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

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 $S_n.$

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.

rtybase
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Stackman
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3 Answers3

5

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 $\sum_{k=1}^n \sin kx$.

RRL
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HINT:

$S_n$ is the partial Fourier series of a "Sawtooth wave".

G Cab
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For convenience, denote $X\lesssim Y$ or $Y\gtrsim X$ for $X\leq CY$ where $C>0$ is some constant that is independent of $n$ and $x$.

One has actually a stronger result: there exists a constant $C$ such that $|S_n(x)|\leq C$ for all $x\in\mathbb{R}$ and all positive integer $n$.

Since $|S_n(x+\pi)| = |S_n(x)|$ for all $x\in\mathbb{R}$, one can reduce the problem to estimates on the interval $[0,\pi]$. Henceforth, we assume that $x\in[0,\pi]$.

One can control the sum $S_n$ by splitting it into two parts: $kx\leq1$ and $kx>1$, where $k=1,2,\cdots,n$. For $kx\leq 1$, we have $|\sin(kx)|\leq kx$ and thus $$ \sum_{1\leq k\leq n\\kx\leq 1}\frac{|\sin(kx)|}{k}\leq \sum_{1\leq k\leq n\\kx\leq 1}\frac{kx}{k} \leq \sum_{1\leq k\leq n\\kx\leq 1}x\leq 1\tag{1} $$ (Note: up to this point, this is essentially rephrasing an argument in the accepted answer. It may be helpful to look at it at different perspectives.)

Now we handle the part where $kx >1$. Applying the trigonometric identity $$ 2\sin a\sin b = \cos(a-b)-\cos(a+b), $$ one has $$ 2\sin\frac{x}{2}\sin(kx) = \cos(\frac{x}{2}-kx)-\cos(\frac{x}{2}+kx) $$ and thus $$ \begin{align} 2\sin\frac{x}{2}\cdot\sum_{1\leq k\leq n\\kx> 1}\frac{\sin(kx)}{k} &=\sum_{1\leq k\leq n\\kx> 1}\frac{\cos(\frac{x}{2}-kx)-\cos(\frac{x}{2}+kx)}{k}\\ &=\sum_{1\leq k\leq n\\kx> 1}\frac{\cos((k-\frac12)x)-\cos((k+\frac12)x)}{k} \end{align} $$ Note that this is very much like a telescoping sum, except that one has the extra denominator $k$. Now we rearrange the sum as follows. Let $k_0$ be smallest $k$ such that $kx>1$. Then $$ \begin{align} &\sum_{1\leq k\leq n\\kx> 1}\frac{\cos((k-\frac12)x)-\cos((k+\frac12)x)}{k} \\=&\frac{\cos((k_0-\frac12)x)}{k_0} +\sum_{k=k_0+1}^n\cos((k-\frac12)x)\left(\frac{1}{k}-\frac{1}{k-1}\right) - \frac{\cos((n+\frac12)x)}{n}\;\tag{2} \end{align} $$ Moreover, $$ \left|\frac{\cos((k_0-\frac12)x)}{k_0}\right|+ \left|\frac{\cos((n+\frac12)x)}{n}\right|\lesssim 1\tag{3} $$ and $$ \left|\sum_{k=k_0+1}^n\cos((k-\frac12)x)\left(\frac{1}{k}-\frac{1}{k-1}\right) \right|\lesssim \sum_{m=1}^\infty\frac{1}{m^2}\tag{4} $$ Also, for $x>\frac{1}{k_0}$, $$ \sin\frac{x}{2}\gtrsim 1\tag{5} $$

Combining (1)--(5) and using the triangle inequality, one obtains $$ |S_n(x)|\lesssim 1\;. $$