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MathematicsShowing $|\sum_{k=1}^n\frac{\sin(kx)}k|<2\sqrt{\pi}$
[+5] [2] user70520
[2013-08-26 04:41:47]
[ sequences-and-series trigonometry inequality summation trigonometric-series ]
[ https://math.stackexchange.com/questions/476279/showing-sum-k-1n-frac-sinkxk2-sqrt-pi ]

For any real $x$ and positive integer $n$, is it true that:

$$\left|\sum_{k=1}^n\frac{\sin(kx)}k\right|<2\sqrt{\pi}\quad ?$$

Please justify.

For $x \in (0,2\pi)$, you could write it as $\int_0^x \sum_{k=1}^n \cos(kt)dt$. Not sure, but this might help. - Prahlad Vaidyanathan
I was thinking about this too. But the integral is kinda messy after you sum the cosines. - user70520
The best it is possible to state is: $$\left|\sum_{n=1}^N \frac{\sin(nx)}{n}\right|\leq\int_{0}^{1}\frac{\sin(\pi x)}{x}\,dx = 1.85194\ldots$$. - Alex Ravsky
[+3] [2013-08-26 05:16:06] Nathaniel Bubis

Hint:

The sum inside the absolute value is the partial sum of the Fourier series of:

$$y = \frac{\pi}{2} - \frac{x}{2}$$ Gibbs phenomenon tells you that the maximum value of the sum is: $$f(0_+) + \pi\cdot0.08949 = \frac{\pi}{2}+0.08949\pi\sim1.85<2\ll2\sqrt{\pi}$$ So, $2\sqrt{\pi}$ is a very loose bound.


(1) ... a partial sum of the Fourier series of ..., you mean. - Robert Israel
@RobertIsrael - but of course :) fixed. - Nathaniel Bubis
Interesting. Too bad I don't know too much about Fourier series. Is there a more elementary approach? - user70520
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[+3] [2013-08-26 05:20:20] Steve Kass

Looks like it. $|\sum_{k=1}^n\frac{\sin(kx)}k|$ is a partial sum of the Fourier series of the function $y=\frac{\pi-|x|}{2}$. Here are the sums for $n$=15 (gold) and $n$=1350 (green) along with the graph of $y=2\sqrt{\pi}$ (blue). Maybe this will help? The Gibbs effect seems to be the only thing that moves the sum anywhere towards $2\sqrt{\pi}$, but does so ever so slightly.


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