{|\sin(kx)|}{k},$$
but that doesn't help, as it doesn't even imply the sequence is bounded. I also tried rewriting a follows
$$S_n=\int\limits^{x}_{0}\sum^{n}_{k=1}\cos(kt)\,dt,$$

but once again I don't see a way to use this to show anything about the boundedness of $S_n.$

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Could you please help me with a hint, only a hint, on how to get started on this problem?

Thank you for your time and appreciate any help or feedback provided.

Answer

By periodicity it is enough find a uniform bound for $x \in [0,\pi]$.

We have

$$\tag{*}\left|\sum_{k=1}^n \frac{\sin kx}{k}\right| = \left|\sum_{k=1}^m \frac{\sin kx}{k}+ \sum_{k=m}^n \frac{\sin kx}{k}\right| \leqslant \sum_{k=1}^m \frac{|\sin kx|}{k}+ \left|\sum_{k=m+1}^n \frac{\sin kx}{k}\right|$$

where $m = \lfloor \frac{1}{x}\rfloor$ and $m \leqslant \frac{1}{x} < m+1$.

Since $|\sin kx| \leqslant k|x| = kx$, we have for the first sum on the RHS of (*),

$$\sum_{k=1}^m \frac{|\sin kx|}{k} \leqslant mx \leqslant 1$$

Hint for remainder of proof

The second sum can be handled using summation by parts and a well-known bound for $\sum_{k=1}^n \sin kx$.